Regular expression to substring
Hi Folks;
I need to extract dynamically substrings from an attribut A.
The varchar2 attribut A is defined like that : "LXXXXX/111111(+),LXXXXX/111111(-),LXXXXXX/111111,etc..." Always the same serie.
I need to store all "111111(+)" "111111(-)" "111111" of the same record in a new attribut named B.
I feel the regular expressions could help me but i'm not a very good...
Thanks for your help . ^^
Try this,
SELECT LTRIM (REGEXP_SUBSTR (attrA,
'/[^,]+',
1,
LEVEL),'/')
FROM T
CONNECT BY LEVEL <= LENGTH (REGEXP_REPLACE ( attrA, '[^/]'))
Example
SQL> WITH T AS (SELECT 'LXXXXX/111111(+),LXXXXX/111111(-),LXXXXXX/111111,' attrA FROM DUAL)
2 SELECT LTRIM (REGEXP_SUBSTR (attrA,
3 '/[^,]+',
4 1,
5 LEVEL),'/') exprssn
6 FROM T
7 CONNECT BY LEVEL <= LENGTH (REGEXP_REPLACE ( attrA, '[^/]'));
EXPRSSN
111111(+)
111111(-)
111111
SQL> G.
Similar Messages
-
How to fetch substring using regular expression
Hi,
I am new to using regular expression and would like to know some basic details of how to use them in Java.
I have a String example= "http://www.google.com/foobar.html#*q*=database&aq=f&aqi=g10&fp=c9fe100d9e542c1e" and would like to get the value of "q" parameter (in bold) using regular expression in java.
For the same example, when we tried using javascript:
match = example.match("/^http:\/\/(?:(?!mail\.)[^\.]+?\.)?google\.[^\?#]+(?:.*[\?#&](?:as_q|q)=([^&]+))?/i}");
document.write('
' + match);
We are getting the output as: http://www.google.com/foobar.html#q=database,*database* where the bold text is the value of "q" parameter.
In Java we are trying to get the value of the q parameter separately or atleast resembles the output given by JavaScript. Please help me resolving this issue.
Regards
PraveenBalusC wrote:
Regex is a cumbersome solution for fixed patterns like URL's. String#substring() in combination with String#indexOf would most likely already suffice.I usually agree, although, in this case, finding the exact parameter might be difficult without a small regex, perhaps:
"\\wq=\\s*"in conjunction with Pattern/Matcher, used similarly to an indexOf() to find the start of the parameter value.
Winston -
Regular expression for 2nd occurance of a substring in a string
Hi,
1)
i want to find the second occurrence of a substring in a string with regular expression so that i can modify that only.
Ex: i have a string like ---> axe,afn,sdk,jdi,afn,mki,mki
in this i want the second occurance of afn and change that one only...
which regular expression i have to use...
Note that ...i have to use regular expression only....no string manipulation methods...(strictly)
2)
How can i apply the multiple regular expressions multiple times on a single string ..i.e in the above instance i have to apply the same 2nd occurrence logic for
substring mki also. for this i have to use a single regular expression string that contains validations for both the sub strings mki and afn.
Thanks in advance,
Venkatjavafreak666 wrote:
Hi,
1)
i want to find the second occurrence of a substring in a string with regular expression so that i can modify that only.
Ex: i have a string like ---> axe,afn,sdk,jdi,afn,mki,mki
in this i want the second occurance of afn and change that one only...
which regular expression i have to use...
Note that ...i have to use regular expression only....no string manipulation methods...(strictly)
2)
How can i apply the multiple regular expressions multiple times on a single string ..i.e in the above instance i have to apply the same 2nd occurrence logic for
substring mki also. for this i have to use a single regular expression string that contains validations for both the sub strings mki and afn.
Thanks in advance,
VenkatWhat do you mean by using a regex to get the index of a second substring? There is not method in Java which uses regex to et the index of a substring.
There are various indexOf(...) methods for this:
String text = "axe,afn,sdk,jdi,afn,mki,mki";
String target = "afn";
int second = text.indexOf(target, text.indexOf(target)+1);
System.out.println("second="+second);Of course you can find the index of a group like this:
Matcher m = Pattern.compile(target+".*?("+target+")").matcher(text);
System.out.println(m.find() ? "index="+m.start(1) : "nothing found");but there is not single method that handles this: you'll have to call the find() and then the start(...) method on the Matcher instance, so the indexOf(...) approach is the favourable one, IMO. -
Searching for a substring using Regular Expression
I have a lengthy String similar to repetetion of the one below
String str="<option value='116813070'>Something1</option><option value='ABCDEF' selected>Something 2</option>"I need to search for the Sub string "<option value='ABCDEF' selected>" (need to get the starting index of sub string) and but the value ABCDEF can be anything numberic with varying length.
Is there any way i can do it using regular expressions(I have no other options than regular expression)?
thanks in advance.If you go through the tutorial then you will find this on the second page:
import java.io.Console;
import java.util.regex.Pattern;
import java.util.regex.Matcher;
public class RegexTestHarness {
public static void main(String[] args){
Console console = System.console();
if (console == null) {
System.err.println("No console.");
System.exit(1);
while (true) {
Pattern pattern =
Pattern.compile(console.readLine("%nEnter your regex: "));
Matcher matcher =
pattern.matcher(console.readLine("Enter input string to search: "));
boolean found = false;
while (matcher.find()) {
console.format("I found the text \"%s\" starting at " +
"index %d and ending at index %d.%n",
matcher.group(), matcher.start(), matcher.end());
found = true;
if(!found){
console.format("No match found.%n");
}It's does everything you need and a bit more. Adapt it to your needs then write a regular expression. Then if you have problems by all means come back and post them up here, but first at least attempt to solve it yourself. -
Get all groups from a regular expression match
Please help me understand how to use Java regular expressions:
I have an expression similar to this:
{noformat}"([^X]+)(X[^X]*)+"{noformat}This should match stuff like "asaasaXdfdfdfXXsdsfd".
How does one access all the matches for the second group (the second groups has a Kleene operator
added so it is not really just one group --- but match.groupCount() is always 2)
Here is roughly the code:
{noformat}java.util.regex.Pattern pattern = {noformat}{noformat}java.util.regex.Pattern.compile({noformat}{noformat}"([^X]+)(X[^X]*)+",{noformat}{noformat}java.util.regex.Pattern.MULTILINE{noformat}{noformat});{noformat}{noformat}java.util.regex.Matcher matcher = pattern.matcher(text);{noformat}{noformat}matcher.find();{noformat}{noformat}int groupcount = matcher.groupCount();{noformat}
Also, without matcher.find() I get an illegalStateException .. which I also get if I use matcher.matches() instead
of matcher.find().
I am obviously missing something here. There is always at least one "X" in the string so shouldn't that pattern always
match the whole string? Since there are often multiple X, shouldnt I get a group for each occurrence of X, followed
by 0 or more other characters?
{noformat}But when I try to match everything by using "^([^X]+)(X[^X]*)+$" I get an "IllegalStateException: No match available" again.{noformat}
What is the correct way to do this?
Edited by: johann_p on May 16, 2008 10:39 AMI am sorry I messed this up. Here is a SSCCE:
import java.util.regex.Pattern;
import java.util.regex.Matcher;
class RegExp1 {
public static void main(String[] args) {
String testString = "first|aaaa | bbbb\n|cccc|ddddd";
Pattern pattern = Pattern.compile("^([^|]+)(\\|[^|]*)+$");
Matcher matcher = pattern.matcher(testString);
matcher.find();
int groupcount = matcher.groupCount();
System.out.println("Found "+groupcount+" groups");
System.out.println("Matcher: "+matcher);
for (int i = 1; i <= groupcount; i++) {
System.out.println("Match "+i+": "+testString.substring(matcher.start(i),matcher.end(i)));
}I figured out a small bug in my first code that explains some of the exception oddities, but my principal question remains:
how do I access all the matches that correspond to the second capturing group?
In the example I would get "first" for Match 1 and "|ddddd" for Match 2, but how do I access all the matches??
Thank you for your help! -
Introduction to regular expressions ...
I'm well aware that there are already some articles on that topic, some people asked me to share some of my knowledge on this topic. Please take a look at this first part and let me know if you find this useful. If yes, I'm going to continue on writing more parts using more and more complicated expressions - if you have questions or problems that you think could be solved through regular expression, please post them.
Introduction
Oracle has always provided some character/string functions in its PL/SQL command set, such as SUBSTR, REPLACE or TRANSLATE. With 10g, Oracle finally gave us, the users, the developers and of course the DBAs regular expressions. However, regular expressions, due to their sometimes cryptic rules, seem to be overlooked quite often, despite the existence of some very interesing use cases. Beeing one of the advocates of regular expression, I thought I'll give the interested audience an introduction to these new functions in several installments.
Having fun with regular expressions - Part 1
Oracle offers the use of regular expression through several functions: REGEXP_INSTR, REGEXP_SUBSTR, REGEXP_REPLACE and REGEXP_LIKE. The second part of each function already gives away its purpose: INSTR for finding a position inside a string, SUBSTR for extracting a part of a string, REPLACE for replacing parts of a string. REGEXP_LIKE is a special case since it could be compared to the LIKE operator and is therefore usually used in comparisons like IF statements or WHERE clauses.
Regular expressions excel, in my opinion, in search and extraction of strings, using that for finding or replacing certain strings or check for certain formatting criterias. They're not very good at formatting strings itself, except for some special cases I'm going to demonstrate.
If you're not familiar with regular expression, you should take a look at the definition in Oracle's user guide Using Regular Expressions With Oracle Database, and please note that there have been some changes and advancements in 10g2. I'll provide examples, that should work on both versions.
Some of you probably already encountered this problem: checking a number inside a string, because, for whatever reason, a column was defined as VARCHAR2 and not as NUMBER as one would have expected.
Let's check for all rows where column col1 does NOT include an unsigned integer. I'll use this SELECT for demonstrating different values and search patterns:
WITH t AS (SELECT '456' col1
FROM dual
UNION
SELECT '123x'
FROM dual
UNION
SELECT 'x123'
FROM dual
UNION
SELECT 'y'
FROM dual
UNION
SELECT '+789'
FROM dual
UNION
SELECT '-789'
FROM dual
UNION
SELECT '159-'
FROM dual
UNION
SELECT '-1-'
FROM dual
SELECT t.col1
FROM t
WHERE NOT REGEXP_LIKE(t.col1, '^[0-9]+$')
;Let's take a look at the 2nd argument of this REGEXP function: '^[0-9]+$'. Translated it would mean: start at the beginning of the string, check if there's one or more characters in the range between '0' and '9' (also called a matching character list) until the end of this string. "^", "[", "]", "+", "$" are all Metacharacters.
To understand regular expressions, you have to "think" in regular expressions. Each regular expression tries to "fit" an available string into its pattern and returns a result beeing successful or not, depending on the function. The "art" of using regular expressions is to construct the right search pattern for a certain task. Using functions like TRANSLATE or REPLACE did already teach you using search patterns, regular expressions are just an extension to this paradigma. Another side note: most of the search patterns are placeholders for single characters, not strings.
I'll take this example a bit further. What would happen if we would remove the "$" in our example? "$" means: (until the) end of a string. Without this, this expression would only search digits from the beginning until it encounters either another character or the end of the string. So this time, '123x' would be removed from the SELECTION since it does fit into the pattern.
Another change: we will keep the "$" but remove the "^". This character has several meanings, but in this case it declares: (start from the) beginning of a string. Without it, the function will search for a part of a string that has only digits until the end of the searched string. 'x123' would now be removed from our selection.
Now there's a question: what happens if I remove both, "^" and "$"? Well, just think about it. We now ask to find any string that contains at least one or more digits, so both '123x' and 'x123' will not show up in the result.
So what if I want to look for signed integer, since "+" is also used for a search expression. Escaping is the name of the game. We'll just use '^\+[0-9]+$' Did you notice the "\" before the first "+"? This is now a search pattern for the plus sign.
Should signed integers include negative numbers as well? Of course they should, and I'll once again use a matching character list. In this list, I don't need to do escaping, although it is possible. The result string would now look like this: '^[+-]?[0-9]+$'. Did you notice the "?"? This is another metacharacter that changes the placeholder for plus and minus to an optional placeholder, which means: if there's a "+" or "-", that's ok, if there's none, that's also ok. Only if there's a different character, then again the search pattern will fail.
Addendum: From this on, I found a mistake in my examples. If you would have tested my old examples with test data that would have included multiple signs strings, like "--", "-+", "++", they would have been filtered by the SELECT statement. I mistakenly used the "*" instead of the "?" operator. The reason why this is a bad idea, can also be found in the user guide: the "*" meta character is defined as 0 to multiple occurrences.
Looking at the values, one could ask the question: what about the integers with a trailing sign? Quite simple, right? Let's just add another '[+-] and the search pattern would look like this: '^[+-]?[0-9]+[+-]?$'.
Wait a minute, what happened to the row with the column value "-1-"?
You probably already guessed it: the new pattern qualifies this one also as a valid string. I could now split this pattern into several conditions combined through a logical OR, but there's something even better: a logical OR inside the regular expression. It's symbol is "|", the pipe sign.
Changing the search pattern again to something like this '^[+-]?[0-9]+$|^[0-9]+[+-]?$' [1] would return now the "-1-" value. Do I have to duplicate the same elements like "^" and "$", what about more complicated, repeating elements in future examples? That's where subexpressions/grouping comes into play. If I want only certain parts of the search pattern using an OR operator, we can put those inside round brackets. '^([+-]?[0-9]+|[0-9]+[+-]?)$' serves the same purpose and allows for further checks without duplicating the whole pattern.
Now looking for integers is nice, but what about decimal numbers? Those may be a bit more complicated, but all I have to do is again to think in (meta) characters. I'll just use an example where the decimal point is represented by ".", which again needs escaping, since it's also the place holder in regular expressions for "any character".
Valid decimals in my example would be ".0", "0.0", "0.", "0" (integer of course) but not ".". If you want, you can test it with the TO_NUMBER function. Finding such an unsigned decimal number could then be formulated like this: from the beginning of a string we will either allow a decimal point plus any number of digits OR at least one digits plus an optional decimal point followed by optional any number of digits. Think about it for a minute, how would you formulate such a search pattern?
Compare your solution to this one:
'^(\.[0-9]+|[0-9]+(\.[0-9]*)?)$'
Addendum: Here I have to use both "?" and "*" to make sure, that I can have 0 to many digits after the decimal point, but only 0 to 1 occurrence of this substrings. Otherwise, strings like "1.9.9.9" would be possible, if I would write it like this:
'^(\.[0-9]+|[0-9]+(\.[0-9]*)*)$'Some of you now might say: Hey, what about signed decimal numbers? You could of course combine all the ideas so far and you will end up with a very long and almost unreadable search pattern, or you start combining several regular expression functions. Think about it: Why put all the search patterns into one function? Why not split those into several steps like "check for a valid decimal" and "check for sign".
I'll just use another SELECT to show what I want to do:
WITH t AS (SELECT '0' col1
FROM dual
UNION
SELECT '0.'
FROM dual
UNION
SELECT '.0'
FROM dual
UNION
SELECT '0.0'
FROM dual
UNION
SELECT '-1.0'
FROM dual
UNION
SELECT '.1-'
FROM dual
UNION
SELECT '.'
FROM dual
UNION
SELECT '-1.1-'
FROM dual
SELECT t.*
FROM t
;From this select, the only rows I need to find are those with the column values "." and "-1.1-". I'll start this with a check for valid signs. Since I want to combine this with the check for valid decimals, I'll first try to extract a substring with valid signs through the REGEXP_SUBSTR function:
NVL(REGEXP_SUBSTR(t.col1, '^([+-]?[^+-]+|[^+-]+[+-]?)$'), ' ')Remember the OR operator and the matching character collections? But several "^"? Some of the meta characters inside a search pattern can have different meanings, depending on their positions and combination with other meta characters. In this case, the pattern translates into: from the beginning of the string search for "+" or "-" followed by at least another character that is not "+" or "-". The second pattern after the "|" OR operator does the same for a sign at the end of the string.
This only checks for a sign but not if there also only digits and a decimal point inside the string. If the search string fails, for example when we have more than one sign like in the "-1.1-", the function returns NULL. NULL and LIKE don't go together very well, so we'll just add NVL with a default value that tells the LIKE to ignore this string, in this case a space.
All we have to do now is to combine the check for the sign and the check for a valid decimal number, but don't forget an option for the signs at the beginning or end of the string, otherwise your second check will fail on the signed decimals. Are you ready?
Does your solution look a bit like this?
WHERE NOT REGEXP_LIKE(NVL(REGEXP_SUBSTR(t.col1,
'^([+-]?[^+-]+|[^+-]+[+-]?)$'),
'^[+-]?(\.[0-9]+|[0-9]+(\.[0-9]*)?)[+-]?$'
)Now the optional sign checks in the REGEXP_LIKE argument can be added to both ends, since the SUBSTR won't allow any string with signs on both ends. Thinking in regular expression again.
Continued in Introduction to regular expressions ... continued.
C.
Fixed some embarrassing typos ... and mistakes.
cdExcellent write up CD. Very nice indeed. Hopefully you'll be completing parts 2 and 3 some time soon. And with any luck, your article will encourage others to do the same....I know there's a few I'd like to see and a few I'd like to have a go at writing too :-)
-
Regular Expressions and comments
Hello,
I've got a problem with regular expressions. I Want to find special words like "todo" in java comments, but wasn't able to manage that satisfactory. I hope someone of you can give me an advice! :-)
example of a comment:
* comment text. todo: what is still todo.
First of all, I tried this:
/\*(.*?)todo(.*?)\*/
but this version seems to ignore the borders of the comment and shows me also parts of the code.
Then I tried this:
/\*([^/]*?)todo(.*?)\*/
this version returns good results, but doesn't notice html formatting like
* <code> .... </code> text. todo: text
So my idea now is to check whether a star precedes the slash, but I don't really know how to combine it.
Or is there another simplier solution? Thanks for your hints!
Greetz JanThanks for your answer, but when I take this expression the programm seems to hang-up. An operation that usually finishs in 3 secs didn't even in 10 minutes. :-( Do you have any idea what could be wrong?
btw. can I take this one:
pattern = Pattern.compile("/\\*(?:[^\\*]+|\\*(?!/))*todo.*?\\*/", Pattern.DOTALL | Pattern.CASE_INSENSITIVE);
matcher = pattern.matcher(text);
text = matcher.replaceAll("");or do I have to take a this instead of replaceAll():
while (matcher.find()) {
text = text.substring(0, matcher.start(1)) + text.substring(matcher.end(1), text.length());
matcher = pattern.matcher(text);
}any hints?
Greetz Jan -
Pattern and Matcher of Regular Expressions
Hello All,
MTMISRVLGLIRDQAISTTFGANAVTDAFWVAFRIPNFLRRLFAEGSFATAFVPVFTEVK
ETRPHADLRELMARVSGTLGGMLLLITALGLIFTPQLAAVFSDGAATNPEKYGLLVDLLR
LTFPFLLFVSLTALAGGALNSFQRFAIPALTPVILNLCMIAGALWLAPRLEVPILALGWA
VLVAGALQLLFQLPALKGIDLLTLPRWGWNHPDVRKVLTLMIPTLFGSSIAQINLMLDTV
IAARLADGSQSWLSLADRFLELPLGVFGVALGTVILPALARHHVKTDRSAFSGALDWGFR
TTLLIAMPAMLGLLLLAEPLVATLFQYRQFTAFDTRMTAMSVYGLSFGLPAYAMLKVLLP
I need some help with the regular expressions in java.
I have encountered a problem on how to retrieve two strings with Pattern and Matcher.
I have written this code to match one substring"MTMISRVLGLIRDQ", but I want to match multiple substrings in a string.
Pattern findstring = Pattern.compile("MTMISRVLGLIRDQ");
Matcher m = findstring.matcher(S);
while (m.find())
outputStream.println("Selected Sequence \"" + m.group() +
"\" starting at index " + m.start() +
" and ending at index " m.end() ".");
Any help would be appreciated.Double post: http://forum.java.sun.com/thread.jspa?threadID=726158&tstart=0
-
Trying to use regular expressions to convert names to Title Case
I'm trying to change names to their proper case for most common names in North America (esp. the U.S.).
Some examples are in the comments of the included code below.
My problem is that *retName = retName.replaceAll("( [^ ])([^ ]+)", "$1".toUpperCase() + "$2");* does not work as I expect. It seems that the toUpperCase method call does not actually do anything to the identified group.
Everything else works as I expect.
I'm hoping that I do not have to iterate through each character of the string, upshifting the characters that follow spaces.
Any help from you RegEx experts will be appreciated.
{code}
* Converts names in some random case into proper Name Case. This method does not have the
* extra processing that would be necessary to convert street addresses.
* This method does not add or remove punctuation.
* Examples:
* DAN MARINO --> Dan Marino
* old macdonald --> Old Macdonald <-- Can't capitalize the 'D" because of Ernst Mach
* ROY BLOUNT, JR. --> Roy Blount, Jr.
* CAROL mosely-BrAuN --> Carol Mosely-Braun
* Tom Jones --> Tom Jones
* ST.LOUIS --> St. Louis
* ST.LOUIS, MO --> St. Louis, Mo <-- Avoid City Names plus State Codes
* This is a work in progress that will need to be updated as new exceptions are found.
public static String toNameCase(String name) {
* Basic plan:
* 1. Strategically create double spaces in front of characters to be capitalized
* 2. Capitalize characters with preceding spaces
* 3. Remove double spaces.
// Make the string all lower case
String retName = name.trim().toLowerCase();
// Collapse strings of spaces to single spaces
retName = retName.replaceAll("[ ]+", " ");
// "mc" names
retName = retName.replaceAll("( mc)", " $1");
// Ensure there is one space after periods and commas
retName = retName.replaceAll("(\\.|,)([^ ])", "$1 $2");
// Add 2 spaces after periods, commas, hyphens and apostrophes
retName = retName.replaceAll("(\\.|,|-|')", "$1 ");
// Add a double space to the front of the string
retName = " " + retName;
// Upshift each character that is preceded by a space
// For some reason this doesn't work
retName = retName.replaceAll("( [^ ])([^ ]+)", "$1".toUpperCase() + "$2");
// Remove double spaces
retName = retName.replaceAll(" ", "");
return retName;
Edited by: FuzzyBunnyFeet on Jan 17, 2011 10:56 AM
Edited by: FuzzyBunnyFeet on Jan 17, 2011 10:57 AMHopefully someone will still be able to provide a RegEx solution, but until that time here is a working method.
Also, if people have suggestions of other rules for letter capitalization in names, I am interested in those too.
* Converts names in some random case into proper Name Case. This method does not have the
* extra processing that would be necessary to convert street addresses.
* This method does not add or remove punctuation.
* Examples:
* CAROL mosely-BrAuN --> Carol Mosely-Braun
* carol o'connor --> Carol O'Connor
* DAN MARINO --> Dan Marino
* eD mCmAHON --> Ed McMahon
* joe amcode --> Joe Amcode <-- Embedded "mc"
* mr.t --> Mr. T <-- Inserted space
* OLD MACDONALD --> Old Macdonald <-- Can't capitalize the 'D" because of Ernst Mach
* old mac donald --> Old Mac Donald
* ROY BLOUNT,JR. --> Roy Blount, Jr.
* ST.LOUIS --> St. Louis
* ST.LOUIS,MO --> St. Louis, Mo <-- Avoid City Names plus State Codes
* Tom Jones --> Tom Jones
* This is a work in progress that will need to be updated as new exceptions are found.
public static String toNameCase(String name) {
* Basic plan:
* 1. Strategically create double spaces in front of characters to be capitalized
* 2. Capitalize characters with preceding spaces
* 3. Remove double spaces.
// Make the string all lower case
String workStr = name.trim().toLowerCase();
// Collapse strings of spaces to single spaces
workStr = workStr.replaceAll("[ ]+", " ");
// "mc" names
workStr = workStr.replaceAll("( mc)", " $1 ");
// Ensure there is one space after periods and commas
workStr = workStr.replaceAll("(\\.|,)([^ ])", "$1 $2");
// Add 2 spaces after periods, commas, hyphens and apostrophes
workStr = workStr.replaceAll("(\\.|,|-|')", "$1 ");
// Add a double space to the front of the string
workStr = " " + workStr;
// Upshift each character that is preceded by a space and remove double spaces
// Can't upshift using regular expressions and String methods
// workStr = workStr.replaceAll("( [^ ])([^ ]+)", "$1"toUpperCase() + "$2");
StringBuilder titleCase = new StringBuilder();
for (int i = 0; i < workStr.length(); i++) {
if (workStr.charAt(i) == ' ') {
if (workStr.charAt(i+1) == ' ') {
i += 2;
while (i < workStr.length() && workStr.charAt(i) == ' ') {
titleCase.append(workStr.charAt(i++));
if (i < workStr.length()) {
titleCase.append(workStr.substring(i, i+1).toUpperCase());
} else {
titleCase.append(workStr.charAt(i));
return titleCase.toString();
{code} -
Introduction to regular expressions ... continued.
After some very positive feedback from Introduction to regular expressions ... I'm now continuing on this topic for the interested audience. As always, if you have questions or problems that you think could be solved through regular expression, please post them.
Having fun with regular expressions - Part 2
Finishing my example with decimal numbers, I thought about a method to test regular expressions. A question from another user who was looking for a way to show all possible combinations inspired me in writing a small package.
CREATE OR REPLACE PACKAGE regex_utils AS
-- Regular Expression Utilities
-- Version 0.1
TYPE t_outrec IS RECORD(
data VARCHAR2(255)
TYPE t_outtab IS TABLE OF t_outrec;
FUNCTION gen_data(
p_charset IN VARCHAR2 -- character set that is used for generation
, p_length IN NUMBER -- length of the generated
) RETURN t_outtab PIPELINED;
END regex_utils;
CREATE OR REPLACE PACKAGE BODY regex_utils AS
-- FUNCTION gen_data returns a collection of generated varchar2 elements
FUNCTION gen_data(
p_charset IN VARCHAR2 -- character set that is used for generation
, p_length IN NUMBER -- length of the generated
) RETURN t_outtab PIPELINED
IS
TYPE t_counter IS TABLE OF PLS_INTEGER INDEX BY PLS_INTEGER;
v_counter t_counter;
v_exit BOOLEAN;
v_string VARCHAR2(255);
v_outrec t_outrec;
BEGIN
FOR max_length IN 1..p_length
LOOP
-- init counter loop
FOR i IN 1..max_length
LOOP
v_counter(i) := 1;
END LOOP;
-- start data generation loop
v_exit := FALSE;
WHILE NOT v_exit
LOOP
-- start generation
v_string := '';
FOR i IN 1..max_length
LOOP
v_string := v_string || SUBSTR(p_charset, v_counter(i), 1);
END LOOP;
-- set outgoing record
v_outrec.data := v_string;
-- now pipe the result
PIPE ROW(v_outrec);
-- increment loop
<<inc_loop>>
FOR i IN REVERSE 1..max_length
LOOP
v_counter(i) := v_counter(i) + 1;
IF v_counter(i) > LENGTH(p_charset) THEN
IF i > 1 THEN
v_counter(i) := 1;
ELSE
v_exit := TRUE;
END IF;
ELSE
-- no further processing required
EXIT inc_loop;
END IF;
END LOOP;
END LOOP;
END LOOP;
END gen_data;
END regex_utils;
/This package is a brute force string generator using all possible combinations of a characters in a string up to a maximum length. Together with the regular expressions, I can now show what combinations my solution would allow to pass. But see for yourself:
SELECT *
FROM (SELECT data col1
FROM TABLE(regex_utils.gen_data('+-.0', 5))
) t
WHERE REGEXP_LIKE(NVL(REGEXP_SUBSTR(t.col1,
'^([+-]?[^+-]+|[^+-]+[+-]?)$'
'^[+-]?(\.[0-9]+|[0-9]+(\.[0-9]*)?)[+-]?$'
;You will see some results, which are perfectly valid for my definition of decimal numbers but haven't been mentioned, like '000' or '+.00'. From now on I will also use this package to verify the solutions I'll present to you and hopefully reduce my share of typos.
Counting and finding certain characters or words in a string can be a tedious task. I'll show you how it's done with regular expressions. I'll start with an easy example, count all spaces in the string "Having fun with regular expressions.":
SELECT NVL(LENGTH(REGEXP_REPLACE('Having fun with regular expressions', '[^ ]')), 0)
FROM dual
;No surprise there. I'm replacing all characters except spaces with a null string. Since REGEXP_REPLACE assumes a NULL string as replacement argument, I can save on adding a third argument, which would look like this:
REGEXP_REPLACE('Having fun with regular expressions', '[^ ]', '')So REPLACE will return all the spaces which we can count with the LENGTH function. If there aren't any, I will get a NULL string, which is checked by the NVL function. If you want you can play around by changing the space character to somethin else.
A variation of this theme could be counting the number of words. Counting spaces and adding 1 to this result could be misleading if there are duplicate spaces. Thanks to regular expressions, I can of course eliminate duplicates.
Using the old method on the string "Having fun with regular expressions" would return anything but the right number. This is, where Backreferences come into play. REGEXP_REPLACE uses them in the replacement argument, a backslash plus a single digit, like this: '\1'. To reference a string in a search pattern, I have to use subexpressions (remember the round brackets?).
SELECT NVL(LENGTH(REGEXP_REPLACE('Having fun with regular expressions', '( )\1*|.', '\1')))
FROM dual
;You may have noticed that I changed from using the "^" as a NOT operator to using the "|" OR operator and the "." any character placeholder. This neat little trick allows to filter all other characters except the one we're looking in the first place. "\1" as backreference is outside of our subexpression since I don't want to count the trailing spaces and is used both in the search pattern and the replacement argument.
Still I'm not satisfied with this: What about leading/trailing blanks, what if there are any special characters, numbers, etc.? Finally, it's time to only count words. For the purpose of this demonstration, I define a word as one or more consecutive letters. If by now you're already thinking in regular expressions, the solution is not far away. One hint: you may want to check on the "i" match parameter which allows for case insensitive search. Another one: You won't need a back reference in the search pattern this time.
Let's compare our solutions than, shall we?
SELECT NVL(LENGTH(REGEXP_REPLACE('Having fun with regular expressions. !',
'([a-z])+|.', '\1', 1, 0, 'i')), 0)
FROM dual;This time I don't use a backreference, the "+" operator (remember? 1 or more) will suffice. And since I want to count the occurences, not the letters, I moved the "+" meta character outside of the subexpression. The "|." trick again proved to be useful.
Case insensitive search does have its merits. It will only search but not transform the any found substring. If I want, for example, extract any occurence of the word fun, I'll just use the "i" match parameter and get this substring, whether it's written as "Fun", "FUN" or "fun". Can be very useful if you're looking for example for names of customers, streets, etc.
Enough about counting, how about finding? What if I want to know the last occurence of a certain character or string, for example the postition of the last space in this string "Where is the last space?"?
Addendum: Thanks to another forum member, I should mention that using the INSTR function can do a reverse search by itself.[i]
WITH t AS (SELECT 'Where is the last space?' col1
FROM dual)
SELECT INSTR(col1, ' ', -1)
FROM DUAL;Now regular expressions are powerful, but there is no parameter that allows us to reverse the search direction. However, remembering that we have the "$" meta character that means (until the) end of string, all I have to do is use a search pattern that looks for a combination of space and non-space characters including the end of a string. Now compare the REGEXP_INSTR function to the previous solution:
SELECT REGEXP_INSTR(t.col1, ' [^ ]*$')
FROM t;So in this case, it'll remain a matter of taste what you want to use. If the search pattern has to look for the last occurrence of another regular expression, this is the way to solve such a requirement.
One more thing about backreferences. They can be used for a sort of primitive "string swapping". If for example you have to transform column values like swapping first and last name, backreferenc is your friend. Here's an example:
SELECT REGEXP_REPLACE('John Doe', '^(.*) (.*)$', '\2, \1')
FROM dual
;What about middle names, for example 'John J. Doe'? Look for yourself, it still works.
You can even use that for strings with delimiters, for example reversing delimited "fields" like in this string '10~20~30~40~50' into '50~40~30~20~10'. Using REVERSE, I would get '05~04~03~02~01', so there has to be another way. Using backreferences however is limited to 9 subexpressions, which limits the following solution a bit, if you need to process strings with more than 9 fields. If you want, you can think this example through and see if your solution matches mine.
SELECT REGEXP_REPLACE('10~20~30~40~50',
'^(.*)~(.*)~(.*)~(.*)~(.*)$',
'\5~\4~\3~\2~\1'
FROM dual;After what you've learned so far, that wasn't too hard, was it? Enough for now ...
Continued in Introduction to regular expressions ... last part..
C.
Fixed some typos and a flawed example ...
cdThank you very much C. Awaiting other parts.... keep going.
One german typo :-)
I'm replacing all characters except spaces mit anull string.I received a functional spec from my Dutch analyst in which it is written
tnsnames voor EDWH:
PCESCRD1 = (DESCRIPTION=(ADDRESS_LIST=(ADDRESS=(PROTOCOL=TCP)
(HOST=blah.blah.blah.com)
(PORT=5227)))
(CONNECT_DATA=(SID=pcescrd1)))
db user: BW_I2_VIEWER / BW_I2_VIEWER_SCRD1Had to look for translators.
Cheers
Sarma. -
Introduction to regular expressions ... last part.
Continued from Introduction to regular expressions ... continued., here's the third and final part of my introduction to regular expressions. As always, if you find mistakes or have examples that you think could be solved through regular expressions, please post them.
Having fun with regular expressions - Part 3
In some cases, I may have to search for different values in the same column. If the searched values are fixed, I can use the logical OR operator or the IN clause, like in this example (using my brute force data generator from part 2):
SELECT data
FROM TABLE(regex_utils.gen_data('abcxyz012', 4))
WHERE data IN ('abc', 'xyz', '012');There are of course some workarounds as presented in this asktom thread but for a quick solution, there's of course an alternative approach available. Remember the "|" pipe symbol as OR operator inside regular expressions? Take a look at this:
SELECT data
FROM TABLE(regex_utils.gen_data('abcxyz012', 4))
WHERE REGEXP_LIKE(data, '^(abc|xyz|012)$')
;I can even use strings composed of values like 'abc, xyz , 012' by simply using another regular expression to replace "," and spaces with the "|" pipe symbol. After reading part 1 and 2 that shouldn't be too hard, right? Here's my "thinking in regular expression": Replace every "," and 0 or more leading/trailing spaces.
Ready to try your own solution?
Does it look like this?
SELECT data
FROM TABLE(regex_utils.gen_data('abcxyz012', 4))
WHERE REGEXP_LIKE(data, '^(' || REGEXP_REPLACE('abc, xyz , 012', ' *, *', '|') || ')$')
;If I wouldn't use the "^" and "$" metacharacter, this SELECT would search for any occurence inside the data column, which could be useful if I wanted to combine LIKE and IN clause. Take a look at this example where I'm looking for 'abc%', 'xyz%' or '012%' and adding a case insensitive match parameter to it:
SELECT data
FROM TABLE(regex_utils.gen_data('abcxyz012', 4))
WHERE REGEXP_LIKE(data, '^(abc|xyz|012)', 'i')
; An equivalent non regular expression solution would have to look like this, not mentioning other options with adding an extra "," and using the INSTR function:
SELECT data
FROM (SELECT data, LOWER(DATA) search
FROM TABLE(regex_utils.gen_data('abcxyz012', 4))
WHERE search LIKE 'abc%'
OR search LIKE 'xyz%'
OR search LIKE '012%'
SELECT data
FROM (SELECT data, SUBSTR(LOWER(DATA), 1, 3) search
FROM TABLE(regex_utils.gen_data('abcxyz012', 4))
WHERE search IN ('abc', 'xyz', '012')
; I'll leave it to your imagination how a complete non regular example with 'abc, xyz , 012' as search condition would look like.
As mentioned in the first part, regular expressions are not very good at formatting, except for some selected examples, such as phone numbers, which in my demonstration, have different formats. Using regular expressions, I can change them to a uniform representation:
WITH t AS (SELECT '123-4567' phone
FROM dual
UNION
SELECT '01 345678'
FROM dual
UNION
SELECT '7 87 8787'
FROM dual
SELECT t.phone, REGEXP_REPLACE(REGEXP_REPLACE(phone, '[^0-9]'), '(.{3})(.*)', '(\1)-\2')
FROM t
;First, all non digit characters are beeing filtered, afterwards the remaining string is put into a "(xxx)-xxxx" format, but not cutting off any phone numbers that have more than 7 digits. Using such a conversion could also be used to check the validity of entered data, and updating the value with a uniform format afterwards.
Thinking about it, why not use regular expressions to check other values about their formats? How about an IP4 address? I'll do this step by step, using 127.0.0.1 as the final test case.
First I want to make sure, that each of the 4 parts of an IP address remains in the range between 0-255. Regular expressions are good at string matching but they don't allow any numeric comparisons. What valid strings do I have to take into consideration?
Single digit values: 0-9
Double digit values: 00-99
Triple digit values: 000-199, 200-255 (this one will be the trickiest part)
So far, I will have to use the "|" pipe operator to match all of the allowed combinations. I'll use my brute force generator to check if my solution works for a single value:
SELECT data
FROM TABLE(regex_utils.gen_data('0123456789', 3))
WHERE REGEXP_LIKE(data, '^(25[0-5]|2[0-4][0-9]|[01]?[0-9]{1,2})$')
; More than 255 records? Leading zeros are allowed, but checking on all the records, there's no value above 255. First step accomplished. The second part is to make sure, that there are 4 such values, delimited by a "." dot. So I have to check for 0-255 plus a dot 3 times and then check for another 0-255 value. Doesn't sound to complicated, does it?
Using first my brute force generator, I'll check if I've missed any possible combination:
SELECT data
FROM TABLE(regex_utils.gen_data('03.', 15))
WHERE REGEXP_LIKE(data,
'^((25[0-5]|2[0-4][0-9]|[01]?[0-9]{1,2})\.){3}(25[0-5]|2[0-4][0-9]|[01]?[0-9]{1,2})$'
; Looks good to me. Let's check on some sample data:
WITH t AS (SELECT '127.0.0.1' ip
FROM dual
UNION
SELECT '256.128.64.32'
FROM dual
SELECT t.ip
FROM t WHERE REGEXP_LIKE(t.ip,
'^((25[0-5]|2[0-4][0-9]|[01]?[0-9]{1,2})\.){3}(25[0-5]|2[0-4][0-9]|[01]?[0-9]{1,2})$'
; No surprises here. I can take this example a bit further and try to format valid addresses to a uniform representation, as shown in the phone number example. My goal is to display every ip address in the "xxx.xxx.xxx.xxx" format, using leading zeros for 2 and 1 digit values.
Regular expressions don't have any format models like for example the TO_CHAR function, so how could this be achieved? Thinking in regular expressions, I first have to find a way to make sure, that each single number is at least three digits wide. Using my example, this could look like this:
WITH t AS (SELECT '127.0.0.1' ip
FROM dual
SELECT t.ip, REGEXP_REPLACE(t.ip, '([0-9]+)(\.?)', '00\1\2')
FROM t
; Look at this: leading zeros. However, that first value "00127" doesn't look to good, does it? If you thought about using a second regular expression function to remove any excess zeros, you're absolutely right. Just take the past examples and think in regular expressions. Did you come up with something like this?
WITH t AS (SELECT '127.0.0.1' ip
FROM dual
SELECT t.ip, REGEXP_REPLACE(REGEXP_REPLACE(t.ip, '([0-9]+)(\.?)', '00\1\2'),
'[0-9]*([0-9]{3})(\.?)', '\1\2'
FROM t
; Think about the possibilities: Now you can sort a table with unformatted IP addresses, if that is a requirement in your application or you find other values where you can use that "trick".
Since I'm on checking INET (internet) type of values, let's do some more, for example an e-mail address. I'll keep it simple and will only check on the
"[email protected]", "[email protected]" and "[email protected]" format, where x represents an alphanumeric character. If you want, you can look up the corresponding RFC definition and try to build your own regular expression for that one.
Now back to this one: At least one alphanumeric character followed by an "@" at sign which is followed by at least one alphanumeric character followed by a "." dot and exactly 3 more alphanumeric characters or 2 more characters followed by a "." dot and another 2 characters. This should be an easy one, right? Use some sample e-mail addresses and my brute force generator, you should be able to verify your solution.
Here's mine:
SELECT data
FROM TABLE(regex_utils.gen_data('a1@.', 9))
WHERE REGEXP_LIKE(data, '^[[:alnum:]]+@[[:alnum:]]+(\.[[:alnum:]]{3,4}|(\.[[:alnum:]]{2}){2})$', 'i'); Checking on valid domains, in my opinion, should be done in a second function, to keep the checks by itself simple, but that's probably a discussion about readability and taste.
How about checking a valid URL? I can reuse some parts of the e-mail example and only have to decide what type of URLs I want, for example "http://", "https://" and "ftp://", any subdomain and a "/" after the domain. Using the case insensitive match parameter, this shouldn't take too long, and I can use this thread's URL as a test value. But take a minute to figure that one out for yourself.
Does it look like this?
WITH t AS (SELECT 'Introduction to regular expressions ... last part. URL
FROM dual
UNION
SELECT 'http://x/'
FROM dual
SELECT t.URL
FROM t
WHERE REGEXP_LIKE(t.URL, '^(https*|ftp)://(.+\.)*[[:alnum:]]+(\.[[:alnum:]]{3,4}|(\.[[:alnum:]]{2}){2})/', 'i')
Update: Improvements in 10g2
All of you, who are using 10g2 or XE (which includes some of 10g2 features) may want to take a look at several improvements in this version. First of all, there are new, perl influenced meta characters.
Rewriting my example from the first lesson, the WHERE clause would look like this:
WHERE NOT REGEXP_LIKE(t.col1, '^\d+$')Or my example with searching decimal numbers:
'^(\.\d+|\d+(\.\d*)?)$'Saves some space, doesn't it? However, this will only work in 10g2 and future releases.
Some of those meta characters even include non matching lists, for example "\S" is equivalent to "[^ ]", so my example in the second part could be changed to:
SELECT NVL(LENGTH(REGEXP_REPLACE('Having fun with regular expressions', '\S')), 0)
FROM dual
;Other meta characters support search patterns in strings with newline characters. Just take a look at the link I've included.
Another interesting meta character is "?" non-greedy. In 10g2, "?" not only means 0 or 1 occurrence, it means also the first occurrence. Let me illustrate with a simple example:
SELECT REGEXP_SUBSTR('Having fun with regular expressions', '^.* +')
FROM dual
;This is old style, "greedy" search pattern, returning everything until the last space.
SELECT REGEXP_SUBSTR('Having fun with regular expressions', '^.* +?')
FROM dual
;In 10g2, you'd get only "Having " because of the non-greedy search operation. Simulating that behavior in 10g1, I'd have to change the pattern to this:
SELECT REGEXP_SUBSTR('Having fun with regular expressions', '^[^ ]+ +')
FROM dual
;Another new option is the "x" match parameter. It's purpose is to ignore whitespaces in the searched string. This would prove useful in ignoring trailing/leading spaces for example. Checking on unsigned integers with leading/trailing spaces would look like this:
SELECT REGEXP_SUBSTR(' 123 ', '^[0-9]+$', 1, 1, 'x')
FROM dual
;However, I've to be careful. "x" would also allow " 1 2 3 " to qualify as valid string.
I hope you enjoyed reading this introduction and hope you'll have some fun with using regular expressions.
C.
Fixed some typos ...
Message was edited by:
cd
Included 10g2 features
Message was edited by:
cdCan I write this condition with only one reg expr in Oracle (regexp_substr in my example)?I meant to use only regexp_substr in select clause and without regexp_like in where clause.
but for better understanding what I'd like to get
next example:
a have strings of two blocks separated by space.
in the first block 5 symbols of [01] in the second block 3 symbols of [01].
In the first block it is optional to meet one (!), in the second block it is optional to meet one (>).
The idea is to find such strings with only one reg expr using regexp_substr in the select clause, so if the string does not satisfy requirments should be passed out null in the result set.
with t as (select '10(!)010 10(>)1' num from dual union all
select '1112(!)0 111' from dual union all --incorrect because of '2'
select '(!)10010 011' from dual union all
select '10010(!) 101' from dual union all
select '10010 100(>)' from dual union all
select '13001 110' from dual union all -- incorrect because of '3'
select '100!01 100' from dual union all --incorrect because of ! without (!)
select '100(!)1(!)1 101' from dual union all -- incorrect because of two occurencies of (!)
select '1001(!)10 101' from dual union all --incorrect because of length of block1=6
select '1001(!)10 1011' from dual union all) --incorrect because of length of block2=4
select '10110 1(>)11(>)0' from dual union all)--incorrect because of two occurencies of (>)
select '1001(>)1 11(!)0' from dual)--incorrect because (!) and (>) are met not in their blocks
--end of test data -
Using Regular Expressions to Find Quoted Text
I have run into a couple problems with the following code.
1) Slash-Star and Slash-Slash commented text must be ignored.
2) It does not detect backslashed quotes, or if that backslash is backslashed.
Can this be accomplished with Regular Expressions, or should I implement this using if/indexOf logic?
Thank You in advance,
Brian
* Finds position of next quoted string in a line
* of source code.
* If no strings exist, then a Pointer position of
* (0,0) is returned.
* @param startPos position to start search from
* @param argText the line of text to search
* @returns next string position
public Pointer getQuotedStringPosition(int startPos, String aString) {
String argText = new String( aString );
Pattern p = Pattern.compile("[\"][^\"]+[\"]");
Matcher m = p.matcher( argText.substring(startPos); );
if( m.find() )
return new Pointer( m.start() + startPos, m.end() + startPos );
else
return new Pointer( 0, 0 ); // indicates nothing was found
}YATArchivist was right about the regular expressions.
I think I've got it but somebody test it if you want. Let me know what you find.
I've included a barebones Position class as well...
import java.util.regex.*;
import java.io.*;
import java.util.*;
@author Joshua A. Logan, Jr.
public class RegexTest
private static final String SLASH_SLASH = "(//.*)";
private static final String SLASH_STAR =
"(/\\*(?:[^\\*]|(?:\\*(?!/)))+(\\*/)?)";
private static final Pattern COMMENT_PATTERN =
Pattern.compile( SLASH_SLASH + "|" + SLASH_STAR );
private static final Pattern QUOTED_STRING_PATTERN =
Pattern.compile( "\" ( (?:(\\\\.) | [^\\\"])*+ ) \"",
Pattern.COMMENTS );
// Breaking the above regular expression down, you'd have:
// " ( (?: (\\ .) | [^\\ "] ) *+ ) "
// ^ ^ ^ ^ ^ ^
// | | | | | |
// 1 2 3 4 5 6
// which matches:
// 1) The starting quote...
// Followed by something that is either:
// 2) some escaped sequence ( e.g. _\n_ or even _\"_ ),
// 3) ...or...
// 4) a character that is neither a _\_ nor a _"_ .
// 5) Keep searching this as much as possible, w/o giving up
// any found text at the end.
// Note: the text found would be in group(1)
// 6) Finally, find the ending quote!!
public static Position [] getQuotedStringPosition( final String text )
Matcher cm = COMMENT_PATTERN.matcher( text ),
qm = QUOTED_STRING_PATTERN.matcher( text );
final int len = text.length();
int startPos = 0;
List positions = new ArrayList();
while ( startPos < len )
if ( cm.find(startPos) )
int commStart = cm.start(),
commEnd = cm.end();
// are we starting @ a comment?
if ( commStart == startPos )
startPos = commEnd;
else if ( qm.find(startPos) )
// Search for unescaped strings in here.
int stringStart = qm.start(1),
stringEnd = qm.end(1);
// Is the quote start after comment start?
if ( stringStart > commStart )
startPos = commEnd; // restart search after comment end...
else if ( (stringEnd > commEnd) ||
(stringEnd < commStart) )
// In this case, the "comment" is actually part of
// the quoted string. We found a match.
positions.add( new Position(text, qm.group(1),
stringStart,
stringEnd) );
int quoteEnd = qm.end();
startPos = quoteEnd;
else
throw new IllegalStateException( "illegal case" );
else
startPos = commEnd;
else
// no comments were found. Search for unescaped strings.
int quoteEnd = len;
if ( qm.find( startPos ) ) {
quoteEnd = qm.end();
positions.add( new Position(text,
qm.group(1),
qm.start(1),
qm.end(1)) );
startPos = quoteEnd;
return positions.isEmpty() ? Position.EMPTY_ARRAY
: (Position[])positions.toArray(
Position.EMPTY_ARRAY);
public static void main( String [] args )
try
BufferedReader br = new BufferedReader(
new InputStreamReader(System.in) );
String input = null;
final String prompt = "\nText (q to quit): ";
System.out.print( prompt );
while ( (input = br.readLine()) != null )
if ( input.equals("q") ) return;
Position [] matches = getQuotedStringPosition( input );
// What does it do?
for ( int i = 0, max = matches.length; i < max; i++ )
System.out.println( "-->" + matches[i] );
System.out.print( prompt );
catch ( Exception e )
System.out.println ( "Exception caught: " + e.getMessage () );
class Position
public Position( String target,
String match,
int start,
int end )
this.target = target;
this.match = match;
this.start = start;
this.end = end;
public String toString()
return "match==" + match + ",{" + start + "," + end + "}";
final String target;
final int start;
final int end;
final String match;
public static final Position [] EMPTY_ARRAY = { };
} -
Regular expressions and string matching
Hi everyone,
Here is my problem, I have a partially decrypted piece string which would appear something like.
Partially deycrpted: the?anage??esideshe?e
Plain text: themanagerresideshere
So you can see that there are a few letter missing from the decryped text. What I am trying to do it insert spaces into the string so that I get:
The ?anage? ?esides he?e
I have a method which splits up the string in substrings of varying lengths and then compares the substring with a word from a dictionary (implemented as an arraylist) and then inserts a space.
The problem is that my function does not find the words in the dictionary because my string is only partially decryped.
Eg: ?anage? is not stored in the dictionary, but the word �manager� is.
So my question is, is there a way to build a regular expression which would match the partially decrypted text with a word from a dictionary (ie - ?anage? is recognised and �manager� from the dictionary).I wrote the following method in order to test the matching using . in my regular expression.
public void getWords(int y)
int x = 0;
for(y=y; y < buff.length(); y++){
String strToCompare = buff.substring(x,y); //where buff holds the partially decrypted text
x++;
Pattern p = Pattern.compile(strToCompare);
for(int z = 0; z < dict.size(); z++){
String str = (String) dict.get(z); //where dict hold all the words in the dictionary
Matcher m = p.matcher(str);
if(m.matches()){
System.out.println(str);
System.out.println(strToCompare);
// System.out.println(buff);
If I run the method where my parameter = 12, I am given the following output.
aestheticism
aestheti.is.
demographics
de.o.ra.....
Which suggests that the method is working correctly.
However, after running for a short time, the method cuts and gives me the error:
PatternSyntaxException:
Null(in java.util.regex.Pattern).
Does anyone know why this would occur? -
Hi Experts,
After going through some documentation on regular expressions in Oracle I have tried to draw some conclusions about the same. As I wasn’t much confident on how the patterns are built, I have tried to interpret them by looking at the output. It’s basically a reverse engineering I have tried to do.
Please let me know if my interpretations are correct. Any additions /suggestions/corrections are most welcome.
Some of the examples may lack conclusions, please ignore those.
select regexp_substr('1PSN/231_3253/ABc','^([[:alnum:]]*)') from dual;
Output: 1PSN
Interpreted as:
^ From the start of the source string
([[:alnum:]]*) zero or more occurrences of alphanumeric characters
select regexp_substr('@@/231_3253/ABc','@*([[:alnum:]]+)') from dual;
Output: 231
Interpreted as:
@* Search for zero or more occurrences of @
([[:alnum:]]+) followed by one or more occurrences of alphanumeric characters
Note: In the above example oracle looks for @(zero times or more) immediately followed by alphanumeric characters.
Since a '/' comes between @ and 231 the o/p is 0 occurences of @ + one or more occurrences of alphanumerics.
select regexp_substr('1@/231_3253/ABc','@+([[:alnum:]]*)') from dual;
Output: @
Interpreted as:
@+ one or more ocurrences of @
([[:alnum:]]*) followed by 0 or more occurrences of alphanumerics
select regexp_substr('1@/231_3253/ABc','@+([[:alnum:]]+)') from dual;
Output: Null
Interpreted as:
@+ one or more occurences of @
([[:alnum:]]+) followed by one or more occurences of aplhanumerics
select regexp_substr('@1PSN/231_3253/ABc125','([[:digit:]]+)$') from dual;
Output: 125
Interpreted as:
([[:digit:]]+) one or more occurences of digits only
$ at the end of the string
select regexp_substr('@1PSN/231_3253/ABc','([^[:digit:]]+)$') from dual;
output: /ABc
Interpreted as:
([^[:digit:]]+)$ one or more occurrences of non-digit literals at the end of the string
'^' inside square brackets marks the negation of the class
Look for http:// followed by a substring of one or more alphanumeric characters and optionally, a period (.)
SELECT REGEXP_SUBSTR('Go to http://www.oracle.com/products and click on database','http://([[:alnum:]]+\.?){3,4}/?') RESULT
FROM dual;
Output: http://www.oracle.com
Interpreted as:
[[:alnum:]]+ one or more occurences of alplanumeric characters
\.? dot optionally (backslash represents escape sequence,? represents optionally)
{3,4} 3 or 4 times
/? followed by forward slash optionally
If you have www.oracle.co.uk; {3,4} extracts it for you as well
Validate email:
select case when
REGEXP_LIKE('[email protected]',
'^([[:alnum:]]+(\_?|\.))([[:alnum:]]*)@([[:alnum:]]+)(.([[:alnum:]]+)){1,2}$') then 'Match Found'
else 'No Match Found'
end
as output from dual;
Interpreted as:
([[:alnum:]]+(\_?|\.)) one or more occurrences of alpha numerics optionally followed by an underscore or dot
([[:alnum:]]*) followed by 0 or more occurrences of alplhanumerics
@ followed by @
([[:alnum:]]+) followed by one or more occurrences of alplhanumerics
(.([[:alnum:]]+)){1,2} followed by a dot followed by alphanumerics from once till max of twice (Ex- .com or .co.uk)
Output: Match Found
Input: [email protected]
Output: Match Found
Input: [email protected]
Output: No Match Found
Truncate the part, ending with digits
select regexp_substr('Yahoo11245@US','^.*[[:digit:]]',1) from dual;
Output: Yahoo11245
select regexp_substr('*Yahoo*11245@US','^.*[[:digit:]]',1) from dual;
Output: *Yahoo*11245
Interpreted as:
.* zero or more occurrences of any characters (dot signifies any character)
Replace 2 to 8 spaces with single space
select regexp_replace('Hello you OPs there','[[:space:]]{2,8}',' ')
from dual;
Search for control characters
select case when
regexp_like('Super' || chr(13) || 'Star' ,'[[:cntrl:]]')
then 'Match Found'
else 'No Match Found'
end
as output from dual;
Output: Match Found
Search for lower case letters only with a string length varying from a min of 3 to max of 12
select case when
regexp_like('terminator' ,'^[[:lower:]]{3,12}$')
then 'Match Found'
else 'No Match Found'
end
as output from dual;
4th character must be a special character
select case when
regexp_like('ter*minator' ,'^...[^[:alnum:]]')
then 'Match Found'
else 'No Match Found'
end
as output from dual;
Ouput: Match Found
Case Sensitive Search
select case when
regexp_like('Republic Of Africa' ,'of','c')
then 'Match Found'
else 'No Match Found'
end
as output from dual;
Output: No match found
c stands for case sensitive
select case when
regexp_like('Republic Of africa' ,'of','i')
then 'Match Found'
else 'No Match Found'
end
as output from dual;
Output: Match Found
i stands for case insensitive
Two consecutive occurences of characters from a to z
select regexp_substr('Republicc Of Africaa' ,'([a-z])\1', 1,1,'i') from dual;
Output: cc
Interpreted as:
([a-z]) character set a-z
\1 consecutive occurence of any character
1 starting from 1st character in the string
1 First occurence
i case insensitive
Three consecutive occurences of characters from 6 to 9
select case when
regexp_like('Patch 10888 applied' ,'([7-9])\1\1')
then 'Match Found'
else 'No Match Found'
end
as output from dual;
Output: Match Found
Phone validator:
select case when
regexp_like('123-44-5555' ,'^[0-9]{3}-[0-9]{2}-[0-9]{4}$')
then 'Match Found'
else 'No Match Found'
end
as output from dual;
Output: Match Found
Input: 111-222-3333
Output: No match found
Interpreted as:
^ start of the string
[0-9]{3} three ocurrences of digits from 0-9
- followed by hyphen
[0-9]{2} two ocurrences of digits from 0-9
- followed by hyphen
[0-9]{4} four ocurrences of digits from 0-9
$ end of the string
************************************************************************Source Links:
http://www.psoug.org/reference/regexp.html
http://www.oracle.com/technology/obe/obe10gdb/develop/regexp/regexp.htm
Edited by: Preta on Feb 25, 2010 4:38 PM
Corrected the example for www.oracle.com
Edited by: Preta Incorported Logan's commentsHi,
It looks like you have a good understanding of how regular expressions work.
You can put comments like the ones in your message directly in the code. For example, your validate e-mail code could be re-written
select case
when REGEXP_LIKE ( '[email protected]'
, '^' || -- Starting from the beginning of the string
'(' || -- Begin \1
'[[:alnum:]]+'|| -- 0 or more alphnumerics
'(\_?|\.)' || -- optional underscore or dot
')' || -- End \1
'([[:alnum:]]*)'|| -- 0 or more alphnumerics
'@' || -- @ sign
'([[:alnum:]]+)'|| -- 1 or more alpanumerics
'(' || -- Begin \5
'\.' || -- dot
'([[:alnum:]]+)'
|| -- 1 or more alphanumerics
')' || -- End \5
'{1,2}' || -- \5 can occur 1 or 2 times
'$' -- End of string
then 'Match Found'
else 'No Match Found'
end as output
from dual;I find this easier to debug and maintain.
There's no denying, it does make the code very long. You be the judge of when to do this.
You use parentheses and \ unnceccessarily sometimes. That's not really an error; if you find they make the code easier to develop and maintain, use them as much as you like.
For example, about the 4th line of the regular expression as I formatted it above:
'(\_?|\.)' || -- optional underscore or dotUnderscore has no special meaning in regular expressions (only in LIKE), so you don't have to escape it.
I might write that line:
'(_|\.)?' || -- optional underscore or dotjust because I think it's clearer.
I think you forgot a \ about 7 lines later:
'\.' || -- dotBe very careful about testing patterns that include literal dots; always make sure that a random character, like ~ , fails in a place where a dot is expected. -
Regular expressions: serious design flaw?
I wondered why sometimes, the replaceAll() method works in my code and sometimes it throws a java.util.regex.PatternSyntaxException. I wrote a little test case to show the problem
private void regularExpressionTest() {
String escapers = "\\([{^$|)?*+."; //characters to escape to avoid PatternSyntaxException
String text = "The quick brown fox jumps over the lazy dog.";
System.out.println("Original: "+text);
String regExp = "o";
String word = "HELLO";
String result = text.replaceAll(regExp, word);
System.out.println("Replace all '"+regExp+"' with "+word+": "+result);
text = "The quick brown {animal} jumps over the lazy dog.";
System.out.println("Original: "+text);
regExp = "{animal}";
// regExp = escapeChars(regExp, escapers); //escape possible regular expression characters
word = "catterpillar";
result = text.replaceAll(regExp, word);
System.out.println("Replace all '"+regExp+"' with "+word+": "+result);
}//regularExpressionTest()The output is:
Original: The quick brown fox jumps over the lazy dog.
Replace all 'o' with HELLO: The quick brHELLOwn fHELLOx jumps HELLOver the lazy dHELLOg.
Original: The quick brown {animal} jumps over the lazy dog.
java.util.regex.PatternSyntaxException: Illegal repetition
{animal}
at java.util.regex.Pattern.error(Pattern.java:1528)
at java.util.regex.Pattern.closure(Pattern.java:2545)
at java.util.regex.Pattern.sequence(Pattern.java:1656)
at java.util.regex.Pattern.expr(Pattern.java:1545)
at java.util.regex.Pattern.compile(Pattern.java:1279)
at java.util.regex.Pattern.<init>(Pattern.java:1035)
at java.util.regex.Pattern.compile(Pattern.java:779)
at java.lang.String.replaceAll(String.java:1663)
at de.icomps.prototypes.Test.regularExpressionTest(Test.java:57)
at de.icomps.prototypes.Test.main(Test.java:34)
Exception in thread "main" As the first replacement works as espected, the second throws an Exception. Possible because '{' is a special character for the regular expression API. In case I know the regular expression, i could escape these special characters. But in my generic server app, the strings are all parameters, so the only way for replaceAll() to work is, to escape all possible special characters in the regular expression.
1. Is there a complete list of all special characters for regular expressions that need to be escaped?
2. Is there a similar replaceAll() method without regular expressions? So that all occurences of a substring can be replaced by another string? (So far, I wrote my own method but this is of course more time consuming for massive string operations.)1. The complete list of specially-recognized characters are listed in the Java 1.4.* API. (Of course, new ones may eventually be added as the regex package matures).
2. It is time consuming to program in general. You should have written your own utility method that goes through a String using indexOf and building up the String in a StringBuffer, and apparently you did. Now you have the utility method...you no longer need to write that method again.
3. Or you could have written some kind of method that automatically escapes the specially-recognized characters...
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