Regular Expression Programmatically from Bean
Hi,
I am on JDeveloper 11.1.2.0.0.
We have few Field validation using Regular Expression.
eg.
Number
[0-9]{1,10}
or
Email
[A-Za-z0-9._%+-]+@[A-Za-z0-9.-]+\\.[A-Za-z]{2,4}
Can we fire it from Bean. I need to move these changes to Value Change Listener or Validater
Thanks,
Kiran
Hi Suresh Frank,
Appreciate your quick response.
When i tried putting in Value Change Listener it worked.
I put below code to show error.
FacesContext context = FacesContext.getCurrentInstance();
FacesMessage message = new FacesMessage(FacesMessage.SEVERITY_ERROR, "Email Format is not correct " + valueChangeEvent.getNewValue(), msg);
context.addMessage(null,message); It throws error in popup window. However after clicking ok it is going to next fields and allowing to submit Form.
My requirement is if value not changed(even though it is in incorrect format) just proceed however if value changed then make sure format is correct and do not allow to proceed.
Can we achieve this using FacesMessage.
Thanks,
Kiran
Similar Messages
-
Regular expressions read from a file?
Hello,
I'm trying to read a bunch of string literals from a file, and then replace their appearances in another string. It works, except when I have special characters in the file, like this:
(hello)
When I use that in Pattern.compile(), it obviously thinks that the parenthesis are not literal. So, I edited my text file and changed the line to:
\\(hello\\)
No change. It will not match the string at all. I've even tried this (just out of deperation):
\\\\(hello\\\\)
It just doesn't seem I can escape it right if the expression is being read from a file.
What am I doing wrong?Another idea: I tried the following
import java.io.*;
public class RegexTest {
public static void main(String[] args) {
try {
File input = new File("pattern.txt");
FileReader in = new FileReader(input);
int c;
StringBuffer buf = new StringBuffer();
try {
c = in.read();
while (c != -1) {
if (!Character.isWhitespace((char)c)) {
buf.append((char) c);
c = in.read();
System.out.println(buf.toString());
} catch (IOException iox) {}
String previous = "(hello) test test test";
System.out.println(previous);
System.out.println(previous.replaceAll(buf.toString(), "test"));
} catch (FileNotFoundException nfnx) {}
}and it works properly.
I found out that if the input line with the "\(hello\)" pattern ends with the usual CRLF the only way to have the expected results is to filter out the extra characters (using the !Character.isWhitespace((char)c) test).
Does this give you any help? -
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 -
I know the purpose of that code , but I don't understand it Sad
SQL> SELECT REGEXP_SUBSTR(
'The final test is is the implementation',
'([[:alnum:]]+)([[:space:]]+)\1') AS substr
FROM dual;
SUBST
is is
I study Regular Expression nowadays , from that link :-
http://www.oracle.com/technology/oramag/webcolumns/2003/techarticles/rischert_regexp_pt1.html
but I'm still need more good resources for ( Regular Expression white paper ) until I can understand all benefits , hence I can take Full Control of it.just one thing to be aware of unrelated to backreferencing:
that Regex is less than perfect for the purposes of matching multiple words together:
SELECT REGEXP_SUBSTR(
'The diagram is isometric',
'([[:alnum:]]+)([[:space:]]+)\1') AS substr
FROM dual;so perhaps something like this:
SELECT REGEXP_SUBSTR(
'The diagram is isometric and and in pencil',
'([[:alnum:]]+)([[:space:]]+)\1\2') AS substr
FROM dual;using the backreference again to find the same space after the second word
but then again, this doesn't work perfectly because what if there's only one space after the second word but two after the first... and what if the duplicate word is at the end of the string without any other characters after it at all!!!!
find someone who can write a perfect regex and you'll have found someone who has too much time on their hands... :-P -
Use regular expressions to extract .llb filename from path
I am trying to be clever (always a dangerous thing) and use a regular expression to extract the name of a library from a filepath converted to a string. Whilst I appreciate there are other ways to do this, regex would appear to be a very powerful neat way, should I be able to get it to work.
i.e if I have a string of the type, C:\applications\versions\library.llb\toplevel.vi, I want to be able to extract library.llb from the string, given that it will be of variable length, may include numbers & spaces and may be within a file hierarchy of variable depth. In other words, I want to extract the portion of the string between the last \ that ends with .llb
The best I have managed so far, is \\+.*llb which returned everything minus the drive letter and the toplevel.vi
Can anyone help me achieve this, or am i better using an alternative method (e.g filepath to array string, search for .llb)
Thanks
Matt
Solved!
Go to Solution.Hi Matt,
attached you'll find two other options.
Mike
Message Edited by MikeS81 on 04-13-2010 01:30 PM
Attachments:
Path.PNG 7 KB -
Regular expressions-how to replace [ and ] characters from a string
Hi,
my input String is "sdf938 [98033]". Now from this given string, i would like to replace the characters occurring within square brackets to empty string, including the square brackets too.
my output String needs to be "sdf938" in this case.. How should I do it using regular expressions? I tried several possible combinations but didn't get the expected results."\\s*\\[[^\\]]+\\]"
-
Matches from regular expression into collection
Hello,
I need to do the following:
I have a long string with some similar repeated data. I would like, using a regular expression, to extracts all matches in a collection. Is there a way of performing this task?
I have look through the owa_pattern package, but as far as I found out, I can extract only a simple match. Here is an exact quote:
"If multiple overlapping strings can match the regular expression, this function takes the longest match. " - http://download.oracle.com/docs/cd/B28359_01/appdev.111/b28419/w_patt.htm
So what can I do if I want to get all the matches?
Thank you in anticipation. Any help would be appreciated.
Best regards,
beroetzI think your need a tokenizer-function.
If the string +:in_str+ is delimited by +:in_delimiter+ you could try this:
SELECT REGEXP_REPLACE(REGEXP_SUBSTR( :in_str || :in_delimiter, '(.*?)' || :in_delimiter, 1, LEVEL ), :in_delimiter, '') TOKEN
BULK COLLECT INTO :my_nested_table
FROM DUAL
CONNECT BY REGEXP_INSTR( :in_str || :in_delimiter, '(.*?)' || :in_delimiter, 1, LEVEL ) > 0
ORDER BY LEVEL ASC;
I wrote a string-to-textarray-tokenizer (and it's pendant) some times ago, being able to cut from certain positions within the string using regular expressions and return the elements into an nested table of varchar2. It looks like:
TYPE pos_arraytype IS TABLE OF POSITIVE ;
TYPE text_arraytype IS TABLE OF VARCHAR2(2000);
FUNCTION stringToTextarray(in_str IN VARCHAR2, in_pos_arr IN pos_arraytype, in_regexp_arr IN text_arraytype DEFAULT NULL, in_trim_strings IN BOOLEAN DEFAULT TRUE)
RETURN text_arraytype ;
in_str is the string to be tokenized
in_pos_arr is a table of positive values of positions in the string to be cut
in_regexp_arr is a table of regular expressions to use at each position declared by in_pos_arr
in_trim_strings is a flag, if the cutted element should be trimmed
using above for example:
in_str = 'Markus van Muster 347651234XY Musterdaam ABCDE'
in_pos_arr = (1, 13, 35, 35, 42)
in_regexp_arr = ('(.?){12}', '([^[:digit:]]?){22}', '[[:digit:]]{4}', '[[:alpha:]]{2}', '(.?){14}')
in_trim_strings = TRUE
RETURN collection ('Markus','van Muster','1234','XY','Musterdaam')
If you need the code, then tell me! I'm looking for....
Cheers,
Martin
Edited by: Nuerni on 17.10.2008 08:49 -
Migrating Regular Expression from apache to Java1.4
Did any one out there migrate from apache regular expressions to the new Java regular expressions? Are the two compatible? How much work is it to do this? What are the pitfalls? Are there any resources on this topic?
Thanks,I did migrate some Jakarta regexp to java 1.4's. If ur old codes just used match() method to see if a string matches the pattern, then search/replace using any decent text edit could quickly migrate all of them. However, if you used lots of the Apache regexp's APIs, the fastest way is to write a wrap class for each class you used in the regexp pacakage, and use the new regexp package to implement all the Apach regexp APIs, then, the only thing you need to do is to replace the import statement to import those wrapper classes. I personally think this is the best and common way to migrate from an old API package to a new one.
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Mining for regular expressions from text
Hi all
I have a situation where I need to develop a set of regular expressions from lists.
I have lists of files uploaded and downloaded, and when the event happened. I need to analyze these and boil them down to a set of regular expressions.
The regular expressions will be used in the future to check the logs for activities. For example:
foo*.zip is downloaded every night. Has it been downloaded tonight.
I am looking for help in analyzing the lists of files and deriving regular expressions to describe them.
Any ideas?
Thanks allcheck the url http://java.sun.com/developer/technicalArticles/releases/1.4regex/index.html
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Regular expression to remove SIDs from list
Hey everyone so I have a script and in it I try to find the current user, as well as the last user. Currently I'm using a regular expression to throw the System account SIDs and other things like that out of the list. However this doesn't seem to be taking
SQL SIDs out of my list ie. ReportServer$LOCAL with the SID S-1-5-80-4264962431-3932693095-1576469926-235475122-2208986020
What's the best way to get only user SIDs?
Here's what I have so far:
$Win32User = Get-WmiObject -Class Win32_UserProfile -ComputerName $Computer
$Win32User = $Win32User | Where-Object {($_.SID -notmatch "^S-1-5-\d[18|19|20]$")}
$Win32User = $Win32User | Sort-Object -Property LastUseTime -Descending
$LastUser = $Win32User | Select-Object -First 1
When I can this it breaks since there is no actual user tied to this SID:
$UserSID = New-Object System.Security.Principal.SecurityIdentifier($LastUser.SID)
$User = $UserSID.Translate([System.Security.Principal.NTAccount])
Thanks for any help!!Start cmd.exe as your domain user and run whoami /user to get your own SID. You will get something like this:
USER INFORMATION
User Name SID
========================= =============================================
DEMOSYSTEM\CustomAccount1 S-1-5-21-3419697060-3810377854-678604692-1000
The last part of the SID, in this case 1000, is called RID. When you create a new user or computer object in your domain, only the RID will be different from your own SID. The RID starts on 1000 and increments as you create new objects.
If you are only interested in user accounts from the same domain as your user, you can use a regex like this, only based on your own SID:
$_.SID -match '^S-1-5-21-3419697060-3810377854-678604692-[\d]{4,10}$' -
Remove regular expression from a string
Hello,
I have a string like this
@1test;'"{input+
Please help me to remove special characters from the string.
Hi Krishna,
DATA : str TYPE STRING VALUE '@1test;"{}]input+',
char,
length TYPE i,
index TYPE i.
length = STRLEN( str ).
WHILE length > index.
char = str+index(1).
WRITE char.
if char CA '+-*/!`@#$%^&()_=[]{};'. " Add/Remove here to include numbers
REPLACE ALL OCCURRENCES OF char in str WITH ''.
REPLACE ALL OCCURRENCES OF '"' in str WITH ''. " characters "{}[] are not comparable
REPLACE ALL OCCURRENCES OF '{' in str WITH ''.
REPLACE ALL OCCURRENCES OF '}' in str WITH ''.
REPLACE ALL OCCURRENCES OF '[' in str WITH ''.
REPLACE ALL OCCURRENCES OF ']' in str WITH ''.
length = STRLEN( str ).
ENDIF.
add 1 to index.
ENDWHILE.
WRITE str.
Add or remove special char from '+-*/!`@#$%^&()_=[]{};' in if part as per your requirement.
Hope it meets your requirement.
Do not forget to mark helpful/correct if ma answer is useful .
Thanks,
Karthik
Hi Krishna,
DATA : str TYPE STRING VALUE '@1test;"{}]input+',
char,
length TYPE i,
index TYPE i.
length = STRLEN( str ).
WHILE length > index.
char = str+index(1).
WRITE char.
if char CA '+-*/!`@#$%^&()_=[]{};'. " Add/Remove here to include numbers
REPLACE ALL OCCURRENCES OF char in str WITH ''.
REPLACE ALL OCCURRENCES OF '"' in str WITH ''. " characters "{}[] are not comparable
REPLACE ALL OCCURRENCES OF '{' in str WITH ''.
REPLACE ALL OCCURRENCES OF '}' in str WITH ''.
REPLACE ALL OCCURRENCES OF '[' in str WITH ''.
REPLACE ALL OCCURRENCES OF ']' in str WITH ''.
length = STRLEN( str ).
ENDIF.
add 1 to index.
ENDWHILE.
WRITE str.
Add or remove special char from '+-*/!`@#$%^&()_=[]{};' in if part as per your requirement.
Hope it meets your requirement.
Do not forget to mark helpful/correct if ma answer is useful .
Thanks,
Karthik
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]*)+$" I get an "IllegalStateException: No match available" again.{noformat}
What is the correct way to do this?
Edited by: johann_p on May 16, 2008 10:39 AM
I am sorry I messed this up. Here is a SSCCE:
import java.util.regex.Pattern;
import java.util.regex.Matcher;
class RegExp1 {
public static void main(String[] args) {
String testString = "first|aaaa | bbbb\n|cccc|ddddd";
Pattern pattern = Pattern.compile("^([^|]+)(\\|[^|]*)+$");
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!
How do I have to define a regular expression to filter out data from file?
Hi all,
I need to extract parts of lines of a ASCII file and didn't get it done with my low knowledge of regular expressions
The file contains hundreds of lines and I am just interested in a few lines, within that lines I just need a part of the data.
One original line looks like that:
TP3| |TP_SMD|Nicht in Stueckliste|~TP TP_SMD TESTPUNKT|-|0|87.770|157.950|0|top|c| |other|TP_SMD|TP_SMD_60RF-TP
Only the bold and underlined information is of interest, I don't need the rest.
I can open that file, read in each line but then I am struggling to pick out only the lines of interest (starting with TP), taking that TP with its number and the coordinates following later on and then writing these shortened lines to a new text file. So the new line should look like that:
TP3; 87.770;157.950;0 (It doesn't matter if the separator will be ; or |)
I thought of using regular expressions - is that the right way or is there a better approach?
Thanks & regards,
gedi, using LabVIEW 8.5
Regards,
gedi
Hi max,
for finding a specific part of a string you can use the "Match Pattern" VI, it is located in the Strings Palette.
Maybe the Extract Numbers.vi example in the examples browser library can help you.
What I did to filter out my data of interest is first to sort out only the columns which I want to have -
then there are still a lot of lines remaining I don't need (this is the thing described above).
The rest I am going to filter out with a (then easy) regular expression with the "Match Pattern" VI.
Regards,
gedi
Regards,
gedi
Regular Expressions from C++ to Java
Does anyone know if it is possible to use the same regular expressions used in C++ with in Java? Here it is a regular expression used in a C++ program:
^[[:digit:]]{4}$"I did a test, but it does not work. It always evaluate to false.
Thanks,
Sid
Does anyone know if it is possible to use the same
regular expressions used in C++ with in Java? C++ doesn't have regexes, so presumably you are using a library (and noting that might help.)
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