Matching zero occurrences using regular expressions

Could someone help me out with a strange regular expression I'm trying to write.
I'm trying to scan through a HTML page, and convert image tags into links. The problem is that the page maybe scanned more than once, so I don't want to convert image tags that have been already done in a previous run.
So I reckon that I want match an image tag, but only if it IS NOT enclosed by a <a href/> tag.
<img src="my.jpg"/>should match, but
<a href="xxxxx"><img src="my.jpg"/></a>Should not match
Any ideas how I go about this?

I recommend something like this: Pattern p = Pattern.compile("<(?:a\\s+href.+?</a>|(img.+?)>)",
 Pattern.CASE_INSENSITIVE | Pattern.DOTALL);
Matcher = p.matcher(page);
while (m.find())
 if (m.group(1) != null)
 // add link tags
}If an image tag is already enclosed in link tags, the first alternative will match the whole thing, and you ignore it. That way, the second alternative can only match a "naked" image tag, which will be stored in group(1).

Similar Messages

Matching substrings between square brackets using regular expressions

Hello,
I am new at Java and have a problem with regular expressions. Let me describe the issue in 3 steps:
1.- I have an english sentence. Some words of the sentence stand between square brackets, for example "I [eat] and [sleep]"
2- I would like to match strings that are in square brackets using regular expressions (java.util.regex.*;) and here is the code I have written for the task
+Pattern findStringinSquareBrackets = Pattern.compile("\\[.*\\]");+
+ Matcher matcherOfWordInSquareBrackets = findStringinSquareBrackets.matcher("I [eat] and [sleep]");+
+//Iteration in the string+
+ while ( matcherOfWordInSquareBrackets.find() )+
+{+
+ System.out.println("Patter found! :"+ outputField.getText().substring(matcherOfWordInSquareBrackets.start(), matcherOfWordInSquareBrackets.end())+""); +
+ }+
3- the result I have after running the code described in 2 is the following: *Patter found!: [eat] and [sleep]*
That is to say that not only words between square brackets are found but also the substring "and". And this is not what I want.
What I would like to have as a result is:
*Patter found!: [eat]*
*Patter found!: [sleep]*
That is to say I want to match only the words between the square brackets and nothing else.
Does somebody know how to do this? Any help would be great.
Best regards,
Abou

You can find the words by looping through the sentence and then return the substring within the indexes.
int start=0;
int end=0;
for(int i=0; i<string.length(); i++)
 if(string.substring(i,i+1).equals("[");
start=i;
if(start!=0)
if(string.substring(i,i+1).equlas("]");
end=i;
return string.substring(start,end+1);
}something like that. This code will only find the firt word however. I do not know much about regex so I cannot help anymore.
Edited by: elasolova on Jun 16, 2009 6:45 AM
Edited by: elasolova on Jun 16, 2009 6:46 AM

Pattern matching using Regular expression

Hi,
I am working on pattern matching using regular expression. I the table, I have 2 columns A and B
A has value 'A499BPAU4A32A386KBCZ4C13C41D20E'
B has value like '*CZ4*M11*7NQ+RDR+RSM-R9A-R9B'
the requirement is that I have to match the columns of B in A. If there is a value with * sign, this must be present in A like 'CZ4' should exit in string A.
The issue I am facing is that there are 2 values with * sign. The code works fine for first match (CZ4) but it does not look further as M11 does not exist in A.
I used the condition
AND instr(A,substr(REGEXP_SUBSTR(B, '*[^*]{3}'),2) ,1)=0
First of all, is this possible to match multiple patterns in one condition?
If yes, please suggest.
Thanks

user2544469 wrote:
Thanks a lot Frank. This query worked wonderful for the test data I have provided however I have some concerns:
- query doesnot include the column BOOK which is a mandatory check.Sorry, that was my mistake. It was a very easy mistake to make, since you posted sample data where it didn't matter. Instead of doing a cross-join between vn and got_must_have_cnt, do an inner join, using book. That means book will have to be in got_must_have_cnt, and all the sub-queries from which it descends. Look for comments that say "March 22".
If you want to treat '+' in test_cat.codes as '*', then the simplest thing is probably just to use REPLACE, so that when the table has '+', you use '*' instead.
WITH got_token_cnt AS
 SELECT cat
 , book -- Added March 22
 , REPLACE (codes, '+', '*') AS codes -- If desired. Changed March 22
 , LENGTH (codes) - LENGTH ( TRANSLATE ( codes
 , 'x*+-'
 , 'x'
 ) AS token_cnt
 FROM test_cat
, cntr AS
 SELECT LEVEL AS n
 FROM ( SELECT MAX (token_cnt) AS max_token_cnt
 FROM got_token_cnt
 CONNECT BY LEVEL <= max_token_cnt
, got_tokens AS
 SELECT t.cat
 , t.book -- Added March 22
 , REGEXP_SUBSTR ( t.codes
 , '[*+-]'
 , 1
 , c.n
 ) AS token_type
 , SUBSTR ( REGEXP_SUBSTR ( t.codes
 , '[*+-][^*+-]*'
 , 1
 , c.n
 , 2
 ) AS token
 FROM got_token_cnt t
 JOIN cntr c ON c.n <= t.token_cnt
, got_must_have_cnt AS
 SELECT cat, book -- Changed March 22
 , COUNT (CASE WHEN token_type = '*' THEN 1 END) AS must_have_cnt
 FROM got_tokens
 GROUP BY cat, book -- Changed March 22
SELECT mh.cat
, vn.vn_no
FROM got_must_have_cnt mh
JOIN vn ON mh.book = vn.book -- Changed March 22
LEFT OUTER JOIN got_tokens gt ON mh.cat = gt.cat
 AND INSTR (vn.codes, gt.token) > 1
GROUP BY mh.cat
, mh.must_have_cnt
, vn.vn_no
HAVING COUNT (CASE WHEN gt.token_type = '*' THEN 1 END) = mh.must_have_cnt
AND COUNT (CASE WHEN gt.token_type = '-' THEN 1 END) = 0
ORDER BY mh.cat
- query is very slow with 60000 records in vn table. Cost is somewhere around 36000.See these threads:
When your query takes too long ...
HOW TO: Post a SQL statement tuning request - template posting
Relational databases were designed to have (at most) one piece of information in each column. If you decide to have multiple items in the same column (as you have a variable number of tokens in the codes column), don't be surprised if that makes things slower and more complicated. Most of the query I posted, and perhaps most of the time needed, is jsut to normalize the data. If you stored the data in a narmalized form, perhaps something like got_tokens, then you wouldn't need the first 3 sub-queries that I posted.
Edited by: Frank Kulash on Mar 22, 2011 12:04 PM

Using regular expressions in java

Does anyone of you know a good source or a tutorial for using regular expressions in java.
I just want to look at some examples....
Thanks

thanks a lot... i have one more query
Boundary matchers
^      The beginning of a line
$      The end of a line
\b      A word boundary
\B      A non-word boundary
\A      The beginning of the input
\G      The end of the previous match
\Z      The end of the input but for the final terminator, if any
\z      The end of the input
if i want to use the $ for comparing with string(text) then how can i use it.
Eg if it is $120 i got a hit
but if its other than that if should not hit.

Using regular expressions for validation in i18n

Can we use regular expressions for validation of inputs in a java application taking care of i18N aspects too. Zip code for different locales are different. Can we use regular expressions to validate zipcode inputs from different locales

hi,
For that shall i have to create individual patterns for matching the inputs from different locales or a single pattern will do in the case of validating phone nos. around the world, zip codes etc. In case different patterns are required, programmer should have a konwledge of difference in patters for different locales.
regards
sdas

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
Praveen

BalusC 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

Checking valid e-mail's using Regular expressions

Hey buddies ,
I desperately need some help here. I need to develop a generic method that wiull use regular expressions and patterns to validate an e-mail.
Does anyonw have any idea on how to do this. And if possible please share some code with me.
Thanks a lot

You can do regular expresions in java using java.util.regex.*:
import java.util.regex.*;
   Pattern p = Pattern.compile("\\S++\\s++");
   Matcher m = p.matcher(sInputLine);
   if(m.find()){
      sUser = m.group().trim();
   }For more info on regular expressions in java, check out the javadocs:
http://java.sun.com/j2se/1.4.2/docs/api/java/util/regex/Pattern.html

A question about using regular expression

Hi,
This is a part of HTML file.
how
are
you
I want to search string between each pair of , tags by using Regular Expression.
For example:
how
are
you
If I use following method
String regx="(.+)";
Matcher m=Pattern.compile(regx).matcher(str);
int currentLoc=0;
while(currentLoc<str.length()){
 if(m.find(currentLoc))
 System.out.println(m.group(1));
 currentLoc=m.end();
}The content between first and last will be searched.
How to solve this problem?

Use a non-greedy match:
(?s)(.+?)(?s) makes the dot match the line terminator (same as setting the dot all option)

Finding URLs using regular expression.

I have an requirement where user will type some text containing URLs like "Please visit this site http://www.google.com/e/qHvQcWco`~!@#$%^&*()-7747. Thank you". This text has to be modified as below before saving it to the database.
"Please visit this site <a href='http://www.google.com/e/qHvQcWco`~!@#$%^&*()-7747'>http://www.google.com/e/qHvQcWco`~!@#$%^&*()-7747</a>. Thank you"
I am using regular expression (http|https)://.+?\\s which marks the end of the url with a white space character.This pattern doesn't work if the URL is located at the end of the string since there will be no space at the end.
For example if the string is "Please visit this site http://www.google.com/e/qHvQcWco`~!@#$%^&*()-7747" the regex will fail.
My acutal problem is to find the URL irrespective its position within the string.
Pattern urlPattern = Pattern.compile("(http|https)://.+?\\s", Pattern.CASE_INSENSITIVE);
Matcher matcher = urlPattern.matcher(plainText);
Map stringIndexMap = new HashMap();
//Searching the input string for urlPattern...
while(matcher.find()) {
String urlString = matcher.group();
//Storing the urls in a hashmap with their indices as keys....
stringIndexMap.put(new Integer(matcher.start()), urlString.trim());
Set keySet = stringIndexMap.keySet();
Iterator it = keySet.iterator();
//Iterating over the hashmap containing urls...
while(it.hasNext()) {
String urlString = (String) stringIndexMap.get(it.next());
* Replacing the url string in the input text with <a href="#" onclick="window.open('<urlString>')"
* using String index
clickableURLString.replace(clickableURLString.indexOf(urlString),
clickableURLString.indexOf(urlString) + urlString.length(),
"<a href=\"#\" onclick=\"window.open('" + urlString
+ "')\">" + urlString + "</a>");
return clickableURLString.toString();

The end of the input is '$' as a regex.
import java.util.regex.*;
public class Prasanna{
public static void main(String[] args){
 String text
= "Please visit this site http://www.google.com/e/qHvQcWco`~!@#$%^&*()-7747";
// String regex = "(http|https)://.+?(?:\\s|$)"; // this works
 String regex = "(http|https)://[^ ]+"; // this also works
 Pattern pat = Pattern.compile(regex, Pattern.CASE_INSENSITIVE);
 Matcher mat = pat.matcher(text);
 while (mat.find()){
 System.out.println(mat.group());
}

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.

Date Validation (yyyy/MM/dd) Using Regular Expression

Hi Friends,
I want to validate date entered by user in yyyy/MM/dd format and for this I want to use Regular Expressions only. Also is there any tool that can be used to generate Regular Expression (for Win2000, Win NT)?
Regards,
Himanshu Rathore

try this
public class Test
     public static void main(String [] args)
          String regex = "\\d{4}/[01]\\d/[0-3]\\d";
          System.out.println("2003/12/11".matches(regex));
          System.out.println("2djd/kj3".matches(regex));
          System.out.println("22/12/12".matches(regex));
          System.out.println("2003/23/05".matches(regex));
          System.out.println("1999/12/51".matches(regex));
          System.out.println("2007/05/07".matches(regex));
}i'm not able to try on it because i only have jdk1.3.1 installed on my computer and these codes
required j2sdk1.4

Validate cfinput using regular expression

Hi,
can somebody help me valditing an cfinput field using regular expressions?
First digit must be a number or "R".
Digit 2-15 can be everything without special characters. Digit 2-15 can also be empty.
I try this, but it doesn't work (Sorry I'm a beginner using regex).
<cfinput type="text" name="field1" required="yes" validate="regular_expression" pattern="[0-9Rr][0-9a-zA-Z]*" maxlength="15">
Thank you in advice!
Claudia

You haven't told your regex to match the entire string, so it will "pass"
any string that has your match anywhere within it, so like as long as it's
got an R or a digit in it: it's OK.
To tell it to match the entire string, anchor it to the start and end of the
string with ^ and $ respectively.
Regular expression syntax: Using special characters
http://help.adobe.com/en_US/ColdFusion/9.0/Developing/WSc3ff6d0ea77859461172e0811cbec0a38f -7ffb.html#WSc3ff6d0ea77859461172e0811cbec0a38f-7fef
Adam

One for the Tekkies: How to get this output using REGULAR EXPRESSIONS?

How to get the below output using REGULAR EXPRESSIONS??
SQL> ed
Wrote file afiedt.buf
1* CREATE TABLE cus___addresses (full_address VARCHAR2(200 BYTE))
SQL> /
Table created.
SQL> PROMPT Address Format is: House #/Housename, street, City, Zip Code, COUNTRY
House #/Housename, street, City, Zip Code, COUNTRY
SQL> INSERT INTO cus___addresses VALUES('1, 3rd street, Lansing, MI 49001, USA');
1 row created.
SQL> INSERT INTO cus___addresses VALUES('3B, fifth street, Clinton, OK 74103, USA');
1 row created.
SQL> INSERT INTO cus___addresses VALUES('Rose Villa, Stanton Grove, Murray, TN 37183, USA');
1 row created.
SQL> SELECT * FROM cus___addresses;
FULL_ADDRESS
1, 3rd street, Lansing, MI 49001, USA
3B, fifth street, Clinton, OK 74103, USA
Rose Villa, Stanton Grove, Murray, TN 37183, USA
SQL> The REG EXP query shouLd output the ZIP codes: i.e. 49001, 74103, 37183 in 3 rows.Edited by: user12240205 on Jun 18, 2012 3:19 AM

Hi,
user12240205 wrote:
... Frank, ʃʃp's method, I understand. But your method, although correct, I find it difficult to understand.
Could you explain how you did this?? What does '.*(\d{5})\D*' and '\1' mean???
Your method is better because it uses only ONE reg expression function. ʃʃp's uses 2.In Oracle 10.2 (I believe) and higher, '\d' is equivalent to '[[:digit:]]', and '\D' is equivalent to '[^[:digit:]]'. I find '\d' and '\D' easier to type, but there's nothing wrong with using '[[:digit:]]' and '[^[:digit:]]'.
'.*' means "0 or more of any character".
'\D*' means "0 or more non-digits".
The whole expression, '.*(\d{5})\D*' means:
a. 0 or more characters (any characters)
b. 5 digits
c. 0 or more non-digits.
'\1' is a Backreference . It means the sub-string that matched the pattern after the 1st '(', up to (but not including) its matching ')'. In this case, that means the sub-string that matched '\d{5}', or b. using the explanation immediately above.
So the entire REGEXP_REPLACE call means "When you see a sub-string consisting of a., follwed immediately by b., followed immedately by c., replace that sub-string with b. alone."

Help in query using regular expression

HI,
I need a help to get the below output using regular expression query. Please help me.
SELECT REGEXP_SUBSTR ('PWRPKG(P/W+P/L+CC)', '[^+]+', 1, lvl) val, lvl
FROM DUAL,(SELECT LEVEL lvl FROM DUAL
CONNECT BY LEVEL <=(SELECT MAX ( LENGTH ('PWRPKG(P/W+P/L+CC)') - LENGTH (REPLACE ('PWRPKG(P/W+P/L+CC)','+',NULL))+ 1) FROM DUAL));
I need the output as
correct result:
==============
val lvl
P/W 1
P/L 2
CC 3
But i tried the above it is not coming the above result. Please help me where i did a mistake.
Thanks in advance

Frank gave you a solution in your other thread. You could simplify it if you are on 11g:
SQL> select * from table_x
2 /
TXT
TECHPKG(INTELLI CC+FRT SONAR)
PWRPKG(P/W+P/L+CC)
select txt,
 regexp_substr(
 txt,
 '(.*\()*([^+)]+)',
 1,
 column_value,
 null,
 2
 ) element,
 column_value element_number
from table_x,
 table(
 cast(
 multiset(
 select level
 from dual
 connect by level <= regexp_count(txt,'\+') + 1
 as sys.OdciNumberList
order by rowid,
 column_value
TXT ELEMENT ELEMENT_NUMBER
TECHPKG(INTELLI CC+FRT SONAR) INTELLI CC 1
TECHPKG(INTELLI CC+FRT SONAR) FRT SONAR 2
PWRPKG(P/W+P/L+CC) P/W 1
PWRPKG(P/W+P/L+CC) P/L 2
PWRPKG(P/W+P/L+CC) CC 3
SQL> SY.

Rplacing space with &nbsb; in html using regular expressions

Hi
I want to replace space with &nbsb; in HTML.
I used the below method to replace space in my html file.
var spacePattern11:RegExp =/(\s)/g;
str= str.replace(spacePattern," "
Here str varaible contains below html file.In this html file i want to replace space present between " What number does this represents" with &nbsb;
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>
<body>
<TEXTFORMAT LEADING="2"></TEXTFORMAT><TEXTFORMAT LEADING="2"> What number does this Roman numeral represents MDCCCXVIII ?</TEXTFORMAT>
</body>
</html>
But by using the above regular expression i am getting like this.
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head><body>
<TEXTFORMAT LEADING="2"></TEXTFORMAT><TEXTFORMAT LEADING="2"> What number does this represents</TEXTFORMAT>
</body>
</html>
Here what happening means it was replacing space with &nbsb; in HTML tags also.But want to replace space with &nbsb; present in the outside of the HTML tags.I want like this using regular expressions in FLEX
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>
<body>What number does this represents</body>
</html>
Hi,Please give me the solution to slove the above problem using regular expressions
Thanks in Advance to all
Regards
ssssssss

sorry i missed some information in above,The modified information was in red color
Hi
I want to replace space with &nbsb; in HTML.
I used the below method to replace space in my html file.
var spacePattern11:RegExp =/(\s)/g;
str= str.replace(spacePattern," "
Here str varaible contains below html file.In this html file i want to replace space present between " What number does this represents" with &nbsb;
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>
<body>
<TEXTFORMAT LEADING="2"></TEXTFORMAT><TEXTFORMAT LEADING="2"> What number does this Roman numeral represents MDCCCXVIII ?</TEXTFORMAT>
</body>
</html>
But by using the above regular expression i am getting like this.
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head><body>
<TEXTFORMAT LEADING="2"></TEXTFORMAT><TEXTFORMAT LEADIN G="2"> What number does this represents</TEXTFORMAT>
</body>
</html>
Here what happening means it was replacing space with &nbsb; in HTML tags also.But want to replace space with &nbsb; present in the outside of the HTML tags.I want like this using regular expressions in FLEX
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>
<body>What&nbsb;number&nbsb;does&nbsb;this&nbsb;represents</body>
</html>
Hi,Please give me the solution to slove the above problem using regular expressions
Thanks in Advance to all
Regards
ssssssss

Matching zero occurrences using regular expressions

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