Regular Expressions - replace inner group

I'm trying to do the following:
1. Given the following string:
Truth, like tumors, require a cut to be removed.
2. Grab the whole sentence and the following:
like tumors
3. Replace "like tumors" with "like posion" and add one line to the sentence like so:
Truth, like posion, require a cut to be removed.
To be free you must endure pain for a season to finally be free from it.
I can grab the sentence with reg expressions, I can even grab the second search (#2) in a group, I'm just wondering if there is an elegant way to replace both at the same time. I could always replace one then search and replace the next, but that seems redundant since I can grab both the first time around. Any help?

according to:
http://javaalmanac.com/egs/java.util.regex/GroupInRep.html?l=find
// Compile regular expression
String patternStr = "\$(\\w+)\$";
String replaceStr = "<$1>";
Pattern pattern = Pattern.compile(patternStr);
// Replace all (\w+) with <$1>
CharSequence inputStr = "a (b c) d (ef) g";
Matcher matcher = pattern.matcher(inputStr);
String output = matcher.replaceAll(replaceStr);
// a (b c) d <ef> gthis should be exactly what you're looking for.

Similar Messages

Regular expression - Replace not like

Hi All,
Is there a regexp pattern to replace anything other than 'oracle' from the below string.
"oracle sdsd oracle xyd fgh oracle idmdh asasas trtrt"
The result will be "oracleoracleoracle"
If I want to write like regexp_replace('oracle sdsd oracle xyd fgh oracle idmdh asasas trtrt',<pattern>), what should be the pattern?
I know how to get the result by nesting regexp and other functions.
But is there any single pattern for this?
Thanks in advance.
Note; This is not a business requirement, trying to learn regexp..

884476 wrote:
Could you please explain what does that pattern mean? We can look at your string as a sequence of substrings where each substring is set of characters (part A) followed by word oracle or, in last substring, by end-of-line (part B). Each of such substrings be want to replace with part B thus removing part A.
Dot (.) means any character. Asterisk (*) means repeated 0 or more times. This is our part A (I'll get back to ? later). Pipe (|) means OR. Dollar sigh ($) means end-of-line. Parenthesis mean grouping. So ((oracle)|$) means string oracle or end-of line. This is our part B. Now back to question mark. By default Oracle regular expressions are "greedy". We say any character repeated any number of times followed by word oracle. Since word oracle itself matched definition of any character repeated any number of times (6 in this case) regexp will match it that way that it will be from the beginning of the string to last occurrence of word oracle - that's why it is called "greedy". Question mark (?) tells regexp not to use greedy matching, therefore '.*?oracle' will stop at first occurrence of word oracle. Now replacement string. Notation \1 is grouping backreference. Group 1 is ((oracle)|$) and, as I already noted, means string oracle or end-of line (whichever was found).
SY.

Regular expressions and capture groups

Hi everyone :)
Is there a way to override the default behaviour of capture groups in regular expressions? More specifically I want to override this:
"The captured input associated with a group is always the subsequence that the group most recently matched."
For example, if I have a string that is this:
* <comment one>
* <comment two>
<some text>
I have a pattern of the form "(.*)(/\\*.*\\*/)(.*)" which will match multi-line comments. I have also specified the flag DOTALL so that the predefined character class '.' matches over line-breaks.
If I apply this pattern to the above string I get comment two being captured, not comment one. This is because of the stipulation that I cited above.
I need to be able to capture only the first match, and prevent the capture group from being overwritten by more recent matches.
Is this possible? Any ideas?
Thanks in advance.
Kind regards,
Ben Deany

Is there a way to override the default behaviour of
capture groups in regular expressions? More
specifically I want to override this:No, but you don't need to.
I have a pattern of the form "(.*)(/\\*.*\\*/)(.*)"
which will match multi-line comments.Comment two is captured by the second group because comment one is eaten by the first group. Use the reluctant quantifier "*?" on the . in the first group instead of the greedy quantifier "*" to get what is apparently the behavior you want. Then the first group will contain nothing, the second group will contain comments one and two, and the third group will contain the following text.
.* is a very powerful thing to use. It will match everything in its path, guzzling text like moonshine at Mardi Gras. The only reason it doesn't match comment two as well is because then the expression as a whole would not match.
The parentheses surrounding the first and third groups are not needed (unless you want to use group methods on them too).

Regular Expression replacement not working

I am trying to use a regular expression to replace non-ascii characters on a file, and I'm afraid I've reached the end of my regex knowledge.
Here is the specific code
'Set the Regular Expression paramaters
Set RegEx = CreateObject("VBScript.Regexp")
RegEx.Global = True
RegEx.Pattern = "[^\u0000-\u007F]"
RegEx.IgnoreCase = True
'Replace the UTF-8 characters
ReplacedText = RegEx.Replace(FileText, "\u0020")
If I understand regular expressions correctly the pattern of "[^\u0000-\u007F]" should replace any character that is not an ascii character, and then replace it with a space (which I understand is "\u0020"). What am I doing wrong?

Simply use
ReplacedText = RegEx.Replace(FileText, " ")
Regards, Hans Vogelaar (http://www.eileenslounge.com)

Regular Expression / replace Function Help

The problem:
cfset myString = "i am a big boy"
cfset outputString = replace("i am a big
boy","i","you","all")
Wrong Output:
you am a byoug boy
Intended Output
you am a big boy
How do I achive that output.

Your first example had only one sentence. That's why I gave
that answer.
Anyway for your real question, you need to use regular
expressions
#rereplace(myString,"i\s","you ","all")# - will give u
you am a big boy you am a big boy you am a big boy you am a
big boy.
\s looks for a whitespace character after the letter i. So
that way it will not change the letter i in big

Regular Expression/Replace - Oracle 7.3

Hi!
I am trying the regular expression SQL functions of 10g to Oracle 7.3 and it seems the older version does not cover this feature yet.
"Aaaa,Bbbb" --> "Aaaa, Bbbb"
REPLACE *",[0-9A-Za-z]"* WITH *", "*
The string pattern is to look for comma-punctuations that is not followed immediately by whitespacess so I can replace this with a comma followed by a whitespace.
Any workaround for this?

Hi,
Welcome to the forum!
kitsune wrote:
Hi!
I am trying the regular expression SQL functions of 10g to Oracle 7.3 and it seems the older version does not cover this feature yet.You're right; regular expressions only work in Oracle 10.1 and higher.
>
>
"Aaaa,Bbbb" --> "Aaaa, Bbbb"
REPLACE *",[0-9A-Za-z]"* WITH *", "*
The string pattern is to look for comma-punctuations that is not followed immediately by whitespacess so I can replace this with a comma followed by a whitespace.
Any workaround for this?You're best bet in Oracle 7.3 would be a user-defined function. That's a very old version; don't expect much.
Do you know anything else about the string? For example, is there some character (say ~) that never occurs in the string? Will there ever be two (or more) whitespace characters after punctuation? What characters do you consider to be whitespace? Which are punctuation? Depending on the answers, you might be able to do something with nested REPLACE and/or TRANSLATE functions.

How to use regular expression replace for this special characters?

hi,
I need to replace the below string, but i couldnt able to do if we use the special charaters '+', '$' . can anyone suggest a way to do this?
select REGEXP_REPLACE('jan + feb 2008','jan + feb 2008', 'feb',1,0,'i') from dual
anwers should be :- feb

you should use escape character \.
the regular expression will look like as follows:
select REGEXP_REPLACE('jan + feb 2008','jan \+ feb 2008', 'feb',1,0,'i') from dual
hope this is what you needed.
cheers,
Davide

Regular expression - Replace a part of an expression

Dear All,
This is not a business requirement, just trying to practice regexp.
Suppose we want to replace a part of a regular expression from a string. As an example, in the string 'THIS Number 124356 Is to Change.This Number 5 Also', I am trying to replace the last digit of each number with $ sign.
Output will be
'THIS Number 12435$ Is to Change.This Number $ Also'.
Will this be possible using a single regexp_replce?
Thanks in advance.

MichaelS wrote:
I am trying to achieve this using a SINGLE regexp_replace.Here we go:
SQL> with t as (
select 'ABC124556def568gh236klJ258' str from dual union all
select 'THIS Number 124356 Is to Change.This Number 5 Also' str from dual
select str, regexp_replace(str, '(\d{0,})\d{1}', '\1$') str2
from t
STR                                                     STR2
ABC124556def568gh236klJ258                              ABC12455$def56$gh23$klJ25$
THIS Number 124356 Is to Change.This Number 5 Also      THIS Number 12435$ Is to Change.This Number $ Also
2 rows selected.
Nice..
Learning for me ..
Never thought of usage like {0,} ... kepping the end value OPEN.
I was trying with \d+?\d, which was not working..

Regular Expression "Replace"

(<PartName>([a-z]))(.*)(</PartName>)
I need to change all copy within the <PartName></PartName> tags with Sentence Case. Above selects all the copy that doesn't have a capital.
To write the above out in find and repalce, I would just use $1$2$3$4 but I need the $2 to be a Capital.
What's crazy is that this would seem to be a fairly common thing to do yet finding an answer is nearly impossible. I even have OReilly's Regular Expression Cookbook book and there's nothing in it about replacing text with changed text.
I've been on this for a day now and it's getting to the point where I could have gone through and done it manually in less time.
Sorry for the rant...

I gave up and just exported out from my xml into Excel and did a formula there...

Regular Expression Replacement

I have a replacement string of the form "[\1\2]='\3'," and
I'd like to insert a carriage at the end of the string. I've tried
\n, \r\n and it doesn't seem to work.

jdeline wrote:
> Chr(10) is newline, Chr(13) is carriage return. BTW, do
we all know what a carriage is? :-)
sure! it's a 4-wheel horse-drawn transportation device!
apparently it
can be returned... presumably if you do not want it any more
or if it
has become outdated... anyone knows WHERE they are returned
to? :)
Azadi Saryev
Sabai-dee.com
Vientiane, Laos
http://www.sabai-dee.com

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:
cd

Can 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

Grouping & Back-references with regular expressions on Replace Text window

I really appreciate the inclusion of the Regular Expressions in the search & replace feature. One thing I am missing is back-references in the replacement expression. For instance, in the unix tools vi or sed, I might do something like this:
s/$firstPart$ $secondPart$ $oldThirdPart$/\2 \1 newThirdPart/g
which would allow me to switch the places of firstPart and secondPart, and totally replace thirdPart. If grouping and back-references are already present in the Replace Text window, how does one correctly invoke them?

duplicate of Grouping & Back-references with regular expressions on Replace Text window

Regular Expressions find and replace

Hi ,
I have a question on using Regular Expressions in Java(java.util.regex).
Problem Description:
I have a string (say for example strHTML) which contains the whole HTML code of a webpage. I want to be able to search for all the image source tags and check whether they are absolute urls to the image source(for eg. <img src="www.google.com/images/logo.gif" >) or relative(for eg. <img src="../images/logo.gif" >).
If they are realtive urls to the image path, then I wish to replace them with their absolute urls throughout the webpage(in this case inside string strHTML).
I have to do it inside a servlet and hence have to use java.
I tried . This is the code. It doesn't match and replace and goes inside an infinite loop i.e probably the pattern matches everything.
//Change all images to actual http addresses FOR example change src="../images/logo.gif" to src="http://www.google.com/../images/logo.gif"
          String ddurl="http://www.google.com/";
String strHTML=" < img src=\"../images/logo.gif\" alt=\"Google logo\">";
Pattern p = Pattern.compile ("(?i)src[\\s]*=[\\s]*[\"\']([./]*.*)[\"\']");
Matcher m = p.matcher (strHTML);
while(m.find())
m.replaceAll(ddurl+m.group(1));
what is wrong in this?
Thanks,
Rajiv

Right, here's the full monte (whatever that means):import java.util.regex.*;
public class Test1
public static void main(String[] args)
    String domain = "http://www.google.com/";
    String strHTML =
      " < img src=\"images/logo.gif\" alt=\"Google logo\">\n" +
      " <img alt=\"Google logo\" src=images/logo.gif >\n" +
      " <IMG SRC=\"/images/logo.gif\" alt=\"Google logo\">\n" +
      " <img alt=\"Google logo\" src=../images/logo.gif>\n" +
      " <img src=http://www.yahoo.com/images/logo.gif alt=\"Yahoo logo\">";
    String regex =
      "(<\\s*img.+?src\\s*=\\s*)   # Capture preliminaries in $1. \n" +
      "(?:                         # First look for URL in quotes. \n" +
      "   ([\"\'])                 #   Capture open quote in $2.   \n" +
      "   (?!http:)                #   If it isn't absolute...     \n" +
      "   /?(.+?)                  #    ...capture URL in $3       \n" +
      "   \\2                      #   Match the closing quote     \n" +
      " |                          # Look for non-quoted URL.      \n" +
      "   (?!http:)                #   If it isn't absolute...     \n" +
      "   /?([^\\s>]+)             #    ...capture URL in $4       \n" +
    Pattern p = Pattern.compile(regex, Pattern.CASE_INSENSITIVE | Pattern.COMMENTS);
    Matcher m = p.matcher(strHTML);
    StringBuffer sbuf = new StringBuffer();
    while (m.find())
      String relURL = m.group(3) != null ? m.group(3) : m.group(4);
      m.appendReplacement(sbuf, "$1\"" + domain + relURL + "\"");
    m.appendTail(sbuf);
    System.out.println(sbuf.toString());
}First off, observe that I'm using free-spacing (or "COMMENTS") mode to make the regex easier to read--all the whitespace and comments will be ignored by the Pattern compiler. I also used the CASE_INSENSITIVE flag instead of an embedded (?i), just to remove some clutter. By the way, your second (?i) was redundant; the first one would remain in effect until "turned off" with a (?-i). Another way to localize a flag's effect by using it within a non-capturing group, e.g., (?i:img).
As jaylogan said, the best way to filter out absolute URL's is by using a negative lookahead, and that's what I've done here. The problem of optional quotes I addressed by trying to match first with quotes, then without. The all-in-one approach might work with URL's, since they can't (AFAIK) contain whitespace anyway, but the alternation method can be used to match any attribute/value pair. It's also, I feel, easier to understand and maintain. Unfortunately, it also means that you can't use replaceAll(), since you have to determine which alternative matched before doing the replacement, but the long version is still pretty simple (especially when you can just copy it from the javadoc for the appendReplacement() method, as I did).

[DW CC] Regular Expression and Back Reference in "Replace with": Escape

I have a problem with escaping a character in "Replace with".
Let's assume the following situation:
Content:
foobar
Search for:
(foo)(bar)
Replace by:
$1zot$2
Result:
foozotbar
Everything is fine.
But I want to insert the number "2" in between foo and bar.
When I use
Replace by: $12$2
The result is:
$12$2
It seems that the regex engine interpretes the "$12" as a whole. And because there's no back reference with the count "12" DW cannot work correctly.
The question:
How can I "escape" the "2"?
Thanks.

Dreamweaver uses JavaScript for its Find and Replace with regex.
I've just checked the Regular Expressions Cookbook by Jan Goyvaerts and Steven Levithan (O'Reilly). It deals with exactly this sort of situation where capturing groups can be ambiguous. This is what it says:
"Java and JavaScript try to be clever with $10 [and above]. If a capturing group with the two-digit number exists in your regular expresssion, both digits are used for the capturing group. If fewer capturing groups exist, only the first digit is used to reference the group, leaving the second as a literal. Thus $23 is the 23rd capturing group, if it exists. Otherwise, it is the second capturing group followed by a literal 3."
In other words, there is no way to escape the 2 in the Replace field. It would appear there's a bug in Dreamweaver's use of capturing groups.

Replace All with Regular Expression

Hi all,
I need a help to replace the String:
to
"<a href=\"1\"">Java Programming</a>"
I was trying to replace with
.replaceAll("\\[\\[*([^\\]]*?)*\\]\\]", "<a href=\"\"></a>")
but I don't know how to separate the parameters values.
Best regards

Hi prometheuzz,
you are the one!!
But how I could do to a Srting like this:
Ah, more requirements...
Things are getting a bit messy now:
import java.util.regex.Matcher;
import java.util.regex.Pattern;
class Main {
    public static void main(String[] args) {
        String line = "any text any text any text any text any text. "+
                      "Click - [[Code: 6 Title: Java Programming]]. "+
                      "any text any text any text any text any text. "+
                      "- [[C�digo: 2 T�tulo: Sun]] text text "+
                      "any text any text any text any text any text.";
        String newLine = line;
        String[] wikiTags = getWikiTags(newLine);
        for(String tag : wikiTags) {
            String newTag = "<a href=\""+get("(Code:|C�digo:)", " ", tag)+
                            "\">"+get("(Title:|T�tulo:)", "$", tag)+"</a>";
            newLine = newLine.replaceFirst("\\[\\["+tag+"\\]\\]", newTag);
        System.out.println(line);
        System.out.println(newLine);
    public static String[] getWikiTags(String text) {
        java.util.List<String> list = new java.util.ArrayList<String>();
        Pattern pattern = Pattern.compile("(?<=\\[\\[)(.*?)(?=\\]\\])");
        Matcher matcher = pattern.matcher(text);
        while(matcher.find()) {
            list.add(matcher.group());
        return list.toArray(new String[list.size()]);
    public static String get(String start, String end, String text) {
        Pattern pattern = Pattern.compile("(?<="+start+"\\s)(.*?)(?="+end+")");
        Matcher matcher = pattern.matcher(text);
        return matcher.find() ? matcher.group() : "#ERROR#";
}Details about regex:
http://java.sun.com/j2se/1.4.2/docs/api/java/util/regex/Pattern.html
http://www.regular-expressions.info/java.html
http://java.sun.com/docs/books/tutorial/essential/regex/
>
Thanks a lot for you helpLike I said, it's a bit of a messy solution. See if you can use (a part of) it.
Good luck.

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