Regex replace only inside tag (need help writing regex)

Question

I need to write regex that replaces a with b but only inside <pre> tag.

Example

a <pre> c a <foo> a d </pre> a

Result

a <pre> c b <foo> b d </pre> a

Please help writing expression for java String.replace function. There is a guarantee that pre tag is not nested.

Cue the torrent of "Don't use regular expressions to parse HTML"... — T.J. Crowder
– T.J. Crowder, Commented Feb 8, 2013 at 8:47
@RuiJarimba trying to write something like <pre>(.*?)a(.*?)</pre> -> $1b$2 but it isn't working — Poma
– Poma, Commented Feb 8, 2013 at 8:52
@Poma please clarify, do you need to change only 'a' characters in <pre> or you need to change whole content in <pre>? — Taky
– Taky, Commented Feb 8, 2013 at 8:54
@T.J.Crowder ok I can replace </?pre> with ` char so it doesn't look like html anymore. — Poma
– Poma, Commented Feb 8, 2013 at 9:23

Adrian Pronk · Accepted Answer · 2013-02-09 05:25:10Z

3

I think the best you can do with String.replace() is something like:

String string = ...
for (;;)
{
    String original = string;
    string = string.replaceFirst("(<pre>.*?)a(.*?</pre>)", "$1b$2");
    if (original.equals(string))
        break;
}

(EDIT: @Bohemian has noted the above regex doesn't work correctly. So it needs to be changed to:
(<pre>(?:(?!</pre>).)*a((?:(?!<pre>).)*</pre>) (untested) to avoid matching outside a <pre>...</pre> section. With this change, we don't need the *? quantifier and can use the more common "greedy" (*) quantifier. This is starting to look a lot like my other answer, which I only really meant as a joke!)

You're better off using a Matcher (following code off the top of my head):

import java.util.regex.Pattern;
import java.util.regex.Matcher;

Pattern pattern = Pattern.compile("(?<=<pre>)(.*?)(?=</pre>)");
Matcher m = pattern.matcher(string);
StringBuffer replacement = new StringBuffer();

while (matcher.find())
{
     matcher.appendReplacement(replacement, "");
     // Careful using unknown text in appendReplacement as any "$n" will cause problems
     replacement.append(matcher.group(1).replace("a", "b"));
}    
matcher.appendTail(replacement);
String result = replacement.toString();

Edit: Changed pattern above so that it does not match surrounding <pre> and </pre>.

edited Feb 9, 2013 at 5:25

answered Feb 8, 2013 at 9:23

Adrian Pronk

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6 Comments

Achintya Jha Over a year ago

You are replacing "a" inside <tag> too.

Poma Over a year ago

@AchintyaJha it was my mistake in example. Yes "a" inside <tag> should be replaced

Bohemian Over a year ago

I think your regex will replace "a" in "<pre>x</pre>a<pre>x</pre>"

Adrian Pronk Over a year ago

@Bohemian: I've edited my answer to indicate that the regex is wrong and provided another one

Bohemian Over a year ago

You could greatly simply your code by using look arounds, and you don't need to be non-greedy (OP guarantees no nesting): (?<=<pre>).*(?=</pre>)". Being non-consuming, lookarounds would mean you don't need to replace the consumed tags.

|

Adrian Pronk · Accepted Answer · 2013-02-08 10:10:34Z

0

Here's a regex that will do the job (I think: I wouldn't bet too much on it passing all tests enter image description here )

String replacement = original.replaceAll(
    "(?<=<pre>(?:(?!</pre>).){0,50})a(?=(?:(?!<pre>).)*</pre>)", 
    "b");

Explanation:

(?<=<pre>(?:(?!</pre>).){0,50}) - look-behind for a preceding <pre> so long as we don't traverse back over </pre> to find it. Java requires a finite maximum length look-behind so we use {0,50} rather than *.
a - The character we want to replace
(?=(?:.(?!<pre>))*</pre>) - Look ahead for </pre> so long as we don't traverse past <pre> to find it.

edited Feb 8, 2013 at 10:10

answered Feb 8, 2013 at 9:57

Adrian Pronk

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Comments

Achintya Jha · Accepted Answer · 2013-02-08 09:23:23Z

-1

Pattern pattern = Pattern.compile("<pre>(.+?)</pre>");
java.util.regex.Matcher matcher = pattern.matcher("a <pre> c a <tag> a d </pre> a");

Try this:

answered Feb 8, 2013 at 9:23

Achintya Jha

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