Question

If I have a regex with a capturing group, e.g. foo(_+f). If I match this against a string and want to replace the first capturing group in all matches with baz so that

foo___f blah foo________f

is converted to:

foobaz blah foobaz

There doesn’t appear to be any easy way to do this using the standard libraries. If I use Matcher.replaceAll() this will replace all matches of the entire pattern and convert the string to

baz blah baz

Obviously I can just iterate through the matches, store the start and end index of each capturing group, then go back and replace them, but is there an easier way?

Thanks,
Don

Answer 1

I think you want something like this?

    System.out.println(
        "foo__f blah foo___f boo___f".replaceAll("(?<=foo)_+f", "baz")
    ); // prints "foobaz blah foobaz boo___f"

Here you simply replace the entire match with "baz", but the match uses lookbehind to ensure that _+f is preceded by foo.

See also

• regular-expressions.info/Lookarounds

---

If lookbehind is not possible (perhaps because the length is not finite), then simply capture even what you’re NOT replacing, and refer to them back in the replacement string.

    System.out.println(
        "fooooo_f boooo_f xxx_f".replaceAll("(fo+|bo+)(_+f)", "$1baz")
    ); // prints "fooooobaz boooobaz xxx_f"

So here we’re effectively only replacing what \2 matches.