Java: How to escape all regex metacharacters in a given String?

Question

I'm looking for a utility method in Java that will escape all regex metacharacters in a given String.

I want to convert this:

foo.bar(baz)

Into this:

foo\.bar\(baz\)

So that I can take any sample string and convert it into a regex-friendly search pattern. Surely one must exist, but I cannot seem to find anything.

(Pattern.quote(String s) offers something similar to what I need, but not the exact same functionality.)

Pattern.quote does do what you want ("take any sample string and convert it into a regex-friendly search pattern") though it escapes the entire string as a whole rather than individually quoting individual meta-characters in the string. foo.bar(baz) then becomes \Qfoo.bar(baz)\E — Alan Krueger
– Alan Krueger, Commented Sep 6, 2017 at 22:38
It doesn't satisfy my particular use case. For example, as the next step, I want to do: text = text.replace(" ","\\s+"); — Bennett Lynch
– Bennett Lynch, Commented Sep 6, 2017 at 23:01
@BennettLynch You should put that into the question. But the answer may be that there's nothing that does what you want. — Barmar
– Barmar, Commented Sep 6, 2017 at 23:33
It might be sufficient to escape every metacharacter, with something like s.replaceAll("[\\\\[\\]{}().?+*^$]", "\\\\$0"). — VGR
– VGR, Commented Sep 7, 2017 at 2:06

Bohemian · Accepted Answer · 2017-09-07 01:55:50Z

2

Pattern.quote(String s) does exactly what you want.

Calling Pattern.quote("foo.bar(baz)") returns "\Qfoo.bar(baz)\E", which matches exactly the same as the pattern "foo\.bar$baz$".

answered Sep 7, 2017 at 1:55

Bohemian♦

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1 Comment

Pattern.quote won’t satisfy the requirement mentioned in Bennett’s comment.