Java regular expression(grouping)

Question

I am try to write a regex to match conditional expressions, for example:

a!=2     1+2<4       f>=2+a

and I try to extract the operator. My current regex is ".+([!=<>]+).+"

But the problem is that the matcher is always trying to match the shortest string possible in the group.

For example, if the expression is a!=2, then group(1) is "=", not "!=" which is what I expect.

So how should I modify this regex to achieve my goal?

`[!=<>]` can represent only one character. Read about [quantifiers](http://docs.oracle.com/javase/tutorial/essential/regex/quant.html) (especially about greedy and reluctant). — Pshemo, Oct 20 '14 at 20:14
Reply to pshemo: sorry it was my typo when typing the question. Now it is corrected — user2829353, Oct 20 '14 at 20:20
Problem stays the same. First `.+` is greedy so it consumes characters which could be matched by `[!=<>]+`. You need to make it reluctant (or instead of `.` you can use negated character set to prevent matching `!=<>`). — Pshemo, Oct 20 '14 at 20:24
BTW, instead of `Reply to pshemo` use `@pshemo` :) this way I will be notified about your comment. — Pshemo, Oct 20 '14 at 20:25

score 1 · Accepted Answer · answered Oct 20 '14 at 20:23

1

You want to match an operator surrounded by something that is not an operator.

An operator, in your definition : [!=<>]

Inversely, not an operator would be : [^!=<>]

Then try :

[^!=<>]+([!=<>]+)[^!=<>]+

answered Oct 20 '14 at 20:23

JM Lord

score 1 · Answer 2 · edited May 23 '17 at 11:57

1

You could also try the reluctant or non-greedy versions (see that other posr for extensive explain). In your example, it would be :

.+?([!=<>]+).+

But this regex could match incorrect comparisons like a <!> b, or a =!><=! b ...

edited May 23 '17 at 11:57

Community

answered Oct 20 '14 at 20:49

Serge Ballesta

score 0 · Answer 3 · answered Oct 20 '14 at 20:27

0

Try this:

.+(!=|[=<>]+).+

your regex is matching a single ! because it is in []

Everything put in brackets will match a single char, which means [!=<>] can match: !, =, <, >

answered Oct 20 '14 at 20:27

Kristijan Iliev

Thank you, but this actually doesn't work. .+ still eats up ! – user2829353 Oct 20 '14 at 20:45
1

No, it's the greediness of the leading `.+` that's causing the problem, as explained in [another answer](http://stackoverflow.com/a/26474663/20938). – Alan Moore Oct 20 '14 at 23:59

3 Answers3