Java Regex: replace with capturing group

Question

I have a string like this:

123456-1/1234/189928/2323 (102457921)

I want to get 102457921. How can I achieve it with regex?

I have tried:

"123456-1/1234/189928/2323 (102457921)".replaceAll("(\\.*\()(\d+)(\))","$2");

But it does not work. Any hints?

split with whitespace and remove braces? – Sara S May 02 '14 at 13:28 — Sara S, May 02 '14 at 13:28
of course, but I'd like a regex solution. – user3111525 May 02 '14 at 13:29 — user3111525, May 02 '14 at 13:29

score 5 · Accepted Answer · answered May 02 '14 at 13:29

5

How about

"123456-1/1234/189928/2323 (102457921)".replaceAll(".*?\\((.*?)\\).*", "$1");

answered May 02 '14 at 13:29

Keppil

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What does `.*?` mean? – user3111525 May 02 '14 at 13:34
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It is the same as `.*`, except that it is _non-greedy_. You can read more [here](http://www.regular-expressions.info/repeat.html). – Keppil May 02 '14 at 13:39

score 1 · Answer 2 · answered May 02 '14 at 13:29

1

Well, you can do this:

"123456-1/1234/189928/2323 (102457921)".replaceAll(".*\((.+)\)","$1");

answered May 02 '14 at 13:29

Amit Joki

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@sshashank124, yes. Changed `+` to `*` – Amit Joki May 02 '14 at 13:31

score 0 · Answer 3 · answered May 02 '14 at 13:29

0

You could do it as:

"123456-1/1234/189928/2323 (102457921)".replaceAll(".*?\(([^)]+)\)","$1");

answered May 02 '14 at 13:29

sshashank124

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score 0 · Answer 4 · answered May 02 '14 at 14:09

0

What about a "double" replaceAll regex simplified one

"123456-1/1234/189928/2323 (102457921)".replaceAll(".*\\(", "").replaceAll("\\).*", "");

answered May 02 '14 at 14:09

Vassilis Blazos

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score 0 · Answer 5 · answered May 02 '14 at 16:59

You can try something like that:

var str = '123456-1/1234/189928/2323 (102457921)';
    console.log(str.replace(/[-\d\/ ]*\((\d+)\)/, "$1"));
    console.log((str.split('('))[1].slice(0, -1));
    console.log((str.split(/\(/))[1].replace(/(\d+)\)/, "$1"));
    console.log((str.split(/\(/))[1].substr(-str.length - 1, 9));
    console.log(str.substring(str.indexOf('(') + 1, str.indexOf(')')));

In other case you must be familiar with the specifics of the input data to generate a suitable regexp.

Java Regex: replace with capturing group

5 Answers5