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
– Sara S

2014-05-02 13:28:28 +00:00
Commented May 2, 2014 at 13:28 — Sara S
– Sara S, Commented May 2, 2014 at 13:28
of course, but I'd like a regex solution.

user3111525
– user3111525

2014-05-02 13:29:48 +00:00
Commented May 2, 2014 at 13:29 — user3111525
– user3111525, Commented May 2, 2014 at 13:29

Keppil · Accepted Answer · 2014-05-02 13:29:13Z

5

How about

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

answered May 2, 2014 at 13:29

Keppil

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2 Comments

user3111525 Over a year ago

What does .*? mean?

Keppil Over a year ago

It is the same as .*, except that it is non-greedy. You can read more here.

Amit Joki · Accepted Answer · 2014-05-02 13:29:31Z

1

Well, you can do this:

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

answered May 2, 2014 at 13:29

Amit Joki

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1 Comment

Amit Joki Over a year ago

@sshashank124, yes. Changed + to *

sshashank124 · Accepted Answer · 2014-05-02 13:29:45Z

0

You could do it as:

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

answered May 2, 2014 at 13:29

sshashank124

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Comments

Vassilis Blazos · Accepted Answer · 2014-05-02 14:09:35Z

0

What about a "double" replaceAll regex simplified one

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

answered May 2, 2014 at 14:09

Vassilis Blazos

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user3578445 · Accepted Answer · 2014-05-02 16:59:59Z

0

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.

answered May 2, 2014 at 16:59

user3578445

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