Java reg expression capture string

Question

I have the following string:

"(1)name1:content1(2)name2:content2(3)name3:content3...(n)namen:contentn"

what I want to do is to capture each of the name_i and content_i, how can I do this? I should mention that name_i is unknown. For example name1 could be "abc", name2 could be "xyz".

What I have tried:

String regex = "\\(\\d\\)(.*):(.*)(?=\\(\\d\\))";
Pattern pattern = Pattern.compile(regex);
Matcher matcher = pattern.matcher(str);
if (matcher.find()) {
  System.out.println(matcher.group(0);
  System.out.println(matcher.group(1);
  System.out.println(matcher.group(2);
}

But the results is not very good. I also tried matcher.mathes(), nothing will be returned.

I suggest a less readable, but a more lenient and efficient regex: "\\(\\d+\\)([^:]+):([^(]*(?:\\((?!\\d+\\))[^(]*)*)" — Wiktor Stribiżew
– Wiktor Stribiżew, Commented Aug 29, 2017 at 10:25

Wiktor Stribiżew · Accepted Answer · 2017-08-29 09:35:48Z

1

You may use

String s = "(1)name1:content1(2)name2:content2(3)name3:content3...(4)namen:content4";
Pattern pattern = Pattern.compile("\\(\\d+\\)([^:]+):([^(]*(?:\\((?!\\d+\\))[^(]*)*)");
Matcher matcher = pattern.matcher(s);
while (matcher.find()){
    System.out.println(matcher.group(1));
    System.out.println(matcher.group(2)); 
}

See the Java demo

Details

\\(\\d+\\) - matches (x) substring where x is 1 or more digits
([^:]+) - Group 1: one or more chars other than :
: - a colon
([^(]*(?:\\((?!\\d+\\))[^(]*)*) - Group 2:
- [^(]* - zero or more chars other than (
- (?:\\((?!\\d+\\))[^(]*)* - zero or more sequences of:
  - \\((?!\\d+\\)) - a ( that is not followed with 1+ digits and )
  - [^(]* - 0+ chars other than (

See the regex demo.

answered Aug 29, 2017 at 9:35

Wiktor Stribiżew

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Comments

Alexey R. · Accepted Answer · 2017-08-29 09:49:21Z

1

This will work if your name and content does not include any non "word"-boundary characters:

public static void test(String input){
    String regexpp = "\\(\\d+\\)(\\w+):(\\w+)";
    Pattern p = Pattern.compile(regexpp);
    Matcher m = p.matcher(input);
    while(m.find()){
        System.out.println("Name: " + m.group(1));
        System.out.println("Content: " + m.group(2));
    }
}

Output:

Name: name1
Content: content1
Name: name2
Content: content2
Name: name3
Content: content3
Name: name99
Content: content99

edited Aug 29, 2017 at 9:49

answered Aug 29, 2017 at 9:23

Alexey R.

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3 Comments

Steven Luo Over a year ago

Thanks for your reply. But name_i is unknown. For example name1 could be "abc", name2 could be "xyz".

Alexey R. Over a year ago

You state the exact string in your question. It is probably better to provide a more representative example not to confuse others.

anubhava Over a year ago

: doesn't need to escaped in regex

Erwin Bolwidt · Accepted Answer · 2017-08-29 09:19:34Z

-1

Your expression matches greedily - your first group eats up the colon first so it won't be possible to match the entire expression. You can use non-greedy matching (using the question mark as in *?) to make your pattern match.

String regex = "\\(\\d\\)(.*?):(.*?)(?=\\(\\d\\))";

answered Aug 29, 2017 at 9:19

Erwin Bolwidt

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1 Comment

Steven Luo Over a year ago

Thanks for you suggestion. It works better than mine. But still only the first group, i.e., name1, content1, is returned in the matcher.group().

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