Java Regex tuning

Question

Can anyone help where I am doing wrong?

My sample text :

{[|Name:A|Class:1|Sex:Male|][|Name:B|Class:2|Sex:Female|][|Name:C|Class:3|Sex:Male|]}

Expected output:

|Name:A|Class:1|Sex:Male|
Name:A
Class:1
Sex:Male
|Name:B|Class:2|Sex:Female|
Name:B
Class:2
Sex:Female
|Name:C|Class:3|Sex:Male|
Name:C
Class:3
Sex:Male

Current output :

|Name:A|Class:1|Sex:Male|
Name:A
Sex:Male
|Name:B|Class:2|Sex:Female|
Name:B
Sex:Female
|Name:C|Class:3|Sex:Male|
Name:C
Sex:Male

My program:

import java.util.regex.Matcher;
import java.util.regex.Pattern;

public class Regex {

    public static void main(String[] args) {

        String example = "{[|Name:A|Class:1|Sex:Male|][|Name:B|Class:2|Sex:Female|][|Name:C|Class:3|Sex:Male|]}";

        Pattern curlyBraces = Pattern.compile("\\[(.*?)\\]");

        Matcher m = curlyBraces.matcher(example);
        while (m.find()) {
            System.out.println(m.group(1));
            String element = m.group(1);
            Pattern pipe = Pattern.compile("\\|(.*?)\\|");
            Matcher mPipe = pipe.matcher(element);
            while (mPipe.find()) {
                System.out.println(mPipe.group(1));
            }
        }
    }
}

Given that you program prints using println, it's extremely unlikely that your output is a single line of text, like you showed, so please format entire question correctly, so we know that you want and what you're actually getting right now. — Andreas
– Andreas, Commented Feb 4, 2016 at 7:24

Tim Pietzcker · Accepted Answer · 2016-02-04 07:49:15Z

1

|Name:A|Class:1|Sex:Male|

because the regex consumes the characters it's matching, and the | between Name:A and Class:1 therefore can only be matched once.

Use lookaround assertions to get around that - they don't consume the text they match:

        Pattern pipe = Pattern.compile("(?<=\\|).*?(?=\\|)");
        Matcher mPipe = pipe.matcher(element);
        while (mPipe.find()) {
            System.out.println(mPipe.group(0));
        }

Another possibility, if you're not expecting empty values, would be to match all "non-pipes" characters:

        Pattern pipe = Pattern.compile("[^|]+");
        Matcher mPipe = pipe.matcher(element);
        while (mPipe.find()) {
            System.out.println(mPipe.group(0));
        }

answered Feb 4, 2016 at 7:49

Tim Pietzcker

337k59 gold badges520 silver badges572 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

JonnyTieM · Accepted Answer · 2016-02-04 10:30:41Z

0

Like Tim Pietzcker already discribed, the | gets conumed by the regex and then it can't find the Class:1.

But you don't need the second regex. Instead of the regex you could use a normal string.split("|"). Should work in your case and is probably faster:

String element = m.group(1);
String[] splitString = element.split("\\|"); // splitString = ["Name:A", "Class:1", "Sex:Male"]

To avoid empty strings in splitString you would have to change your first regex Pattern to "\\[\\|(.*?)\\|\\]".

edited Feb 4, 2016 at 10:30

answered Feb 4, 2016 at 8:02

JonnyTieM

1772 silver badges12 bronze badges

4 Comments

Tim Pietzcker Over a year ago

That gives you ["", "Name:A", "Class:1", "Sex:Male"] though - only trailing empty splits are removed by default.

JonnyTieM Over a year ago

You're right, though you could adjust the first Pattern to "\\[\\|(.*?)\\|\\]" and then you wouldn't have the issue with the empty splits. I'll update the answer.

Alan Moore Over a year ago

Java's String#split() treats its argument as a regex, so you still have to escape the pipe.

JonnyTieM Over a year ago

@AlanMoore Thanks, totally forgot that. Updated the answer.

Collectives™ on Stack Overflow

Java Regex tuning

2 Answers 2

Comments

4 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

Comments

4 Comments

Your Answer

Sign up or log in

Post as a guest

Related