How do I find multiple substrings from one string using regex in Java?

Question

I want to find every instance of a number, followed by a comma (no space), followed by any number of characters in a string. I was able to get a regex to find all the instances of what I was looking for, but I want to print them individually rather than all together. I'm new to regex in general, so maybe my pattern is wrong?

This is my code:

String test = "1 2,A 3,B 4,23";
Pattern p = Pattern.compile("\\d+,.+");
Matcher m = p.matcher(test);
while(m.find()) {
  System.out.println("found: " + m.group());
}

This is what it prints:

found: 2,A 3,B 4,23

This is what I want it to print:

found: 2,A
found: 3,B
found: 4,23

Thanks in advance!

Evgeniy Dorofeev · Accepted Answer · 2013-03-31 05:17:15Z

3

try this regex

    Pattern p = Pattern.compile("\\d+,.+?(?= |$)");

answered Mar 31, 2013 at 5:17

Evgeniy Dorofeev

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pickypg · Accepted Answer · 2013-03-31 05:17:17Z

3

You could take an easier route and split by space, then ignore anything without a comma:

String values = test.split(' ');

for (String value : values) {
    if (value.contains(",") {
        System.out.println("found: " + value);
    }
}

answered Mar 31, 2013 at 5:17

pickypg

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3 Comments

Archer Over a year ago

This does seem easier, but I have to add some more conditions later making regex more suitable. Thank you though!

MD. Sahib Bin Mahboob Over a year ago

Yout beat me by 4 mins @pickypg

MD. Sahib Bin Mahboob Over a year ago

@Archer Then i believe it would be better to add this conditions to the question. So people can keep it in mind before answering.

Ted Hopp · Accepted Answer · 2013-03-31 05:17:34Z

1

What you apparently left out of your requirements statement is where "any number of characters" is supposed to end. As it stands, it ends at the end of the string; from your sample output, it seems you want it to end at the first space.

Try this pattern: "\\d+,[^\\s]*"

answered Mar 31, 2013 at 5:17

Ted Hopp

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