Regex Referencing Java

Question

I have a String and a Regex.

the string will look like DH18

and I have my regex ([a-zA-Z]*)([0-9]*) which splits up the letters and the words

I only want the digits, so that which is contained in the second curly brackets. How do I go about referencing this.

I want a statement like

if ($1 == 18)
   blah;

Sergey Kalinichenko · Accepted Answer · 2013-12-10 12:55:23Z

6

You can use a non-capturing group or a lookbehind for the part that you do not want:

(?:[a-zA-Z]*)([0-9]*)  // Non-capturing group
(?<=[a-zA-Z]*)([0-9]*) // Lookbehind

Note that regex returns groups only as strings, even when the group represents a number. Therefore, the check $1 == 18 needs to use equals():

if (matcher.group(1).equals("18")) {
    ...
}

answered Dec 10, 2013 at 12:55

Sergey Kalinichenko

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2 Comments

shyam Over a year ago

Why do you need a non-capturing group at all? Wouldn't [a-zA-Z]*([0-9]*) suffice?

Sergey Kalinichenko Over a year ago

@shyam I assume that the letters part needs to be present for a match to be valid, otherwise there's no need for the OP to mention it at all.

harsh · Accepted Answer · 2013-12-10 12:59:44Z

1

With ([a-zA-Z]*)([0-9]*) you have to pick $2 instead of $1 to get digits as $0 is the complete match, and capturing parts start from $1 .

if($2.equals("18"))

answered Dec 10, 2013 at 12:59

harsh

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Comments

Satheesh Cheveri · Accepted Answer · 2013-12-10 12:57:00Z

0

Pattern p = Pattern.compile("-?\\d+");
    Matcher m = p.matcher("DH18");
    while (m.find()) {
        System.out.println(m.group()); //will print 18
    }

answered Dec 10, 2013 at 12:57

Satheesh Cheveri

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