Java Regular Expression throwing error

Question

I am trying to capture a certain number chunk from within a text. Suppose the text is Kane is 12345 feet high. I want to capture 12345. I am trying to use this:

import java.util.regex.Pattern; 
import java.util.regex.Matcher; 
String text = "Kane is 12345 feet high";
Pattern p = Pattern.compile("Kane is (\\d+) feet high");
Matcher m = p.matcher(text);
String s0 = m.group(0);

However I am getting a Match not found error. What am I doing wrong here? I mean, in Perl, this perfectly prints out 12345:

$foo = "Kane is 12345 feet high";
$foo =~ /Kane is (\d+) feet high/;
print $1;

You need to use matches for an exact match of the whole regex or find to find the regex in a sub-string. Unfortunately it doesn't give you a more useful error like Match not found, did you forget to call matches() or find() ;) — Peter Lawrey
– Peter Lawrey, Commented Jan 9, 2013 at 10:29
Also reading the API documentation is actually a good practice for a programmer. — m0skit0
– m0skit0, Commented Jan 9, 2013 at 10:35

Marko Topolnik · Accepted Answer · 2013-01-09 11:15:33Z

8

Just instantiating Matcher isn't enough: you must call m.matches() and the standard practice would be to put it in an if:

if (m.matches()) s0 = m.group(1);

edited Jan 9, 2013 at 11:15

answered Jan 9, 2013 at 10:26

Marko Topolnik

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1 Comment

Alan Moore Over a year ago

+1, but I think you should be using group(1), not group(0). I know the OP used group(0), but he also said he was trying to extract the number.

PermGenError · Accepted Answer · 2013-01-09 10:26:32Z

6

you have to invoke Matcher.find() before calling Matcher.group()

String text = "Kane is 12345 feet high";
Pattern p = Pattern.compile("Kane is (\\d+) feet high");
Matcher m = p.matcher(text);
if(m.find()){
String s0 = m.group(0);

}

Suppose the text is Kane is 12345 feet high. I want to capture 12345

i think, your Regex should be like below in order to just capture the number in your text.

    Pattern p = Pattern.compile("\\d+");

answered Jan 9, 2013 at 10:26

PermGenError

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3 Comments

NedStarkOfWinterfell Over a year ago

No, only \d+ won't suffice, for there may be numeric chunks in Kane as well, like Kane142..So in that case I will need to capture the 2nd group.

PermGenError Over a year ago

@Cupidvogel ohk.. i tought you only needed literal digits, :P

NedStarkOfWinterfell Over a year ago

Yeah, that would look like it from my question..:P

Ayaz Ali Khatri · Accepted Answer · 2013-01-09 10:38:01Z

1

Try This:

public static String stripNonDigits(final String input){
  final StringBuilder sb = new StringBuilder();
  for(int i = 0; i < input.length(); i++){
    final char c = input.charAt(i);
    if(c > 47 && c < 58){
        sb.append(c);
    }
}
return sb.toString();

} Test Code:

public static void main(final String[] args){
final String input = "0-123-abc-456-xyz-789";
final String result = stripNonDigits(input);
System.out.println(result);
 }

Output:

0123456789

Might be helpfull thanks

answered Jan 9, 2013 at 10:38

Ayaz Ali Khatri

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1 Comment

Alan Moore Over a year ago

I'm sure that works fine, but it doesn't answer the question. What the OP wants to know is why his own code doesn't work.

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