Java: Need to extract a number from a string

Question

I have a string containing a number. Something like "Incident #492 - The Title Description".
I need to extract the number from this string.
Tried

Pattern p = Pattern.compile("\\d+");  
Matcher m = p.matcher(theString);  
String substring =m.group();

By getting an error

java.lang.IllegalStateException: No match found

What am I doing wrong?
What is the correct expression?
I'm sorry for such a simple question, but I searched a lot and still not found how to do this (maybe because it's too late here...)

Can you assert that there will not be other numbers in the String? — Tunaki
– Tunaki, Commented Aug 29, 2015 at 21:34

Sergey Kalinichenko · Accepted Answer · 2015-08-29 21:38:00Z

3

You are getting this exception because you need to call find() on the matcher before accessing groups:

Matcher m = p.matcher(theString);  
while (m.find()) {
    String substring =m.group();
    System.out.println(substring);
}

Demo.

edited Aug 29, 2015 at 21:38

answered Aug 29, 2015 at 21:35

Sergey Kalinichenko

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2 Comments

Andreas Over a year ago

Or matches() if theString is not allowed extraneous characters.

Sergey Kalinichenko Over a year ago

@Andreas That's right. OP's string allows other characters, though.

Makoto · Accepted Answer · 2015-08-29 21:44:15Z

1

There are two things wrong here:

The pattern you're using is not the most ideal for your scenario, it's only checking if a string only contains numbers. Also, since it doesn't contain a group expression, a call to group() is equivalent to calling group(0), which returns the entire string.
You need to be certain that the matcher has a match before you go calling a group.

Let's start with the regex. Here's what it looks like now.

Regular expression visualization

Debuggex Demo

That will only ever match a string that contains all numbers in it. What you care about is specifically the number in that string, so you want an expression that:

Doesn't care about what's in front of it
Doesn't care about what's after it
Only matches on one occurrence of numbers, and captures it in a group

To that, you'd use this expression:

.*?(\\d+).*

Regular expression visualization

Debuggex Demo

The last part is to ensure that the matcher can find a match, and that it gets the correct group. That's accomplished by this:

if (m.matches()) {
    String substring = m.group(1);
    System.out.println(substring);
}

All together now:

Pattern p = Pattern.compile(".*?(\\d+).*");
final String theString = "Incident #492 -  The Title Description";
Matcher m = p.matcher(theString);
if (m.matches()) {
    String substring = m.group(1);
    System.out.println(substring);
}

edited Aug 29, 2015 at 21:44

answered Aug 29, 2015 at 21:39

Makoto

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2 Comments

Sergey Kalinichenko Over a year ago

The pattern OP is using is 100% correct, as long as you are not trying to use matches() on it. He does not need the stuff in front or behind the \\d+ part, and he does not need groups beyond the default group zero, unless he wishes to capture other parts of the string.

Makoto Over a year ago

@dasblinkenlight: I'll rephrase it then - it's not the most ideal regex to use, then. It will work so long as you're using find, but when I author regexes, I like it to be clear that it's going to match specific things.

schtever · Accepted Answer · 2015-08-29 21:35:26Z

0

You need to invoke one of the Matcher methods, like find, matches or lookingAt to actually run the match.

answered Aug 29, 2015 at 21:35

schtever

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