Java Regex Finding digits in a String

Question

I am trying to find digits in a string. I know that in finding a digit is done by \d but when I try it on a sample text like the following:

127.0.0.1 - - [11/Dec/2012:11:57:36 -0500] "GET http:// localhost/ HTTP/1.1" 503 418 "-" "Mozilla/5.0 (X11; Linux x86_64) AppleWebKit/537.11 (KHTML, like Gecko) Chrome/23.0.1271.95 Safari/537.11"

using my java code

Pattern test = Pattern.compile("\\d");
testLine = in.readLine(); // basically the text above 
// extract date and time log in and number of times a user has hit the page
numTimesAccess++; // increment number of lines in a count   
System.out.println(test.matcher(testLine).group());
System.out.println(test.matcher(testLine).start());
System.out.println(test.matcher(testLine).end());

I get an error exception stating that No Match is found. Is something wrong with my regular expression patter or in the way I am trying to access the text matching the patterns.

Thanks I got it now I appreciate everyone's help

jonathan1987
– jonathan1987

2012-12-12 16:35:53 +00:00
Commented Dec 12, 2012 at 16:35 — jonathan1987
– jonathan1987, Commented Dec 12, 2012 at 16:35

PermGenError · Accepted Answer · 2012-12-12 16:27:55Z

7

Firstly, you should call Matcher.find() before you invoke Matcher.group()

use "\\d+" as regex if you consider 127 as a whole single digit.

        Pattern p = Pattern.compile("\\d+");
        Matcher m = p.matcher(s);
        while(m.find()){
        System.out.println(m.group() + " " + m.start() + " " + m.end());
        }

edited Dec 12, 2012 at 16:27

answered Dec 12, 2012 at 16:07

PermGenError

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4 Comments

jonathan1987 Over a year ago

I tried that and still nothing. By the way I am running this code on ubuntu linux if anything.

PermGenError Over a year ago

@jonathan1987 can you post the code that you've tried and also mention the exception as well

jonathan1987 Over a year ago

sure. Pattern test = Pattern.compile("\\d"); testLine = in.readLine(); // basically the text above System.out.println(test.matcher(testLine).group()); and my exception thrown: Exception in thread "main" java.lang.IllegalStateException: No match found at java.util.regex.Matcher.group(Matcher.java:468) at java.util.regex.Matcher.group(Matcher.java:428) at logfilevarnish.ParseLogFile.main(ParseLogFile.java:23)

PermGenError Over a year ago

@jonathan1987 you should call Matcher.find() before you call group(), check my edit

jlordo · Accepted Answer · 2012-12-12 16:20:54Z

0

If you really want to find single digits, you need this:

Pattern pattern = Pattern.compile("\\d");
Matcher matcher = pattern.matcher(testline);
while (matcher.find()) {
    System.out.println(matcher.group());
}

if you want to find non-floating-point numbers, change the regex to "\\d+".

edited Dec 12, 2012 at 16:20

answered Dec 12, 2012 at 16:09

jlordo

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3 Comments

jonathan1987 Over a year ago

I tried that I figured I take out the while look and just run it once. The first time matcher.find() was true and then when I printed it out, I still got an exception thrown.

jlordo Over a year ago

if you only want the first match, replace the while with if

jlordo Over a year ago

you get the exception, because you always create a new Matcher instead of using just one.

Community · Accepted Answer · 2017-05-23 11:45:14Z

0

Try using \d* instead of \d+. Have a look at this post-: Finding a Number in a string.

edited May 23, 2017 at 11:45

CommunityBot

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answered Dec 12, 2012 at 16:39

Sreyan

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Comments

Serg Burlaka · Accepted Answer · 2019-03-01 15:05:40Z

0

just add and use the simple ktx:

 fun String.digits() = 
      Pattern.compile("\\d+").matcher(this).run { if (find()) group() else "" }!!

edited Mar 1, 2019 at 15:05

answered Mar 1, 2019 at 15:00

Serg Burlaka

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