Java regex, IllegalStateException: No match found

Question

I spent a bit of time looking for the answer, but I couldn't find it, so, here's the question:

I'm getting this error:

    StandardWrapperValve[jsp]: Servlet.service() for servlet jsp threw  
    exceptionjava.lang.IllegalStateException: No match found
        at java.util.regex.Matcher.group(Matcher.java:536)
        at extra.regex.checkRegex(regex.java:21)

And here's the code

    public class regex {
        public static String checkRegex(String check, String regex)
        {
            Pattern p = Pattern.compile(regex);
            Matcher m = p.matcher(check);
            m.matches();
            return m.group(1);
        }
    }

I'm calling checkRegex() with this:

    String query = request.getQueryString();//this query looks like "sel=45"
    String regulex = "#^(.+)#"; //even this guy returns no matches

    //out.println(query); //it is is not null
    String result = regex.checkRegex(query, regulex);

    out.println(result);

I've seen a few questions here, but all were about missing m.matches() or m.find(), but I've triple checked that.

This is explained in the Javadoc: docs.oracle.com/javase/7/docs/api/java/util/regex/… — Oliver Charlesworth
– Oliver Charlesworth, Commented Aug 1, 2015 at 16:55
@SotiriosDelimanolis matches() returns a boolean, an extra check needed? @Reimeus, I've checked that via regex101, unfortunately, it matches completely — Alex Shereshovets
– Alex Shereshovets, Commented Aug 1, 2015 at 16:57
"I've seen a few questions here, but all were about missing m.matches() or m.find()" do you know what is purpose of these methods? Do you know difference between them? — Pshemo
– Pshemo, Commented Aug 1, 2015 at 16:57

Jens · Accepted Answer · 2015-08-01 16:55:19Z

3

You shoud use m.group() in an if block:

     if(m.matches()) {
        return m.group(1);
      } else {
        return null;
      }

answered Aug 1, 2015 at 16:55

Jens

69.6k15 gold badges107 silver badges133 bronze badges

Sign up to request clarification or add additional context in comments.

6 Comments

Alex Shereshovets Over a year ago

Yes, that fixed the error, but the value is still null

Jens Over a year ago

@AlexShereshovets and what schould the result should be?

Alex Shereshovets Over a year ago

with this regex - it should return the whole query, I know, this is pointless, but it takes nothing to change the expression)

Pshemo Over a year ago

@AlexShereshovets Post example text you want to parse and show part which you want to find. Also #^ doesn't make sense because ^ represents start of string, which but if it is placed right after # it means that it will never match start of string. This simply can't work.

Alex Shereshovets Over a year ago

@Pshemo for example, I want to find 121123123234 in sel=121123123234, so I will use this #[^=]+=([0-9]+)# or #sel=([0-9]+)#, but I tried to make it simpler and throw away cases when my regex could be wrong

|

Shar1er80 · Accepted Answer · 2015-08-01 17:44:39Z

1

Let's break down your pattern:

"#^(.+)#"

#^ - Literal string
(.+) - Any character (may or may not match line terminators) 1 or more times
# - Literal string

So why does sel=45 fail to match? The String does not begin with "#^" and does not end with "#".

It would seem you want to capture data in between "#" that may or may not be there (You need to clarify this). The following pattern accomplishes that:

"#?([^#]+)#?"

This will match sel=45 and will match #sel=45#

Example:

public static void main(String[] args) throws Exception {
    String regulex = "#?([^#]+)#?"; //even this guy returns no matches

    System.out.println(checkRegex("#sel=45#", regulex));
    System.out.println(checkRegex("sel=45", regulex));
}

public static String checkRegex(String check, String regex) {
    Pattern p = Pattern.compile(regex);
    Matcher m = p.matcher(check);
    return m.matches() ? m.group(1): null;
}

Results:

sel=45
sel=45

answered Aug 1, 2015 at 17:44

Shar1er80

9,0512 gold badges24 silver badges31 bronze badges

1 Comment

Alex Shereshovets Over a year ago

Oh, I used # as a delimiter, but it was not required, and this messed it all up. Thank you

ka4eli · Accepted Answer · 2015-08-01 16:55:40Z

0

This regex expression

  String regulex = "^#(.+)#";

will match

    String query = "#sel=45#";

This combination doesn't make sense #^. ^ indicates the beginning of string, nothing can be before it. And if you query does not contain # it won't match regulex.

answered Aug 1, 2015 at 16:55

ka4eli

5,4443 gold badges25 silver badges43 bronze badges

Collectives™ on Stack Overflow

Java regex, IllegalStateException: No match found

3 Answers 3

6 Comments

1 Comment

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

6 Comments

1 Comment

Comments

Your Answer

Sign up or log in

Post as a guest

Related