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Jan 6th, 2010

regex to parse input string

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I want to parse a string into something like an array:

ftp://user:pass@host:portpath
ftp://anon:1234@111.222.333.444:9999/path1/path2/

I read the java docs about using Pattern and Matcher but I keep getting an "IllegalStateException: No match found" error.

java Syntax (Toggle Plain Text)
import java.util.regex.Matcher;
import java.util.regex.Pattern;
 
public class test {
 
	public static void main(String[] args) {
		String s = "ftp://anon:bah@111.222.333.444:9999/path1/path2/";
		Pattern p = Pattern.compile("^ftp://(/S+):(/S+)@(/S+):(/d+)(/S+)$");
		Matcher m = p.matcher(s);
 
		for (int i=0; i<m.groupCount(); i++)
			System.out.println("Group"+i+": "+m.group(i));
 
	}
}
import java.util.regex.Matcher;
import java.util.regex.Pattern;

public class test {

	public static void main(String[] args) {
		String s = "ftp://anon:bah@111.222.333.444:9999/path1/path2/";
		Pattern p = Pattern.compile("^ftp://(/S+):(/S+)@(/S+):(/d+)(/S+)$");
		Matcher m = p.matcher(s);
		
		for (int i=0; i<m.groupCount(); i++)
			System.out.println("Group"+i+": "+m.group(i));

	}
}

"IllegalStateException: No match found" when it goes to print...

Help would be appreciated. Thnx in adv.

Last edited by TheWhite; Jan 6th, 2010 at 8:54 pm.

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TheWhite

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Jan 7th, 2010

Re: regex to parse input string

1. Matcher.group() throws IllegalStateException - If no match has yet been attempted, or if the previous match operation failed.
In your case you haven't attempted to match before calling group(). So the exception.
2. If attempted match fails group() still throws IllegalStateExc...
Hope the following clarifies:

Java Syntax (Toggle Plain Text)
package com.kash.test;
 
import java.util.regex.Matcher;
import java.util.regex.Pattern;
 
public class Main {
 
	public static void main(String[] args) {
		String s = "ftp://anon:bah@111.222.333.444:9999/path1/path2/";
		String pattern = "^ftp://(/S+):(/S+)@(/S+):(/d+)(/S+)$";
		Pattern p = Pattern.compile(pattern);
		Matcher m = p.matcher(s);
 
		/*
		 * ..call m.matches() or m.lookingAt() or m.find() here..
		 */
 
		if (m.matches()) {
			for (int i = 0; i < m.groupCount(); i++)
				System.out.println("Group" + i + ": " + m.group(i));
		} else {
			System.out.println("\"" + s + "\" did not match \"" + pattern + "\"");
			System.out.println("so if I call m.group() it'll throw IllegalStateException !");
		}
	}
 
}
package com.kash.test;

import java.util.regex.Matcher;
import java.util.regex.Pattern;

public class Main {

	public static void main(String[] args) {
		String s = "ftp://anon:bah@111.222.333.444:9999/path1/path2/";
		String pattern = "^ftp://(/S+):(/S+)@(/S+):(/d+)(/S+)$";
		Pattern p = Pattern.compile(pattern);
		Matcher m = p.matcher(s);

		/*
		 * ..call m.matches() or m.lookingAt() or m.find() here..
		 */

		if (m.matches()) {
			for (int i = 0; i < m.groupCount(); i++)
				System.out.println("Group" + i + ": " + m.group(i));
		} else {
			System.out.println("\"" + s + "\" did not match \"" + pattern + "\"");
			System.out.println("so if I call m.group() it'll throw IllegalStateException !");
		}
	}

}

thekashyap

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Jan 7th, 2010

Re: regex to parse input string

I already know the reason no m.group() throws the exception is because no matches were found.

My problem is the pattern which I think should work against that string, but doesn't. Can someone help me figure out the correct pattern?

TheWhite

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Jan 7th, 2010

Re: regex to parse input string

java Syntax (Toggle Plain Text)
package com.kash.test;
 
import java.util.regex.Matcher;
import java.util.regex.Pattern;
 
public class Main {
 
	public static void main(String[] args) {
		String s = "ftp://anon:bah@111.222.333.444:9999/path1/path2/";
		String pattern = "^ftp://(\\S+):(\\S+)@(\\S+):(\\d+)(\\S+)$";
		Pattern p = Pattern.compile(pattern);
		Matcher m = p.matcher(s);
 
		/*
                 * ------------THIS COMMENT IS STILL VALID------------
		 * ..call m.matches() or m.lookingAt() or m.find() here..
		 */
 
		if (m.matches()) {
			for (int i = 0; i < m.groupCount(); i++)
				System.out.println("Group" + i + ": " + m.group(i));
		} else {
			System.out.println("\"" + s + "\" did not match \"" + pattern + "\"");
			System.out.println("so if I call m.group() it'll throw IllegalStateException !");
		}
	}
 
}
package com.kash.test;

import java.util.regex.Matcher;
import java.util.regex.Pattern;

public class Main {

	public static void main(String[] args) {
		String s = "ftp://anon:bah@111.222.333.444:9999/path1/path2/";
		String pattern = "^ftp://(\\S+):(\\S+)@(\\S+):(\\d+)(\\S+)$";
		Pattern p = Pattern.compile(pattern);
		Matcher m = p.matcher(s);

		/*
                 * ------------THIS COMMENT IS STILL VALID------------
		 * ..call m.matches() or m.lookingAt() or m.find() here..
		 */

		if (m.matches()) {
			for (int i = 0; i < m.groupCount(); i++)
				System.out.println("Group" + i + ": " + m.group(i));
		} else {
			System.out.println("\"" + s + "\" did not match \"" + pattern + "\"");
			System.out.println("so if I call m.group() it'll throw IllegalStateException !");
		}
	}

}

Output produced:

Java Syntax (Toggle Plain Text)
Group0: ftp://anon:bah@111.222.333.444:9999/path1/path2/
Group1: anon
Group2: bah
Group3: 111.222.333.444
Group4: 9999
Group0: ftp://anon:bah@111.222.333.444:9999/path1/path2/
Group1: anon
Group2: bah
Group3: 111.222.333.444
Group4: 9999

Last edited by thekashyap; Jan 7th, 2010 at 6:12 am. Reason: added full code

thekashyap

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Jan 7th, 2010

Re: regex to parse input string

ahhh I should have looked more carefully, slashes were wrong way

.

Also, its missing the last group which should be /path1/path2/ any suggestions on that?

TheWhite

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Jan 7th, 2010

Re: regex to parse input string

It turns out m.groupCount() does not include group 0, so the for loop needs to be bound by m.groupCount() + 1.

Thanks everyone.

TheWhite

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