Solved

Java regex needed

Posted on 2011-09-28
10
172 Views
Last Modified: 2012-05-12
Hi,

How can I use the attached code to extract "B004UJLUQU" from "dp/B004UJLUQU"?

The attached code returns the first matched string, and has been very helpful in the past such that I don't want to change it.

Thanks
public static String extractRegex(String source, String regex){
		Pattern re = Pattern.compile(regex, Pattern.DOTALL|Pattern.MULTILINE|Pattern.CASE_INSENSITIVE);
		Matcher m = re.matcher(source);
		while (m.find())
			return m.group(0);
		return null;
	}

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0
Comment
Question by:wsyy
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10 Comments
 
LVL 86

Expert Comment

by:CEHJ
ID: 36717956
Something like the following should work
public static String extract(String s) {
        return s.replaceAll(".*?/(.*)", "$1");
    }

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0
 
LVL 47

Expert Comment

by:for_yan
ID: 36717958
Why wouldnot you use "B004UJLUQU" as the pattern:

public static String extractRegex(String source, String regex){
		Pattern re = Pattern.compile("B004UJLUQU", Pattern.DOTALL|Pattern.MULTILINE|Pattern.CASE_INSENSITIVE);
		Matcher m = re.matcher(source);
		while (m.find())
			return m.group(0);
		return null;
	}

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0
 
LVL 47

Expert Comment

by:for_yan
ID: 36718013
I tested this:

                 String source = "dp/B004UJLUQU";

//public static String extractRegex(String source, String regex){
		Pattern re = Pattern.compile("/(.+)$", Pattern.DOTALL|Pattern.MULTILINE|Pattern.CASE_INSENSITIVE);
		Matcher m = re.matcher(source);
		while (m.find())
			System.out.println(m.group(1));
	//	return null;

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Output:
B004UJLUQU

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0
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LVL 47

Expert Comment

by:for_yan
ID: 36718038
This is as a method, though I changed group(0) to group(1):
  public static String extractRegex(String source, String regex){
		Pattern re = Pattern.compile(regex, Pattern.DOTALL|Pattern.MULTILINE|Pattern.CASE_INSENSITIVE);
		Matcher m = re.matcher(source);
		while (m.find())
			return m.group(1);
		return null;
	}


		   public static void main(String[] args){





                 String source = "dp/B004UJLUQU";
                  String regex = "/(.+)$";
           System.out.println(extractRegex( source, regex));

}

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Output:

B004UJLUQU

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0
 
LVL 47

Accepted Solution

by:
for_yan earned 50 total points
ID: 36718094
Now if you want to keep the method intact (with group(0)) that's how you do it:
    public static String extractRegex(String source, String regex){
		Pattern re = Pattern.compile(regex, Pattern.DOTALL|Pattern.MULTILINE|Pattern.CASE_INSENSITIVE);
		Matcher m = re.matcher(source);
		while (m.find())
			return m.group(0);
		return null;
	}


          String source = "dp/B004UJLUQU";
                  String regex = "(?<=/).+$";

               System.out.println(extractRegex( source, regex));

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B004UJLUQU

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0
 

Author Comment

by:wsyy
ID: 36718229
CEH, I don't want to change group(0) in the method.

for_yan, the last method you provided works. can you please explain it a little bit?

I guess that (?<=/) removes the "/", but I don't understand the ".+$" part.
0
 
LVL 47

Expert Comment

by:for_yan
ID: 36718233
.+ this maens any charcter any tnumber of times but at least once
$ - means the end of the string
0
 
LVL 47

Expert Comment

by:for_yan
ID: 36718245
So it looks for the slash and then picks up anything after the slash to th end of the string
0
 
LVL 47

Expert Comment

by:for_yan
ID: 36718256
and this is called positive lookbehind (?<=/) - it checks that ther is a slash before , but does not include it into match
0
 

Author Closing Comment

by:wsyy
ID: 36718335
Many thanks!
0

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