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Regex

Posted on 2011-09-18
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Medium Priority
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Last Modified: 2012-05-12
how do you specify a pattern for

* space * & large

meaning any number of other characters before and after "space", followed by "large"
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Question by:bhomass
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18 Comments
 
LVL 47

Expert Comment

by:for_yan
ID: 36557627
        Pattern p = Pattern.compile(".*space.*large");

        Matcher m1 = p.matcher("sdfsdfspace  large");

        if(m1.find())System.out.println("matched");

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matched

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0
 
LVL 35

Expert Comment

by:Terry Woods
ID: 36557641
Note for_yan's pattern still requires the 2 words to be on the same line. Did you want to capture or replace part of the string at all, or just detect that it matched
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LVL 35

Expert Comment

by:Terry Woods
ID: 36557644
And are you working with Java, or Perl, or something else?
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LVL 47

Expert Comment

by:for_yan
ID: 36557671

This is matcvhing even if there is end of line:
                             String sep = System.getProperty("line.separator");

     //  Pattern p = Pattern.compile(".*space.*large");
          Pattern p = Pattern.compile(".*space.*large", Pattern.DOTALL);

        Matcher m1 = p.matcher("sdf" +sep +  " sdfspace " + sep + "  large");

        if(m1.find())System.out.println("matched");

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matched

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0
 
LVL 47

Expert Comment

by:for_yan
ID: 36557687



Another way to specify that end of line should also be acceptable.
the code below works (when we use ("(?s:.*space.*large)") instead of ("(.*space.*large)") )
 ?s: is a "single-line mode" and this should work not only in Java - as opposed to Pattern.DOTALL which is obviously
Java specific
                             String sep = System.getProperty("line.separator");

      Pattern p = Pattern.compile("(?s:.*space.*large)");
        // p = Pattern.compile("(.*space.*large)");

         // Pattern p = Pattern.compile(".*space.*large", Pattern.DOTALL);

        Matcher m1 = p.matcher("sdf" +sep +  " sdfspace " + sep + "  large");

        if(m1.find())System.out.println("matched");

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 matched 

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0
 

Author Comment

by:bhomass
ID: 36557708
I follow for_yan almost all the way. please explain to me what is ?s:

I actually need this to be generally applicable, language independent, but I will test it out using java whenever possible.
0
 
LVL 47

Expert Comment

by:for_yan
ID: 36557711

The same thing but more informative output:

                             String sep = System.getProperty("line.separator");

      Pattern p = Pattern.compile("(?s:.*space.*large)");
        // p = Pattern.compile("(.*space.*large)");

         // Pattern p = Pattern.compile(".*space.*large", Pattern.DOTALL);

        Matcher m1 = p.matcher("sdf" +sep +  " sdfspace " + sep + "  large");

        if(m1.find())System.out.println(("sdf" +sep +  " sdfspace " + sep + "  large").substring(m1.start(),m1.end()));

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Output:
sdf
 sdfspace 
  large

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LVL 47

Expert Comment

by:for_yan
ID: 36557716
(?s:here your regex) - this ?s: means single-line mode which means that dot also matches new line charcter
0
 
LVL 47

Expert Comment

by:for_yan
ID: 36557717
but you should be careful - if those are many lines in the file - I guess in this single-line mode
it will match all lines before

0
 

Author Comment

by:bhomass
ID: 36557718
is there a way to specify the distance between "space" and large is less than 30 char?
0
 
LVL 35

Assisted Solution

by:Terry Woods
Terry Woods earned 400 total points
ID: 36557720
I haven't seen this syntax for the single-line pattern modifier before:
      Pattern p = Pattern.compile("(?s:.*space.*large)");
Does it just apply to the group it's in?

I'm more familiar with this:
      Pattern p = Pattern.compile("(?s).*space.*large");

And I see for_yan has just explained it, so I won't write any more!
0
 
LVL 47

Expert Comment

by:for_yan
ID: 36557723

Actually, I think this regex should be language indpendent - I don't think there is anything java specific here:
"(?s:.*space.*large)"
0
 
LVL 35

Expert Comment

by:Terry Woods
ID: 36557726
is there a way to specify the distance between "space" and large is less than 30 char?

Yes:
      Pattern p = Pattern.compile("(?s).*space.{0,29}large");
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LVL 47

Accepted Solution

by:
for_yan earned 560 total points
ID: 36557727


 Pattern p = Pattern.compile("(?s:.*space.{1,30}large)");
0
 
LVL 35

Expert Comment

by:Terry Woods
ID: 36557728
And actually, the leading .* is redundant, so this would be fine too:

      Pattern p = Pattern.compile("(?s)space.{0,29}large");
0
 
LVL 47

Expert Comment

by:for_yan
ID: 36557730
Yes 0,29 iif you ewant strict and are ok witth no space
0
 
LVL 35

Expert Comment

by:Terry Woods
ID: 36557733
specify the distance between "space" and large is less than 30 char?
      Pattern p = Pattern.compile("(?s)space.{0,29}large");  // also allows no gap between the words
      Pattern p = Pattern.compile("(?s)space.{1,29}large");

specify the distance between "space" and large is less than or equal to 30 char?
      Pattern p = Pattern.compile("(?s)space.{0,30}large");  // also allows no gap between the words
      Pattern p = Pattern.compile("(?s)space.{1,30}large");
0
 

Author Comment

by:bhomass
ID: 36557740
great answers
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