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Java Regex Question

Posted on 2008-06-24
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Last Modified: 2013-11-24
I'd like to be able to see if a pattern exists.
If so, I'd like to see if ANOTHER pattern encapsulates it.  It won't always, but if it does, I'd like to detect it.

blah blah {{Pattern1[[Value]]}} blah blah [[Value2]].

So if the [[]] is found, cool.  I can do this with pattern matcher.  But I don't know how to see if it's inside of an additional pattern?  So in the above, if [[ ]] is found great.  Is there a {{}} around it?

Thanks!
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Question by:ecuguru
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5 Comments
 
LVL 84

Expert Comment

by:ozo
ID: 21861362
       System.out.println("blah blah {{Pattern1[[Value]]}} blah blah [[Value2]]".matches(".*Pattern1\\[\\[Value\\]\\].*"));
        System.out.println("blah blah {{Pattern1[[Value]]}} blah blah [[Value2]]".matches(".*\\{\\{Pattern1\\[\\[Value]]}}.*"));
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LVL 37

Expert Comment

by:zzynx
ID: 21863281
>> if [[ ]] is found great.  Is there a {{}} around it?

String regEx = ".*\\{\\{.*\\[\\[.*]]}}.*";
System.out.println("blah blah {{Pattern1[[Value]]}} blah blah [[Value2]]".matches(regEx)); // Match
System.out.println("blah blah Pattern1[[Value]] blah blah [[Value2]]".matches(regEx));         // No Match
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LVL 7

Expert Comment

by:SprudeVI
ID: 22087102
Use groups to determine what you have found. In reg. expressions groups are enclosed in brackets. Later you can use the "String Matcher.group(int grNumber)" method to get the string matched by the group.
import java.util.regex.Matcher;
import java.util.regex.Pattern;
 
public class Test {
  public static void main(String[] args)  {
    String s = "blah blah {{Pattern1[[Value]]}} blah blah [[Value2]]";
    
    Pattern pattern = Pattern.compile(
      "(Pattern1\\[\\[(.+?)\\]\\])|(\\[\\[(.+?)\\]\\])");
    //   ^        ^^^   ^            ^      ^
    //   |         |    |            |      |
    //   |         |    |            |      +-- group 4
    //   |         |    |            +-- group 3  
    //   |         |    +-- group 2  
    //   |         +-- rect. brackets have to be quoted  
    //   +-- group 1
    Matcher matcher = pattern.matcher(s);
    while (matcher.find()) {
      if (matcher.group(2) != null) {
        System.out.println("Value found inside pattern!");
        System.out.println("> Pattern: " + matcher.group(1));
        System.out.println("> Value: " + matcher.group(2));
        System.out.println();
      } else if (matcher.group(3) != null) {
        System.out.println("Value found without pattern!");
        System.out.println("> Value: " + matcher.group(4));
        System.out.println();
      }
    }
  }
}

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Accepted Solution

by:
SprudeVI earned 2000 total points
ID: 22087108

import java.util.regex.Matcher;
import java.util.regex.Pattern;
 
public class Test {
  public static void main(String[] args)  {
    String s = "blah blah {{Pattern1[[Value]]}} blah blah [[Value2]]";
    
    Pattern pattern = Pattern.compile(
      "(Pattern1\\[\\[(.+?)]])|(\\[\\[(.+?)]])");
    //   ^      ^^^   ^        ^      ^
    //   |       |    |        |      |
    //   |       |    |        |      +-- group 4
    //   |       |    |        +-- group 3  
    //   |       |    +-- group 2  
    //   |       +-- rect. brackets have to be quoted  
    //   +-- group 1
    Matcher matcher = pattern.matcher(s);
    while (matcher.find()) {
      if (matcher.group(2) != null) {
        System.out.println("Value found inside pattern!");
        System.out.println("> Pattern: " + matcher.group(1));
        System.out.println("> Value: " + matcher.group(2));
        System.out.println();
      } else if (matcher.group(3) != null) {
        System.out.println("Value found without pattern!");
        System.out.println("> Value: " + matcher.group(4));
        System.out.println();
      }
    }
  }
}

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Author Closing Comment

by:ecuguru
ID: 31470407
Perfect, sorry about the delay in assigning.
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