Small regexp tip

To find the shortest match in a using using regular expressions you can use a “Reluctant” qualifier like “.+?” instead of simply “.+” (greedy qualifier) which will find the longest match.

Wow – did that make my day!! 🙂

4 thoughts on “Small regexp tip”

  1. Hi Mikkel,

    you possibly mean .* instead of .+

    .* : 0, 1 or many (as much as possible)
    .+ : 1 or many
    .+? : 0, 1 or many (see it as (.+)? : (1 or many) or 0)

    Ciao
    Thomas

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  2. No I actually mean it as a write it – consider some input such as:

    Section
  Contents1
End Section
Section
  Contents2
End Section

    If I wanted to use regexp to split the input into two parts with a section in each I would need to use “.+?” and not “.+” to indicate the actual contents between the “Section” and “End Section” parts. If I do the result will be what you expect. I a don’t the result will only be a single match with the entire input.

    Try running the below code:

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

public class Lekkim {

  public static void main(String[] args) {
    String input = "Sectionn  Contents1nEnd Sectionn" +
       "nSectionn  Contents2nEnd Sectionn";

    Pattern p1 = Pattern.compile("(Section.+End Section)", Pattern.DOTALL);
    Pattern p2 = Pattern.compile("(Section.+?End Section)", Pattern.DOTALL);

    System.out.println("Input is:n" + input);
    System.out.println("*** *** *** *** ***");

    Matcher m = p1.matcher(input);
    while (m.find()) {
      System.out.println("Match:n" + m.group());
    }

    System.out.println("*** *** *** *** ***");

    m = p2.matcher(input);
    while (m.find()) {
      System.out.println("Match:n" + m.group());
    }

  }
}

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  4. I ran into problems with reluctant qualifiers recently – it seems as if older versions of Python did not implement regex properly. It nearly drove me nuts trying to figure out what was going on. That being said, once I found out about them I went about happily revising old code…

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