Regular Expressions + Including one space in pattern

Question

I'm trying to figure out how to write a pattern to match to the following: "3Z 5Z". The numbers in this can vary, but the Z's are constant. The issue I'm having is trying to include the white space... Currently I have this as my pattern

 pattern = @"\b*Z\s*Z\b";

The '*' represent the wildcard for the number preceding the "Z", but it doesn't seem to want to work with the space in it. For example, I can use the following pattern successfully for matching to the same thing without the space (i.e. 3Z5Z)

pattern = @"\b*Z*Z\b";

I am writing this program in .NET 4.0 (C#). Any help is much appreciated!

EDIT: This pattern is part of a larger string, for example: 3Z 10Z lock 425"

score 5 · Accepted Answer · edited Apr 25 '14 at 16:05

5

Try this:

pattern = @"\b\d+Z\s+\d+Z\b";

Explanation:

"
\b    # Assert position at a word boundary
\d    # Match a single digit 0..9
   +     # Between one and unlimited times, as many times as possible, giving back as needed (greedy)
Z     # Match the character “Z” literally
\s    # Match a single character that is a “whitespace character” (spaces, tabs, line breaks, etc.)
   +     # Between one and unlimited times, as many times as possible, giving back as needed (greedy)
\d    # Match a single digit 0..9
   +     # Between one and unlimited times, as many times as possible, giving back as needed (greedy)
Z     # Match the character “Z” literally
\b    # Assert position at a word boundary
"

By the way:

\b*

Should throw an exception. \b is a word anchor. You can't quantify it.

edited Apr 25 '14 at 16:05

Alan Moore

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answered Dec 01 '11 at 18:51

FailedDev

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Hi FailedDev - thanks for the suggestion but it's still not returning a Match.. Maybe you can spot a mistake - I now have: pattern = @"\b\d+Z\s+\d+Z\b"; (edit - removed astrix but still does not work) – keynesiancross Dec 01 '11 at 19:00
@keynesiancross **\d+*** this should also throw an exception. Remove the asterisk after the +. – FailedDev Dec 01 '11 at 19:02
Sorry - removed that astrix (see edit above) but still does not provide match... Also added edit to OP – keynesiancross Dec 01 '11 at 19:06
Turns out I'm just an idiot and was setting my breakpoint before the method runs.. You're code above works great. Thanks a lot – keynesiancross Dec 01 '11 at 19:11

score 1 · Answer 2 · answered Dec 01 '11 at 18:54

Try this code.

using System;
using System.Text.RegularExpressions;

namespace ConsoleApplication1
{
  class Program
  {
    static void Main(string[] args)
    {
      string txt="3Z 5Z";

      string re1="(\\d+)";  // Integer Number 1
      string re2="(Z)"; // Any Single Character 1
      string re3="( )"; // Any Single Character 2
      string re4="(\\d+)";  // Integer Number 2
      string re5="(Z)"; // Any Single Character 3

      Regex r = new Regex(re1+re2+re3+re4+re5,RegexOptions.IgnoreCase|RegexOptions.Singleline);
      Match m = r.Match(txt);
      if (m.Success)
      {
            String int1=m.Groups[1].ToString();
            String c1=m.Groups[2].ToString();
            String c2=m.Groups[3].ToString();
            String int2=m.Groups[4].ToString();
            String c3=m.Groups[5].ToString();
            Console.Write("("+int1.ToString()+")"+"("+c1.ToString()+")"+"("+c2.ToString()+")"+"("+int2.ToString()+")"+"("+c3.ToString()+")"+"\n");
      }
      Console.ReadLine();
    }
  }
}

FYI, the `Singleline` option has no effect here. All it does is allow the dot to match any character *including linefeed*, and there are no linefeeds in this regex. — Alan Moore, Apr 25 '14 at 16:13

score 1 · Answer 3 · answered Dec 01 '11 at 18:59

1

I addition to other posts I would add characters of the Begin and End of string.

patter = "^\d+Z\s\d+Z$"

answered Dec 01 '11 at 18:59

Uriil

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True but OP had word anchors, so probably this string is part of a bigger string. – FailedDev Dec 01 '11 at 19:01

Regular Expressions + Including one space in pattern

3 Answers3