c# parse and replace content in string

Question

This is my first attempt to use regular expression.

What I want to archieve is to convert this string:

" <Control1 x:Uid="1"  />

  <Control2 x:Uid="2"  /> "

to

" <Control1 {1}  />

  <Control2 {2}  /> "

Basically, convert x:Uid="n" to {n}, where n represents an integer.

What I thought it would work (of course it doesn't) is like this:

  string input = " <Control1 x:Uid="1"  />
                   <Control2 x:Uid="2"  /> ";

  string pattern = "\b[x:Uid=\"[\d]\"]\w+";
  string replacement = "{}";
  Regex rgx = new Regex(pattern);
  string result = rgx.Replace(input, replacement);

Or

  Regex.Replace(input, pattern, delegate(Match match)
  {
       // do something here
       return result
  });

I'm struggling to define the pattern and replacement string. I'm not sure if I'm in the right direction to solve the problem.

You may ommit the \b and \w as well as the braces around x:Uid. Try this instead: x:Uid=\"(\d)\" in order to retrieve the integers and then set replacement to {$1} — MakePeaceGreatAgain
– MakePeaceGreatAgain, Commented Oct 17, 2013 at 11:26
Just to make sure: Uids can only be exactly one digit long? Because \d matches exactly one digit. — Tim Pietzcker
– Tim Pietzcker, Commented Oct 17, 2013 at 11:29

Tim Pietzcker · Accepted Answer · 2013-10-17 11:26:06Z

3

Square brackets define a character class, which you don't want here. Instead you want to use a capturing group:

string pattern = @"\bx:Uid=""(\d)""";
string replacement = "{$1}";

Note the use of a verbatim string to make sure the \b is interpreted as a word boundary anchor instead of a backspace character.

answered Oct 17, 2013 at 11:26

Tim Pietzcker

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5 Comments

MakePeaceGreatAgain Over a year ago

You´ll have to escape the " within your regex by using \"

Hans Keﬆing Over a year ago

@HimBromBeere - he already escaped it, by using "" (in a verbatim string, preceded by that @)

flindeberg Over a year ago

Shouldn't \d+ be used instead? Even thought the author did not state so I have a feeling it might be more than one digit in the real implementation :)

Tim Pietzcker Over a year ago

@flindeberg: Thus my comment/question above.

flindeberg Over a year ago

@TimPietzcker I totally missed that :)

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