Replace chars if not match
Asked Answered
H

2

51

I'm looking for a regular expression that removes illegal characters. But I don't know what the characters will be.

For example:

In a process, I want my string to match ([a-zA-Z0-9/-]*). So I would like to replace all characters that don't match the regexp above.

Halfpint answered 16/12, 2010 at 11:29 Comment(1)
Possible duplicate of #3847794Sheldon
K
91

That would be:

[^a-zA-Z0-9/-]+

[^ ] at the start of a character class negates it - it matches characters not in the class.

See also: Character Classes

Kex answered 16/12, 2010 at 11:31 Comment(0)
S
0

Thanks to Kobi's answer I've created a helper method to strips unaccepted characters .

The allowed pattern should be in Regex format, expect them wrapped in square brackets. A function will insert a tilde after opening squere bracket. I anticipate that it could work not for all RegEx describing valid characters sets,but it works for relatively simple sets, that we are using.

/// <summary>
/// Replaces  not expected characters.
/// </summary>
/// <param name="text"> The text.</param>
/// <param name="allowedPattern"> The allowed pattern in Regex format, expect them wrapped in brackets</param>
/// <param name="replacement"> The replacement.</param>
/// <returns></returns>
/// //        https://mcmap.net/q/345888/-replace-chars-if-not-match.
//https://mcmap.net/q/354899/-replace-remove-characters-that-do-not-match-the-regular-expression-net
//[^ ] at the start of a character class negates it - it matches characters not in the class.
//Replace/Remove characters that do not match the Regular Expression
static public string ReplaceNotExpectedCharacters( this string text, string allowedPattern,string replacement )
{
  allowedPattern = allowedPattern.StripBrackets( "[", "]" );
  //[^ ] at the start of a character class negates it - it matches characters not in the class.
  var result = Regex.Replace(text, @"[^" + allowedPattern + "]", replacement);
  return result; //returns result free of negated chars
}
Sheldon answered 28/10, 2012 at 2:54 Comment(0)

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