Fastest method to replace all instances of a chara

Question

What is the fastest way to replace all instances of a string/character in a string in JavaScript? A while, a for-loop, a regular expression?

Answer 1

@Gumbo adding extra answer - user.email.replace(/foo/gi,"bar");

/foo/g - Refers to the all string to replace matching the case sensitive

/foo/gi - Refers to the without case sensitive and replace all For Eg: (Foo, foo, FoO, fOO)

DEMO

Answer 2

You can use the following:

newStr = str.replace(/[^a-z0-9]/gi, '_');

or

newStr = str.replace(/[^a-zA-Z0-9]/g, '_');

This is going to replace all the character that are not letter or numbers to ('_'). Simple change the underscore value for whatever you want to replace it.

Answer 3

I tried a number of these suggestions after realizing that an implementation I had written of this probably close to 10 years ago actually didn't work completely (nasty production bug in an long-forgotten system, isn't that always the way?!)... what I noticed is that the ones I tried (I didn't try them all) had the same problem as mine, that is, they wouldn't replace EVERY occurrence, only the first, at least for my test case of getting "test....txt" down to "test.txt" by replacing ".." with "."... maybe I missed so regex situation? But I digress...

So, I rewrote my implementation as follows. It's pretty darned simple, although I suspect not the fastest but I also don't think the difference will matter with modern JS engines, unless you're doing this inside a tight loop of course, but that's always the case for anything...

function replaceSubstring(inSource, inToReplace, inReplaceWith) {

  var outString = inSource;
  while (true) {
    var idx = outString.indexOf(inToReplace);
    if (idx == -1) {
      break;
    }
    outString = outString.substring(0, idx) + inReplaceWith +
      outString.substring(idx + inToReplace.length);
  }
  return outString;

}

Hope that helps someone!

Answer 4

Use the replace() method of the String object.

As mentioned in the selected answer, the /g flag should be used in the regex, in order to replace all instances of the substring in the string.

Answer 5

// Find, Replace, Case
// i.e "Test to see if this works? (Yes|No)".replaceAll('(Yes|No)', 'Yes!');
// i.e.2 "Test to see if this works? (Yes|No)".replaceAll('(yes|no)', 'Yes!', true);
String.prototype.replaceAll = function(_f, _r, _c){ 

  var o = this.toString();
  var r = '';
  var s = o;
  var b = 0;
  var e = -1;
  if(_c){ _f = _f.toLowerCase(); s = o.toLowerCase(); }

  while((e=s.indexOf(_f)) > -1)
  {
    r += o.substring(b, b+e) + _r;
    s = s.substring(e+_f.length, s.length);
    b += e+_f.length;
  }

  // Add Leftover
  if(s.length>0){ r+=o.substring(o.length-s.length, o.length); }

  // Return New String
  return r;
};

Answer 6

I think the real answer is that it completely depends on what your inputs look like. I created a JsFiddle to try a bunch of these and a couple of my own against various inputs. No matter how I look at the results, I see no clear winner.

• RegExp wasn't the fastest in any of the test cases, but it wasn't bad either.
• Split/Join approach seems fastest for sparse replacements.

• This one I wrote seems fastest for small inputs and dense replacements:

  function replaceAllOneCharAtATime(inSource, inToReplace, inReplaceWith) {
      var output="";
      var firstReplaceCompareCharacter = inToReplace.charAt(0);
      var sourceLength = inSource.length;
      var replaceLengthMinusOne = inToReplace.length - 1;
      for(var i = 0; i < sourceLength; i++){
          var currentCharacter = inSource.charAt(i);
          var compareIndex = i;
          var replaceIndex = 0;
          var sourceCompareCharacter = currentCharacter;
          var replaceCompareCharacter = firstReplaceCompareCharacter;
          while(true){
              if(sourceCompareCharacter != replaceCompareCharacter){
              output += currentCharacter;
              break;
          }
          if(replaceIndex >= replaceLengthMinusOne) {
              i+=replaceLengthMinusOne;
              output += inReplaceWith;
              //was a match
              break;
          }
          compareIndex++; replaceIndex++;
          if(i >= sourceLength){
              // not a match
              break;
          }
          sourceCompareCharacter = inSource.charAt(compareIndex)
              replaceCompareCharacter = inToReplace.charAt(replaceIndex);
          }   
          replaceCompareCharacter += currentCharacter;
      }
      return output;
  }