我不知道最快的是什么，但我知道最易读的是什么——最短最简单的是什么。即使它比其他溶液慢一点，也值得使用。

所以使用:

 "string".replace("a", "b");
 "string".replace(/abc?/g, "def");

享受好的代码而不是更快的速度(嗯…1/100000秒不是差)和丑陋的差。；）

仅从速度问题考虑，我相信上面链接中提供的区分大小写的示例将是目前为止最快的解决方案。

var token = "\r\n";
var newToken = " ";
var oldStr = "This is a test\r\nof the emergency broadcasting\r\nsystem.";
newStr = oldStr.split(token).join(newToken);

newStr是 “这是对紧急广播系统的一次测试。”

使用String对象的replace()方法。

正如所选答案中提到的，应该在正则表达式中使用/g标志，以便替换字符串中子字符串的所有实例。

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...

因此，我重写了我的实现如下。这是非常简单的，虽然我怀疑不是最快的，但我也不认为现代JS引擎的区别是重要的，除非你在一个紧密的循环中做这件事，但这总是任何事情的情况…

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;

}

希望这能帮助到别人!

// 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;
};

我认为真正的答案是，这完全取决于你的输入是什么样的。我创建了一个JsFiddle来尝试这些方法，并针对各种输入创建了自己的JsFiddle。无论我如何看待这些结果，我都看不到明显的赢家。

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; }