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。”

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; }

最简单的方法是使用带有g标志的正则表达式来替换所有实例:

str.replace(/foo/g, "bar")

这将把字符串str中出现的所有foo替换为bar。如果你只有一个字符串，你可以像这样将它转换为RegExp对象:

var pattern = "foobar",
    re = new RegExp(pattern, "g");

你可以使用以下方法:

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

or

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

这将替换所有不是字母或数字的字符('_')。简单地更改下划线值的任何你想替换它。

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。；）

// 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;
};