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。

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; }

var mystring = 'This is a string';
var newString = mystring.replace(/i/g, "a");

newString现在是'Thas as a strange '

像这样使用正则表达式对象

var regex = new RegExp(' ' '， 'g'); STR = STR .replace(regex， '\ ");

它将取代所有出现的“into”。

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");