替换单词数组

您可以这样做：

var x = 'The old car is just whatever';
var arr = [{ old: /old/gi, new: 'new' },
           { old: /whatever/gi, new: 'whatevernew' },
           { old: /car/gi, new: 'boat' }];
for (var ii = 0; ii < arr.length; ii++) {
    x = x.replace(arr[ii].old, arr[ii].new);
}

lbu和RobG使用回调提供了一种有趣的方法。这里有一个更通用的版本，其中有一个函数，它只将要替换的数据结构和要替换的内容作为参数。

function multiReplace(str, params) {
    var regStr = "";
    for (var i in params) {
        regStr += "(" + i + ")|";    // build regEx string
    }
    regStr = regStr.slice(0, -1);   // remove extra trailing |
    return(str.replace(new RegExp(regStr, "gi"), function(a) {
        return(params[a]);
    }));
}

var test = 'This old car is just whatever and really old';
var replaceParam = {"old": "new", "whatever": "something", "car": "boat"};
var result = multiReplace(test, replaceParam);
alert(result);

还有一把小提琴在演奏：http://jsfiddle.net/jfriend00/p8wKH/

试试这个：

var regexes = { 'new': /old/gi, 'whatevernew': /whatever/gi, 'boat': /car/gi };
$.each(regexes, function(newone, regex){
  x = x.replace(regex, newone);
});

或者这个：

var regexes = { 'old':'new', 'whatever':'whatevernew', 'car':'boat'};
$.each(regexes, function(oldone, newone){
  x = x.replace(new RegExp(oldone, 'gi'), newone);
});

这里还有几个：

// Multi RegExp version
var replaceSeveral = (function() {
  var data = {old:'new', whatever: 'whatevernew', car: 'boat'};
  return function(s) {
    var re;
    for (var p in data) {
      if (data.hasOwnProperty(p)) {
        re = new RegExp('(^|''b)' + p + '(''b|$)','ig');
        s = s.replace(re, '$1' + data[p] + '$2');
      }
    }
    return s;
  }
}());
// Replace function version
var comparitor = (function() {
  var data = {old:'new', whatever: 'whatevernew', car: 'boat'};
  return function (word) {
    return data.hasOwnProperty(word.toLowerCase())? data[word] : word;
  }
}());

var s = 'old this old whatever is a car';
alert(
  s 
  + ''n' + replaceSeveral(s)
  + ''n' + s.replace(/'w+/ig, comparitor)
);

Ibu的回调解决方案非常干净，但可以进一步简化：

x = x.replace(/'b(?:old|whatever|car)'b/gi,
        function (m0) {
            return {'old': 'new',
                    'car': 'boat',
                    'whatever': 'something'}[m0];
        });

这种使用对象文字的技术非常有效。我修改了正则表达式，通过添加单词边界只匹配整个单词(以避免将gold更改为gnew等(。

编辑：经过仔细检查，我发现jfriend00的解决方案使用了相同的技术(并且被推广为更有用(。

replace方法支持使用回调函数：

var x = 'This old car is just whatever';
var y = x.replace(/(old)|(whatever)|(car)/gi,function (a) {
    var str = "";
    switch(a){
      case "old":
         str = "new";
         break;
      case "whatever":
         str = "something";
         break;
      case "car":
         str = "boat";
         break;
      default: 
         str= "";
    }
    return str;
});
alert(y);
// Y will print "This new car is just something"