我使用@Nimphious优秀的第二种方法，发现偶尔返回的字符串是数字-而不是字母数字。 我使用的解决方案是测试使用!isNaN，并再次使用递归调用该函数。 何苦呢?我使用这个函数来创建对象键，如果所有的键都是字母数字，那么所有的键都可以正常排序，但如果你使用 数字作为键与字母数字(字符串)混合在一起循环遍历对象将产生与原始顺序不同的顺序。

function newRandomString(length, chars) {
  var mask = '';
  if (chars.indexOf('a') > -1) mask += 'abcdefghijklmnopqrstuvwxyz';
  if (chars.indexOf('A') > -1) mask += 'ABCDEFGHIJKLMNOPQRSTUVWXYZ';
  if (chars.indexOf('#') > -1) mask += '0123456789';
  if (chars.indexOf('$') > -1) mask += '0123456789';

  var result = '';
  for (var i = length; i > 0; --i) result += mask[Math.floor(Math.random() * 
  mask.length)];
  /*    
        we need a string not a number !isNaN(result)) will return true if '1234' or '3E77'
        because if we're looping through object keys (created by newRandomString()) and 
        a number is used and all the other keys are strings then the number will 
        be first even if it was the 2nd or third key in object
  */
  //use recursion to try again
  if(!isNaN(result)){
    console.log('found a number....:'+result);
    return newRandomString(length, chars)
  }else{
    return result;
  }
};

var i=0;
while (i < 1000) {
  var a = newRandomString(4, '#$aA');
  console.log(i+' - '+a);
  //now we're using recursion this won't occur
  if(!isNaN(a)){
    console.log('=============='+i+' - '+a);
  }
  i++;
}

console.log('3E77:'+!isNaN('3E77'));//true
console.log('1234:'+!isNaN('1234'));//true
console.log('ab34:'+!isNaN('ab34'));//false

当我看到这个问题时，我想到了我必须生成uuid的时候。我不能把代码的功劳，因为我确信我在stackoverflow上找到了它。如果你不想在字符串中使用破折号，那就去掉破折号。函数如下:

function generateUUID() {
    var d = new Date().getTime();
    var uuid = 'xxxxxxxx-xxxx-4xxx-yxxx-xxxxxxxxxxxx'.replace(/[xy]/g,function(c) {
        var r = (d + Math.random()*16)%16 | 0;
        d = Math.floor(d/16);
        return (c=='x' ? r : (r&0x7|0x8)).toString(16);
    });
    return uuid.toUpperCase();
}

小提琴:http://jsfiddle.net/nlviands/fNPvf/11227/

我只是发现了一个非常好的优雅的解决方案:

Math.random().toString(36).slice(2)

这个实现的注意事项:

This will produce a string anywhere between zero and 12 characters long, usually 11 characters, due to the fact that floating point stringification removes trailing zeros. It won't generate capital letters, only lower-case and numbers. Because the randomness comes from Math.random(), the output may be predictable and therefore not necessarily unique. Even assuming an ideal implementation, the output has at most 52 bits of entropy, which means you can expect a duplicate after around 70M strings generated.

var randomString = function(length) {
  var str = '';
  var chars ='0123456789ABCDEFGHIJKLMNOPQRSTUVWXTZabcdefghiklmnopqrstuvwxyz'.split(
      '');
  var charsLen = chars.length;
  if (!length) {
    length = ~~(Math.random() * charsLen);
  }
  for (var i = 0; i < length; i++) {
    str += chars[~~(Math.random() * charsLen)];
  }
  return str;
};

如果你只想允许特定的字符，你也可以这样做:

function randomString(length, chars) {
    var result = '';
    for (var i = length; i > 0; --i) result += chars[Math.floor(Math.random() * chars.length)];
    return result;
}
var rString = randomString(32, '0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ');

下面是一个演示的jsfiddle: http://jsfiddle.net/wSQBx/

另一种方法是使用一个特殊的字符串，告诉函数使用什么类型的字符。你可以这样做:

function randomString(length, chars) {
    var mask = '';
    if (chars.indexOf('a') > -1) mask += 'abcdefghijklmnopqrstuvwxyz';
    if (chars.indexOf('A') > -1) mask += 'ABCDEFGHIJKLMNOPQRSTUVWXYZ';
    if (chars.indexOf('#') > -1) mask += '0123456789';
    if (chars.indexOf('!') > -1) mask += '~`!@#$%^&*()_+-={}[]:";\'<>?,./|\\';
    var result = '';
    for (var i = length; i > 0; --i) result += mask[Math.floor(Math.random() * mask.length)];
    return result;
}

console.log(randomString(16, 'aA'));
console.log(randomString(32, '#aA'));
console.log(randomString(64, '#A!'));

小提琴:http://jsfiddle.net/wSQBx/2/

或者，像下面描述的那样使用base36方法，你可以这样做:

function randomString(length) {
    return Math.round((Math.pow(36, length + 1) - Math.random() * Math.pow(36, length))).toString(36).slice(1);
}

32个字符:

for(var c = ''; c.length < 32;) c += Math.random().toString(36).substr(2, 1)