一个取长度的简单函数

getRandomToken(len: number): string {
  return Math.random().toString(36).substr(2, len);
}

如果你传递6，它会生成6位字母数字

很好，很简单，而且不局限于一定数量的字符:

let len = 20, str = "";
while(str.length < len) str += Math.random().toString(36).substr(2);
str = str.substr(0, len);

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;
};

beans建议的另一种答案变体

(Math.random()*1e32).toString(36)

通过改变乘数1e32，你可以改变随机字符串的长度。

这是一个简单的代码来生成随机字符串字母。 看看这段代码是如何工作的。 去(lenthOfStringToPrint);-使用此函数生成最终字符串。

var letters = {
  1: ["q","w","e","r","t","y","u","i","o","p","a","s","d","f","g","h","j","k","l","z","x","c","v","b","n","m"],
  2: ["Q","W","E","R","T","Y","U","I","O","P","A","S","D","F","G","H","J","K","L","Z","X","C","V","B","N","M"]
},i,letter,final="";
random = (max,min) => {
  return Math.floor(Math.random()*(max-min+1)+min);
}
function go(length) {
  final="",letter="";
  for (i=1; i<=length; i++){
    letter = letters[random(0,3)][random(0,25)];
    final+=letter;
  }
  return final;
}

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