我想要一个5个字符的字符串,由从集合[a-zA-Z0-9]中随机选取的字符组成。

用JavaScript实现这一点的最佳方法是什么?


当前回答

将字符作为thisArg放在map函数中会创建一个“单行”:

Array.apply(null, Array(5))
.map(function(){ 
    return this[Math.floor(Math.random()*this.length)];
}, "abcdefghijklmnopqrstuvwxyz")
.join('');

其他回答

function generate(length) {
  var letters = ["a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z","A","B","C","D","E","F","G","H","I","J","K","L","M","N","O","P","Q","R","S","T","U","V","W","X","Y","Z","0","1","2","3","4","5","6","7","8","9"];
  var IDtext = "";
  var i = 0;
  while (i < length) {
    var letterIndex = Math.floor(Math.random() * letters.length);
    var letter = letters[letterIndex];
    IDtext = IDtext + letter;
    i++;
  }
  console.log(IDtext)
}

我没有找到支持小写和大写字符的干净解决方案。

仅小写支持:

Math.random().toString(36).substr(2,5)

基于该解决方案,支持小写和大写:

Math.random().toString(36).substr(2,5).split(“”).map(c=>Math.randm()<0.5?c.toUpperCase():c).jjoin(“”);

更改substr(2,5)中的5以调整到所需的长度。

下面这个怎么样。。。这将产生真正随机的值:

function getRandomStrings(length) {
  const value = 'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789';
  const randoms = [];
  for(let i=0; i < length; i++) {
     randoms.push(value[Math.floor(Math.random()*value.length)]);
  }
  return randoms.join('');
}

但如果您在ES6中寻找一个较短的语法:

const getRandomStrings = length => Math.random().toString(36).substr(-length);

这个小巧的小把戏怎么样?

var possible = 'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789';
var stringLength = 5;

function pickRandom() {
    return possible[Math.floor(Math.random() * possible.length)];
}

var randomString = Array.apply(null, Array(stringLength)).map(pickRandom).join('');

您需要Array.apply来将空数组变为未定义的数组。

如果您正在为ES2015进行编码,那么构建阵列会简单一些:

var randomString = Array.from({ length: stringLength }, pickRandom).join('');

这将在变量c中存储5个字母数字字符。

for(var c = ''; c.length < 5;) c += Math.random().toString(36).substr(2, 1)