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

我喜欢doubletap的Math.random().toString(36).substring(7)答案的简洁,但并不是因为它有很多像hack-likecrack正确指出的冲突。它生成了11个字符字符串,但在100万个样本中,重复率为11%。

这里有一个更长(但仍然很短)、更慢的替代方案,在100万个样本空间中只有133个副本。在极少数情况下,字符串仍将短于11个字符:

Math.abs(Math.random().toString().split('')
    .reduce(function(p,c){return (p<<5)-p+c})).toString(36).substr(0,11);

其他回答

我知道每个人都已经做好了,但我想以最轻量级的方式(轻代码,而不是CPU)尝试一下:

函数rand(长度,电流){电流=电流?当前:“”;返回长度?rand(--length,“0123456789ABCDEFGHIJKLMNOPQRSTUVWXTZabcdefghiklmnopqrstuvxyz”.charAt(Math.floor(Math.random()*60))+电流):电流;}console.log(rand(5));

这需要一点时间来理解,但我认为它确实显示了javascript的语法是多么棒。

",,,,,".replace(/,/g,function (){return "AzByC0xDwEv9FuGt8HsIrJ7qKpLo6MnNmO5lPkQj4RiShT3gUfVe2WdXcY1bZa".charAt(Math.floor(Math.random()*62))});

生成安全的随机字母数字Base-62字符串:

函数生成UID(长度){return window.btoa(String.fromCharCode(…window.crypto.getRandomValues(新Uint8Array(长度*2))).replace(/[+/]/g,“”).substring(0,长度);}console.log(生成UID(22));//“yFg3Upv2cE9cKOXd7hHwWp”console.log(生成UID(5));//“YQGzP”

函数randomstring(L){var s=“”;var randomchar=函数(){var n=数学地板(Math.random()*62);如果(n<10)返回n//1-10如果(n<36)返回String.fromCharCode(n+55)//A-Z型return String.fromCharCode(n+61)//a-z型}而(s.length<L)s+=randomchar();返回s;}console.log(随机字符串(5));

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)
}