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

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

其他回答

这一个结合了许多给出的答案。

var randNo=Math.floor(Math.random()*100)+2+“”+new Date().getTime()+Math.floof(Math.rrandom()*100)+2+(Math.rand().toString(36).replace(/[^a-zA-Z]+/g,'').substr(0,5));console.log(randNo);

我用了一个月,效果很好。

正如这里的几个人所指出的,将Math.random()的结果直接传递给.string(36)有几个问题。

它的随机性很差。生成的字符数量各不相同,平均而言取决于Javascript中浮点数如何工作的棘手细节。如果我试图生成11个或更少的字符,但不能生成超过11个字符,这似乎是有效的。而且它不灵活。允许或禁止某些字符是不容易的。

对于任何使用lodash的人,我有一个紧凑的解决方案,它没有这些问题:

_.range(11).map(i => _.sample("abcdefghijklmnopqrstuvwxyz0123456789")).join('')

如果要允许某些字符(例如大写字母)或禁止某些字符(如l和1等不明确的字符),请修改上面的字符串。

我已经制作了一个字符串原型,它可以生成一个给定长度的随机字符串。

如果你想要特殊字符,你也可以解密,你可以避免一些。

/**
 * STRING PROTOTYPE RANDOM GENERATOR
 * Used to generate a random string
 * @param {Boolean} specialChars
 * @param {Number} length
 * @param {String} avoidChars
 */
String.prototype.randomGenerator = function (specialChars = false, length = 1, avoidChars = '') {
    let _pattern = 'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789';
    _pattern += specialChars === true ? '(){}[]+-*/=' : '';
    if (avoidChars && avoidChars.length) {
        for (let char of avoidChars) {
            _pattern = _pattern.replace(char, '');
        }
    }
    let _random = '';
    for (let element of new Array(parseInt(length))) {
        _random += _pattern.charAt(Math.floor(Math.random() * _pattern.length));
    }
    return _random;
};

您可以这样使用:

// Generate password with specialChars which contains 10 chars and avoid iIlL chars
var password = String().randomGenerator(true, 10, 'iIlL');

希望有帮助。

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)
}

这是doubletap答案的稍微改进版本。当Math.random()返回0、0.5、0.25、0.125等时,它会考虑gertas对案例的评论。

((Math.random()+3*Number.MIN_VALUE)/Math.PI).toString(36).slice(-5)

它防止零被传递给toString,我将最小的浮点值添加到Math.random()。它确保传递给toString的数字有足够的数字,通过除以一个几乎不合理的数字。