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?
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
随机unicode字符串
此方法将返回一个随机字符串,其中包含任何受支持的unicode字符,这不是OP要求的100%,而是我想要的:
function randomUnicodeString(length){
return Array.from({length: length}, ()=>{
return String.fromCharCode(Math.floor(Math.random() * (65536)))
}).join('')
}
根本原因
这是谷歌搜索“随机字符串javascript”时的最高结果,但OP只要求a-zA-Z0-9。
其他回答
如果您无法键入字符集,使用String.fromCharCode和范围内的Math.random可以在任何Unicode代码点范围内创建随机字符串。例如,如果您想要17个随机藏文字符,可以输入ranstr(17,0xf00,0xfff),其中(0xf00,0xff)对应于藏文Unicode块。在我的实现中,如果不指定代码点范围,生成器将输出ASCII文本。函数ranchar(a,b){a=(a==未定义?0:a);b=(b===未定义?127:b);return String.fromCharCode(Math.floor(Math.random()*(b-a)+a));}函数transtr(len,a,b){a=a||32;var结果=“”;对于(var i=0;i<len;i++){结果+=ranchar(a,b)}返回结果;}//以下是随机Unicode块的一些示例console.log('拉丁语基本块:'+transtr(10,0x000,0x007f))console.log('拉丁语-1增补块:'+transtr(10,0x080,0x0ff))console.log('货币符号块中:'+transtr(10,0x20a0,0x20cf))console.log('在类字母符号块中:'+transtr(10,0x2100,0x214f))console.log('在Dingbats块中:'+transtr(10,0x2700,0x27bf))
这是我用过的。这是一对夫妇的组合。我在循环中使用它,它生成的每个ID都是唯一的。它可能不是5个字符,但它保证是唯一的。
var newId =
"randomid_" +
(Math.random() / +new Date()).toString(36).replace(/[^a-z]+/g, '');
带有es6排列运算符的更新版本:
[…数组(30)].map(()=>Math.random().toString(36)[2]).join(“”)
30是一个任意数字,您可以选择所需的任何令牌长度36是可以传递给numeric.toString()的最大基数,表示所有数字和a-z小写字母2用于从如下所示的随机字符串中选择第三个索引:“0.mfbiohx64i”,我们可以取0之后的任何索引。
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。