我使用这个版本. 它是安全和简单的. 它不是为了产生格式化的Uids,它只是为了产生你需要的电缆的随机线条。

export function makeId(length) {
  let result = '';
  const characters = 'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789';
  const charactersLength = characters.length;

  for (let i = 0; i < length; i++) {
    let letterPos = parseInt(crypto.getRandomValues(new Uint8Array(1))[0] / 255 * charactersLength - 1, 10)
    result += characters[letterPos]
  }
  return result;
}

[编辑 2021-10-16 以反映 RFC4122 符合 UUID 的最新最佳做法]

大多数读者在这里会想使用 uuid 模块,它已被测试和支持。

如果其中没有一个为您工作,则有这个方法(基于这个问题的原始答案):函数 uuidv4() {返回([1e7]+-1e3+-4e3+-8e3+-1e11)。代替(/[018]/g,c =>(c ^ crypto.getRandomValues(新Uint8Array(1))[0] & 15 >> c / 4).toString(16) ; } console.log(uuidv4());

注意:依靠 Math.random() 的任何 UUID 发电机的使用是由于这里最好的解释而受到强烈的拒绝(包括此答案的以前版本中显示的剪辑) TL;DR:基于 Math.random() 的解决方案不提供良好的独特性保证。

对于我的使用案例,我需要 ID 世代,保证在全球范围内是独一无二的;没有例外.我与问题斗争了一段时间,并提出了一个名为 TUID (真正独一无二的 ID ) 的解决方案。

  // RFC 4122
  //
  // A UUID is 128 bits long
  //
  // String representation is five fields of 4, 2, 2, 2, and 6 bytes.
  // Fields represented as lowercase, zero-filled, hexadecimal strings, and
  // are separated by dash characters
  //
  // A version 4 UUID is generated by setting all but six bits to randomly
  // chosen values
  var uuid = [
    Math.random().toString(16).slice(2, 10),
    Math.random().toString(16).slice(2, 6),

    // Set the four most significant bits (bits 12 through 15) of the
    // time_hi_and_version field to the 4-bit version number from Section
    // 4.1.3
    (Math.random() * .0625 /* 0x.1 */ + .25 /* 0x.4 */).toString(16).slice(2, 6),

    // Set the two most significant bits (bits 6 and 7) of the
    // clock_seq_hi_and_reserved to zero and one, respectively
    (Math.random() * .25 /* 0x.4 */ + .5 /* 0x.8 */).toString(16).slice(2, 6),

    Math.random().toString(16).slice(2, 14)].join('-');

这个应用程序的100万次执行只需要32.5秒,这是我从未在浏览器中看到的最快(唯一的解决方案没有漏洞/漏洞)。

/**
 * Generates a GUID string.
 * @returns {string} The generated GUID.
 * @example af8a8416-6e18-a307-bd9c-f2c947bbb3aa
 * @author Slavik Meltser.
 * @link http://slavik.meltser.info/?p=142
 */
function guid() {
    function _p8(s) {
        var p = (Math.random().toString(16)+"000000000").substr(2,8);
        return s ? "-" + p.substr(0,4) + "-" + p.substr(4,4) : p ;
    }
    return _p8() + _p8(true) + _p8(true) + _p8();
}

console.time('t');
for (var i = 0; i < 10000000; i++) {
    guid();
};
console.timeEnd('t');

算法:

Math.random() 函数返回 0 和 1 之间的十数数,在十数分数点(例如 0.4363923368509859 )之后有 16 个数字。 然后我们采取这个数字并将其转换为 16 个基线(从上面的示例中,我们将获得 0.6fb7687f)。 Math.random(.toString(16 ) 然后我们切断了 0 个预定(.0.6fb7687f => 6fb7687f ) 并获得八个六分数字符的序列。

链接到我的博客上的这篇文章

享受吧:)

我想了解布罗法的答案,所以我扩展了它并添加了评论:

var uuid = function () {
    return 'xxxxxxxx-xxxx-4xxx-yxxx-xxxxxxxxxxxx'.replace(
        /[xy]/g,
        function (match) {
            /*
            * Create a random nibble. The two clever bits of this code:
            *
            * - Bitwise operations will truncate floating point numbers
            * - For a bitwise OR of any x, x | 0 = x
            *
            * So:
            *
            * Math.random * 16
            *
            * creates a random floating point number
            * between 0 (inclusive) and 16 (exclusive) and
            *
            * | 0
            *
            * truncates the floating point number into an integer.
            */
            var randomNibble = Math.random() * 16 | 0;

            /*
            * Resolves the variant field. If the variant field (delineated
            * as y in the initial string) is matched, the nibble must
            * match the mask (where x is a do-not-care bit):
            *
            * 10xx
            *
            * This is achieved by performing the following operations in
            * sequence (where x is an intermediate result):
            *
            * - x & 0x3, which is equivalent to x % 3
            * - x | 0x8, which is equivalent to x + 8
            *
            * This results in a nibble between 8 inclusive and 11 exclusive,
            * (or 1000 and 1011 in binary), all of which satisfy the variant
            * field mask above.
            */
            var nibble = (match == 'y') ?
                (randomNibble & 0x3 | 0x8) :
                randomNibble;

            /*
            * Ensure the nibble integer is encoded as base 16 (hexadecimal).
            */
            return nibble.toString(16);
        }
    );
};