基于加密 API 的 2017-06-28 的 Broofa 的 TypeScript 版本:

function genUUID() {
    // Reference: https://stackoverflow.com/a/2117523/709884
    return ("10000000-1000-4000-8000-100000000000").replace(/[018]/g, s => {
        const c = Number.parseInt(s, 10)
        return (c ^ crypto.getRandomValues(new Uint8Array(1))[0] & 15 >> c / 4).toString(16)
    })
}

原因:

使用 + 之间的数字( )和数字不有效 从行到数字的转换必须是明确的

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);
        }
    );
};

这也适用于Node.js,如果您更换了Let Buffer = 新 Uint8Array(); crypto.getRandomValues 与Let Buffer = crypto.randomBytes(16)

它应该在性能中打击最常见的表达解决方案。

const hex = '0123456789ABCDEF' let generateToken = function() { let buffer = new Uint8Array(16) crypto.getRandomValues(buffer) buffer[6] = 0x40 (buffer[6] & 0xF) buffer[8] = 0x80 (buffer[8] & 0xF) let segments = [] for (let i = 0; i < 16; ++i) { segments.push(hex[(buffer[i] >> 4 & 0xF)]) segments.push(hex[(buffer[i] >> 0 & 0xF)])) if (i ==

表演图(每个人都喜欢它们):JSbench

在这里,你可以找到一个非常小的功能,产生UUID。

最后一个版本是:

function b(
  a                  // Placeholder
){
  var cryptoObj = window.crypto || window.msCrypto; // For Internet Explorer 11
  return a           // If the placeholder was passed, return
    ? (              // a random number from 0 to 15
      a ^            // unless b is 8,
      cryptoObj.getRandomValues(new Uint8Array(1))[0]  // in which case
      % 16           // a random number from
      >> a/4         // 8 to 11
      ).toString(16) // in hexadecimal
    : (              // or otherwise a concatenated string:
      [1e7] +        // 10000000 +
      -1e3 +         // -1000 +
      -4e3 +         // -4000 +
      -8e3 +         // -80000000 +
      -1e11          // -100000000000,
      ).replace(     // Replacing
        /[018]/g,    // zeroes, ones, and eights with
        b            // random hex digits
      )
}

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。

基于布罗法的作品,我添加了一些更多的随机性,通过添加时间表到 math.random():

function uuidv4() {
    return 'xxxxxxxx-xxxx-4xxx-yxxx-xxxxxxxxxxxx'.replace(/[xy]/g, function (c) {
        var r = parseFloat('0.' + Math.random().toString().replace('0.', '') + new Date().getTime()) * 16 | 0,
            v = c == 'x' ? r : (r & 0x3 | 0x8);
        return v.toString(16);
    });
}