使用 Blobs 的单行解决方案。

window.URL.createObjectURL(new Blob([])).substring(31);

最终值(31)取决于URL的长度。

编辑:

像Rinogo所建议的那样,一个更小而普遍的解决方案:

URL.createObjectURL(new Blob([])).substr(-36);

以下是基于 RFC 4122 的某些代码,第 4.4 节(从真实随机或偏随机号创建 UUID 的算法)。

function createUUID() {
    // http://www.ietf.org/rfc/rfc4122.txt
    var s = [];
    var hexDigits = "0123456789abcdef";
    for (var i = 0; i < 36; i++) {
        s[i] = hexDigits.substr(Math.floor(Math.random() * 0x10), 1);
    }
    s[14] = "4";  // bits 12-15 of the time_hi_and_version field to 0010
    s[19] = hexDigits.substr((s[19] & 0x3) | 0x8, 1);  // bits 6-7 of the clock_seq_hi_and_reserved to 01
    s[8] = s[13] = s[18] = s[23] = "-";

    var uuid = s.join("");
    return uuid;
}

这创建了一个版本 4 UUID (创建于假冒随机数字):

function uuid()
{
   var chars = '0123456789abcdef'.split('');

   var uuid = [], rnd = Math.random, r;
   uuid[8] = uuid[13] = uuid[18] = uuid[23] = '-';
   uuid[14] = '4'; // version 4

   for (var i = 0; i < 36; i++)
   {
      if (!uuid[i])
      {
         r = 0 | rnd()*16;

         uuid[i] = chars[(i == 19) ? (r & 0x3) | 0x8 : r & 0xf];
      }
   }

   return uuid.join('');
}

下面是产生的 UUID 的样本:

682db637-0f31-4847-9cdf-25ba9613a75c
97d19478-3ab2-4aa1-b8cc-a1c3540f54aa
2eed04c9-2692-456d-a0fd-51012f947136

您可以使用 npm 包指南、GUID 发电机和验证器。

例子:

Guid.raw();
// -> '6fdf6ffc-ed77-94fa-407e-a7b86ed9e59d'

注意: 此包已被削弱. 使用 uuid 相反。

例子:

const uuidv4 = require('uuid/v4');
uuidv4(); // ⇨ '10ba038e-48da-487b-96e8-8d3b99b6d18a'

简单的JavaScript作为这个问题的最佳答案的组合。

var crypto = window.crypto || window.msCrypto || null; // IE11 fix var Guid = Guid || (function() { var EMPTY = '00000000-0000-0000-0000-000000000000'; var _padLeft = function(paddingString, width, replacementChar) { return paddingString.length >= width ? paddingString : _padLeft(replacementChar + paddingString, width, replacementChar || ' '); }; var _s4 = function(number) { var hexadecimalResult = number.toString(16); return _padLeft(hexadecimalResult, 4, '0'); }; var _cryptoGuid = function() { var buffer = new window.Uint16Array(8); crypto.getRandomValues(buffer); return [_s4(buffer[0]) + _s4(buffer[1]), _s4(buffer[2]), _s4(buffer[3]), _s4(buffer[4]), _s4(buffer[5]) + _s4(buffer[6]) + _s4(buffer[7])].join('-'); }; var _guid = function() { var currentDateMilliseconds = new Date().getTime(); return 'xxxxxxxx-xxxx-4xxx-yxxx-xxxxxxxxxxxx'.replace(/[xy]/g, function(currentChar) { var randomChar = (currentDateMilliseconds + Math.random() * 16) % 16 | 0; currentDateMilliseconds = Math.floor(currentDateMilliseconds / 16); return (currentChar === 'x' ? randomChar : (randomChar & 0x7 | 0x8)).toString(16); }); }; var create = function() { var hasCrypto = crypto != 'undefined' && crypto !== null, hasRandomValues = typeof(window.crypto.getRandomValues) != 'undefined'; return (hasCrypto && hasRandomValues) ? _cryptoGuid() : _guid(); }; return { newGuid: create, empty: EMPTY }; })(); // DEMO: Create and show GUID console.log('1. New Guid: ' + Guid.newGuid()); // DEMO: Show empty GUID console.log('2. Empty Guid: ' + Guid.empty);

使用:

主持人( )

“c6c2d12f-d76b-5739-e551-07e6de5b0807”

公平 公平

“00000000-0000-0000-0000-00000000”

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);
        }
    );
};