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);
        }
    );
};

使用:

let uniqueId = Date.now().toString(36) + Math.random().toString(36).substring(2);

document.getElementById(“独一无二”。innerHTML = Math.random().toString(36).substring(2) + (新日期()).getTime().toString(36); <div id="unique"> </div>

您可以通过添加更多数字来增加这个数字,但要创建100%独特的ID,您将需要使用全球计数器。

let u = Date.now().toString(16) + Math.random().toString(16) + '0'.repeat(16);
let guid = [u.substr(0,8), u.substr(8,4), '4000-8' + u.substr(13,3), u.substr(16,12)].join('-');

let u = Date.now().toString(16)+Math.random().toString(16)+'0'.repeat(16); let guid = [u.substr(0,8), u.substr(8,4), '4000-8' + u.substr(13,3), u.substr(16,12)].join('-'); document.getElementById("unique").innerHTML = guid; <div id="unique"> </div>

var uuid = function() {
    var buf = new Uint32Array(4);
    window.crypto.getRandomValues(buf);
    var idx = -1;
    return 'xxxxxxxx-xxxx-4xxx-yxxx-xxxxxxxxxxxx'.replace(/[xy]/g, function(c) {
        idx++;
        var r = (buf[idx>>3] >> ((idx%8)*4))&15;
        var v = c == 'x' ? r : (r&0x3|0x8);
        return v.toString(16);
    });
};

这个版本是基于Briguy37的答案和一些Bitwise运营商从泡沫中提取Nibble大小的窗户。

它应该遵守RFC类型4(随机)方案,因为我上次遇到麻烦与Java的UUID不符合的UUID。

我发现这个脚本有用于在JavaScript中创建GUID

HTTPS://github.com/addui/GUIDJS

var myGuid = GUID();

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”

                  Desktop   Android
           broofa: 1617ms   12869ms
               e1:  636ms    5778ms
               e2:  606ms    4754ms
               e3:  364ms    3003ms
               e4:  329ms    2015ms
               e5:  147ms    1156ms
               e6:  146ms    1035ms
               e7:  105ms     726ms
             guid:  962ms   10762ms
generateQuickGuid:  292ms    2961ms
  - Note: 500k iterations, results will vary by browser/CPU.

对于一个解释,让我们从Brofa的代码开始:

因此,它取代 x 与任何随机六分数数字, y 与随机数据(除了强迫顶部两个比特到 10 个 RFC 特征),而 regex 不匹配 - 或 4 个字符,所以他不需要处理它们。

下一个优化是另一个经典. 因为我们只处理四个字符串的输出,让我们将字符串的数量切成一半,并在每个字符串中处理八个字符串。 这是荒谬的,因为我们仍然必须处理RFC相应的字符串位置,但它不是太困难。

编辑: http://jcward.com/UUID.js - UUID.generate()