如何在JavaScript中创建GUID(全球独特识别器)?GUID/UUID应该至少有32个字符,并且应该保持在ASCII范围内,以避免在通过它们时遇到麻烦。

我不确定在所有浏览器上有哪些习惯,如何“随机”和种植内置的随机号码发电机等。


当前回答

我正在使用下面的功能:

function NewGuid()
{
    var sGuid = "";
    for (var i=0; i<32; i++)
    {
        sGuid += Math.floor(Math.random()*0xF).toString(0xF);
    }
    return sGuid;
}

其他回答

易于使用一个简单的 uuid 包 https://www.npmjs.com/package/uuid

const { v4: uuidv4 } = require('uuid');
uuidv4(); // ⇨ '1b9d6bcd-bbfd-4b2d-9b5d-ab8dfbbd4bed'

GitHub上的JavaScript项目 - https://github.com/LiosK/UUID.js

参见 RFC 4122 http://www.ietf.org/rfc/rfc4122.txt. 功能 产生 RFC 4122 符合 UUIDs. 版本 4 UUIDs (UUIDs 从随机数字) 和版本 1 UUIDs (基于时间的 UUIDs) 可用. UUID 对象允许各种访问 UUID 包括访问 UUID 字段。

对于想要 RFC 4122 版本 4 符合速度考虑的解决方案的人(少数电话到 Math.random()):

var rand = Math.random; 函数 UUID() { var nbr, randStr = "; do { randStr += (nbr = rand()).toString(16).substr(3, 6); } 同时(randStr.length < 30); 返回( randStr.substr(0, 8) + "-" + randStr.substr(8, 4) + "-4" + randStr.substr(12, 3) + "-" + ((nbr*4♰0) +8).toString(16) + // [89ab] randStr.substr(15, 3) + "-" + rand

上面的功能应该在速度和随机性之间保持适当的平衡。

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);
        }
    );
};