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'

这可能对某人有用......

var d = new Date().valueOf();
var n = d.toString();
var result = '';
var length = 32;
var p = 0;
var chars = '0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ';

for (var i = length; i > 0; --i){
    result += ((i & 1) && n.charAt(p) ? '<b>' + n.charAt(p) + '</b>' : chars[Math.floor(Math.random() * chars.length)]);
    if(i & 1) p++;
};

HTTPS://jsfiddle.net/j0evrdf1/1/

基于布罗法的答案,我有自己的答案:

这里是使用 crypto.getRandomValues 的加密更强的版本。

函数 uuidv4() { const a = crypto.getRandomValues(新 Uint16Array(8)); let i = 0; return '00-0-4-1-000'.replace(/[^-]/g, s => (a[i++] + s * 0x10000 >> s).toString(16).padStart(4, '0') ); } console.log(uuidv4());

这里是使用 Math.random 的更快版本,使用几乎相同的原则:

函数 uuidv4() { return '00-0-4-1-000'.replace(/[^-]/g, s => ((Math.random() + ~~s) * 0x10000 >> s).toString(16).padStart(4, '0') ); } console.log(uuidv4());

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);
        }
    );
};

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();