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'

对于那些在Windows上使用JavaScript的人(例如,Windows Script Host(WSH),CScript,和HTA)。一个可以使用ActiveX。

WScript.Echo((new ActiveXObject("Scriptlet.TypeLib")).Guid)

请注意,这个答案仅适用于我列出的技术,它不会在任何浏览器中工作,甚至不是Microsoft Edge! 因此,你的距离与这个答案不同。

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);
        }
    );
};

来自Sagi shkedy的技术博客:

function generateGuid() {
  var result, i, j;
  result = '';
  for(j=0; j<32; j++) {
    if( j == 8 || j == 12 || j == 16 || j == 20)
      result = result + '-';
    i = Math.floor(Math.random()*16).toString(16).toUpperCase();
    result = result + i;
  }
  return result;
}

还有其他方法涉及使用 ActiveX 控制,但远离这些!

我以为值得指出的是,没有GUID发电机可以保证独特的钥匙(查看维基百科文章)。总是有冲突的可能性。

這只是一個概念,大多數人肯定可以以多種方式改進,但並不像我認為那樣緩慢。

总的来说,这个代码包含在毫秒内编码的黑客时间表(有某些黑客它给了12个数字,所以代码甚至在2527-06-24,但不是在5138-11-16之后工作),这意味着它是可编码的。

我知道这可能会缩短,性能提高,但我对结果感到高兴(除了MAC地址)。

目前Nanoseconds = () => { return nodeMode? process.hrtime.bigint() : BigInt(Date.now() * 1000000); } nodeFindMacAddress = () => { // 提取 MAC 地址 const 界面 = require('os').network 界面(); let result = null; for (index in interfaces) { let entry = interfaces[index]; entry.forEach(item => { if (item.mac!== '00:00:00:00:00:00') { result = '-'