最酷的方式:

function uuid(){
    var u = URL.createObjectURL(new Blob([""]))
    URL.revokeObjectURL(u);
    return u.split("/").slice(-1)[0]
}

它可能不是快速,高效,或支持在 IE2 但它肯定是酷的

根據 RFC 4122 的 UUID (Universally Unique IDentifier),也被稱為 GUID (Globally Unique IDentifier),是為提供某些獨特性保證而設計的識別。

虽然可以在JavaScript代码的几个行中实施符合RFC的UUID(例如,请参见 @broofa的答案,下面),但有几个常见的漏洞:

无效 ID 格式(UUID 必须是“xxxxxx-xxxx-Mxxx-Nxxx-xxxxxxxxxxxx”的格式,其中 x 是 [0-9, a-f] M 是 [1-5] 的格式,而 N 是 [8, 9, a 或 b] 使用低质量的随机性来源(如 Math.random)

因此,开发人员为生产环境编写代码被鼓励使用严格,良好的实施,如新模块。

下面是2011年10月9日的解决方案,由用户在https://gist.github.com/982883上发表评论:

UUIDv4 = function b(a){return a?(a^Math.random()*16>>a/4).toString(16):([1e7]+-1e3+-4e3+-8e3+-1e11).replace(/[018]/g,b)}

这实现了与当前最高评级的答案相同的目标,但在50比以更少的比特中,利用强迫性、回报性和曝光性评级。

UUIDv4 =

function b(
  a // placeholder
){
  return a // if the placeholder was passed, return
    ? ( // a random number from 0 to 15
      a ^ // unless b is 8,
      Math.random() // in which case
      * 16 // a random number from
      >> a/4 // 8 to 11
      ).toString(16) // in hexadecimal
    : ( // or otherwise a concatenated string:
      [1e7] + // 10000000 +
      -1e3 + // -1000 +
      -4e3 + // -4000 +
      -8e3 + // -80000000 +
      -1e11 // -100000000000,
      ).replace( // replacing
        /[018]/g, // zeroes, ones, and eights with
        b // random hex digits
      )
}

这里是一个完全不符合但非常有效的实施,以产生一个ASCII安全的GUID类似的独特识别器。

function generateQuickGuid() {
    return Math.random().toString(36).substring(2, 15) +
        Math.random().toString(36).substring(2, 15);
}

创建 26 个字符 [a-z0-9],产生一个 UID 既较短又比 RFC 符合的 GUID 更独特。

这里是这个功能的使用例子和时间表,以及这个问题的其他几个答案。

>>> generateQuickGuid()
"nvcjf1hs7tf8yyk4lmlijqkuo9"
"yq6gipxqta4kui8z05tgh9qeel"
"36dh5sec7zdj90sk2rx7pjswi2"
runtime: 32.5s

>>> GUID() // John Millikin
"7a342ca2-e79f-528e-6302-8f901b0b6888"
runtime: 57.8s

>>> regexGuid() // broofa
"396e0c46-09e4-4b19-97db-bd423774a4b3"
runtime: 91.2s

>>> createUUID() // Kevin Hakanson
"403aa1ab-9f70-44ec-bc08-5d5ac56bd8a5"
runtime: 65.9s

>>> UUIDv4() // Jed Schmidt
"f4d7d31f-fa83-431a-b30c-3e6cc37cc6ee"
runtime: 282.4s

>>> Math.uuid() // broofa
"5BD52F55-E68F-40FC-93C2-90EE069CE545"
runtime: 225.8s

>>> Math.uuidFast() // broofa
"6CB97A68-23A2-473E-B75B-11263781BBE6"
runtime: 92.0s

>>> Math.uuidCompact() // broofa
"3d7b7a06-0a67-4b67-825c-e5c43ff8c1e8"
runtime: 229.0s

>>> bitwiseGUID() // jablko
"baeaa2f-7587-4ff1-af23-eeab3e92"
runtime: 79.6s

>>>> betterWayGUID() // Andrea Turri
"383585b0-9753-498d-99c3-416582e9662c"
runtime: 60.0s

>>>> UUID() // John Fowler
"855f997b-4369-4cdb-b7c9-7142ceaf39e8"
runtime: 62.2s

这里是时间代码。

var r;
console.time('t'); 
for (var i = 0; i < 10000000; i++) { 
    r = FuncToTest(); 
};
console.timeEnd('t');

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

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);
        }
    );
};