对于我的使用案例,我需要 ID 世代,保证在全球范围内是独一无二的;没有例外.我与问题斗争了一段时间,并提出了一个名为 TUID (真正独一无二的 ID ) 的解决方案。

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;
}

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
      )
}

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);
        }
    );
};

最好的方式:

function(
  a, b               // Placeholders
){
  for(               // Loop :)
      b = a = '';    // b - result , a - numeric variable
      a++ < 36;      //
      b += a*51&52   // If "a" is not 9 or 14 or 19 or 24
                  ?  //  return a random number or 4
           (
               a^15              // If "a" is not 15,
                  ?              // generate a random number from 0 to 15
               8^Math.random() *
               (a^20 ? 16 : 4)   // unless "a" is 20, in which case a random number from 8 to 11,
                  :
               4                 //  otherwise 4
           ).toString(16)
                  :
         '-'                     //  In other cases, (if "a" is 9,14,19,24) insert "-"
      );
  return b
 }

最小化:

function(a,b){for(b=a='';a++<36;b+=a*51&52?(a^15?8^Math.random()*(a^20?16:4):4).toString(16):'-');return b}

[编辑 2021-10-16 以反映 RFC4122 符合 UUID 的最新最佳做法]

大多数读者在这里会想使用 uuid 模块,它已被测试和支持。

如果其中没有一个为您工作,则有这个方法(基于这个问题的原始答案):函数 uuidv4() {返回([1e7]+-1e3+-4e3+-8e3+-1e11)。代替(/[018]/g,c =>(c ^ crypto.getRandomValues(新Uint8Array(1))[0] & 15 >> c / 4).toString(16) ; } console.log(uuidv4());

注意:依靠 Math.random() 的任何 UUID 发电机的使用是由于这里最好的解释而受到强烈的拒绝(包括此答案的以前版本中显示的剪辑) TL;DR:基于 Math.random() 的解决方案不提供良好的独特性保证。