在这里,你可以找到一个非常小的功能,产生UUID。

最后一个版本是:

function b(
  a                  // Placeholder
){
  var cryptoObj = window.crypto || window.msCrypto; // For Internet Explorer 11
  return a           // If the placeholder was passed, return
    ? (              // a random number from 0 to 15
      a ^            // unless b is 8,
      cryptoObj.getRandomValues(new Uint8Array(1))[0]  // 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
      )
}

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'

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);
        }
    );
};

它提供简单,快速的生成RFC4122 UUIDS。

特性:

创建 RFC4122 版本 1 或版本 4 UUIDs 在 Node.js 和浏览器中运行. 在支持平台上加密强大的随机 # 发行。

使用 NPM 安装:

npm install uuid

或者通过浏览器使用 uuid:

下载原始文件(uuid v1): https://raw.githubusercontent.com/kelektiv/node-uuid/master/v1.js 下载原始文件(uuid v4): https://raw.githubusercontent.com/kelektiv/node-uuid/master/v4.js

想要更小? 查看此: https://gist.github.com/jed/982883

使用:

// Generate a v1 UUID (time-based)
const uuidV1 = require('uuid/v1');
uuidV1(); // -> '6c84fb90-12c4-11e1-840d-7b25c5ee775a'

// Generate a v4 UUID (random)
const uuidV4 = require('uuid/v4');
uuidV4(); // -> '110ec58a-a0f2-4ac4-8393-c866d813b8d1'

// Generate a v5 UUID (namespace)
const uuidV5 = require('uuid/v5');

// ... using predefined DNS namespace (for domain names)
uuidV5('hello.example.com', v5.DNS)); // -> 'fdda765f-fc57-5604-a269-52a7df8164ec'

// ... using predefined URL namespace (for, well, URLs)
uuidV5('http://example.com/hello', v5.URL); // -> '3bbcee75-cecc-5b56-8031-b6641c1ed1f1'

// ... using a custom namespace
const MY_NAMESPACE = '(previously generated unique uuid string)';
uuidV5('hello', MY_NAMESPACE); // -> '90123e1c-7512-523e-bb28-76fab9f2f73d'

ECMAScript 2015(ES6):

import uuid from 'uuid/v4';
const id = uuid();

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
      )
}

[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。