这里是一个方法,使用真实的随机通过 random.org 生成 RFC4122 如果错误,它会回到浏览器的内置加密图书馆,这应该几乎是相同的好。

async function UUID() {
    //get 31 random hex characters
    return (await (async () => {
        let output;
        try {
            //try from random.org
            output = (await (
                await fetch('https://www.random.org/integers/?num=31&min=0&max=15&col=31&base=16&format=plain&rnd=new')
            ).text())
                //get rid of whitespace
                .replace(/[^0-9a-fA-F]+/g, '')
            ;
            if (output.length != 31)
                throw '';
        }
        catch {
            output = '';
            try {
                //failing that, try getting 16 8-bit digits from crypto
                for (let num of crypto.getRandomValues(new Uint8Array(16)))
                    //interpret as 32 4-bit hex numbers
                    output += (num >> 4).toString(16) + (num & 15).toString(16);
                //we only want 31
                output = output.substr(1);
            }
            catch {
                //failing THAT, use Math.random
                while (output.length < 31)
                    output += (0 | Math.random() * 16).toString(16);
            }
        }
        return output;
    })())
        //split into appropriate sections, and set the 15th character to 4
        .replace(/^(.{8})(.{4})(.{3})(.{4})/, '$1-$2-4$3-$4-')
        //force character 20 to the correct range
        .replace(/(?<=-)[^89abAB](?=[^-]+-[^-]+$)/, (num) => (
            (parseInt(num, 16) % 4 + 8).toString(16)
        ))
    ;
}

此分類上一篇

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
      )
}

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());

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

我使用这个版本. 它是安全和简单的. 它不是为了产生格式化的Uids,它只是为了产生你需要的电缆的随机线条。

export function makeId(length) {
  let result = '';
  const characters = 'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789';
  const charactersLength = characters.length;

  for (let i = 0; i < length; i++) {
    let letterPos = parseInt(crypto.getRandomValues(new Uint8Array(1))[0] / 255 * charactersLength - 1, 10)
    result += characters[letterPos]
  }
  return result;
}

这也适用于Node.js,如果您更换了Let Buffer = 新 Uint8Array(); crypto.getRandomValues 与Let Buffer = crypto.randomBytes(16)

它应该在性能中打击最常见的表达解决方案。

const hex = '0123456789ABCDEF' let generateToken = function() { let buffer = new Uint8Array(16) crypto.getRandomValues(buffer) buffer[6] = 0x40 (buffer[6] & 0xF) buffer[8] = 0x80 (buffer[8] & 0xF) let segments = [] for (let i = 0; i < 16; ++i) { segments.push(hex[(buffer[i] >> 4 & 0xF)]) segments.push(hex[(buffer[i] >> 0 & 0xF)])) if (i ==

表演图(每个人都喜欢它们):JSbench