基于加密 API 的 2017-06-28 的 Broofa 的 TypeScript 版本:

function genUUID() {
    // Reference: https://stackoverflow.com/a/2117523/709884
    return ("10000000-1000-4000-8000-100000000000").replace(/[018]/g, s => {
        const c = Number.parseInt(s, 10)
        return (c ^ crypto.getRandomValues(new Uint8Array(1))[0] & 15 >> c / 4).toString(16)
    })
}

原因:

使用 + 之间的数字( )和数字不有效 从行到数字的转换必须是明确的

ES6 样品

const guid=()=> {
  const s4=()=> Math.floor((1 + Math.random()) * 0x10000).toString(16).substring(1);     
  return `${s4() + s4()}-${s4()}-${s4()}-${s4()}-${s4() + s4() + s4()}`;
}

这个是基于日期,并添加一个随机的补充到“保证”的独特性。

它很好地用于CSS识别器,总是返回类似的东西,并且很容易被黑客攻击:

uid-139410573297741

var getUniqueId = function (prefix) {
            var d = new Date().getTime();
            d += (parseInt(Math.random() * 100)).toString();
            if (undefined === prefix) {
                prefix = 'uid-';
            }
            d = prefix + d;
            return d;
        };

只有在任何人落下谷歌正在寻找一个小用途图书馆的情况下,ShortId满足了这个问题的所有要求,它允许指定允许的字符和长度,并保证不序列,不重复的线条。

为了使这一点更为真实的答案,该图书馆的核心使用下列逻辑来制作其简短的ID:

function encode(lookup, number) {
    var loopCounter = 0;
    var done;

    var str = '';

    while (!done) {
        str = str + lookup( ( (number >> (4 * loopCounter)) & 0x0f ) | randomByte() );
        done = number < (Math.pow(16, loopCounter + 1 ) );
        loopCounter++;
    }
    return str;
}

/* Generates the short id */
function generate() {

    var str = '';

    var seconds = Math.floor((Date.now() - REDUCE_TIME) * 0.001);

    if (seconds === previousSeconds) {
        counter++;
    } else {
        counter = 0;
        previousSeconds = seconds;
    }

    str = str + encode(alphabet.lookup, version);
    str = str + encode(alphabet.lookup, clusterWorkerId);
    if (counter > 0) {
        str = str + encode(alphabet.lookup, counter);
    }
    str = str + encode(alphabet.lookup, seconds);

    return str;
}

我没有编辑这只是反映这个方法的最基本部分,所以上面的代码包含一些额外的逻辑从图书馆. 如果你对它正在做的一切好奇,请看看来源: https://github.com/dylang/shortid/tree/master/lib

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
      )
}

下面的版本是布罗法的答案的调整,但更新,包括一个“真实”随机功能,使用加密图书馆在可用的地方,以及Alea()功能作为落后。

  Math.log2 = Math.log2 || function(n){ return Math.log(n) / Math.log(2); }
  Math.trueRandom = (function() {
  var crypt = window.crypto || window.msCrypto;

  if (crypt && crypt.getRandomValues) {
      // If we have a crypto library, use it
      var random = function(min, max) {
          var rval = 0;
          var range = max - min;
          if (range < 2) {
              return min;
          }

          var bits_needed = Math.ceil(Math.log2(range));
          if (bits_needed > 53) {
            throw new Exception("We cannot generate numbers larger than 53 bits.");
          }
          var bytes_needed = Math.ceil(bits_needed / 8);
          var mask = Math.pow(2, bits_needed) - 1;
          // 7776 -> (2^13 = 8192) -1 == 8191 or 0x00001111 11111111

          // Create byte array and fill with N random numbers
          var byteArray = new Uint8Array(bytes_needed);
          crypt.getRandomValues(byteArray);

          var p = (bytes_needed - 1) * 8;
          for(var i = 0; i < bytes_needed; i++ ) {
              rval += byteArray[i] * Math.pow(2, p);
              p -= 8;
          }

          // Use & to apply the mask and reduce the number of recursive lookups
          rval = rval & mask;

          if (rval >= range) {
              // Integer out of acceptable range
              return random(min, max);
          }
          // Return an integer that falls within the range
          return min + rval;
      }
      return function() {
          var r = random(0, 1000000000) / 1000000000;
          return r;
      };
  } else {
      // From https://web.archive.org/web/20120502223108/http://baagoe.com/en/RandomMusings/javascript/
      // Johannes Baagøe <baagoe@baagoe.com>, 2010
      function Mash() {
          var n = 0xefc8249d;

          var mash = function(data) {
              data = data.toString();
              for (var i = 0; i < data.length; i++) {
                  n += data.charCodeAt(i);
                  var h = 0.02519603282416938 * n;
                  n = h >>> 0;
                  h -= n;
                  h *= n;
                  n = h >>> 0;
                  h -= n;
                  n += h * 0x100000000; // 2^32
              }
              return (n >>> 0) * 2.3283064365386963e-10; // 2^-32
          };

          mash.version = 'Mash 0.9';
          return mash;
      }

      // From http://baagoe.com/en/RandomMusings/javascript/
      function Alea() {
          return (function(args) {
              // Johannes Baagøe <baagoe@baagoe.com>, 2010
              var s0 = 0;
              var s1 = 0;
              var s2 = 0;
              var c = 1;

              if (args.length == 0) {
                  args = [+new Date()];
              }
              var mash = Mash();
              s0 = mash(' ');
              s1 = mash(' ');
              s2 = mash(' ');

              for (var i = 0; i < args.length; i++) {
                  s0 -= mash(args[i]);
                  if (s0 < 0) {
                      s0 += 1;
                  }
                  s1 -= mash(args[i]);
                  if (s1 < 0) {
                      s1 += 1;
                  }
                  s2 -= mash(args[i]);
                  if (s2 < 0) {
                      s2 += 1;
                  }
              }
              mash = null;

              var random = function() {
                  var t = 2091639 * s0 + c * 2.3283064365386963e-10; // 2^-32
                  s0 = s1;
                  s1 = s2;
                  return s2 = t - (c = t | 0);
              };
              random.uint32 = function() {
                  return random() * 0x100000000; // 2^32
              };
              random.fract53 = function() {
                  return random() +
                      (random() * 0x200000 | 0) * 1.1102230246251565e-16; // 2^-53
              };
              random.version = 'Alea 0.9';
              random.args = args;
              return random;

          }(Array.prototype.slice.call(arguments)));
      };
      return Alea();
  }
}());

Math.guid = function() {
    return 'xxxxxxxx-xxxx-4xxx-yxxx-xxxxxxxxxxxx'.replace(/[xy]/g, function(c)    {
      var r = Math.trueRandom() * 16 | 0,
          v = c == 'x' ? r : (r & 0x3 | 0x8);
      return v.toString(16);
  });
};