我们可以使用代替和crypto.getRandomValues以获得这样的输出:

xxxxxxxxxxxxxxxxxxxxxxx

此分類上一篇

如果我们正在寻找一个 opti 解决方案,我们必须用一个序列(32 )取代 crypto.getRandomValues(新 Uint8Array(1))[0]。

const uuidv4 = () =>
  ([1e7] + -1e3 + -4e3 + -8e3 + -1e11).replace(/[018]/g, c =>
    (c ^ crypto.getRandomValues(new Uint8Array(1))[0] & 15 >> c / 4).toString(16)
  );

console.log(uuidv4());

要得到这个代码:

函数 uuidv4() { 让比特 = window.crypto.getRandomValues(新 Uint8Array(32)); const randomBytes = () => (比特 = bytes.slice(1)) && bytes[0]; 返回([1e7] + -1e3 + -4e3 + -8e3 + -1e11); 取代(/[018]/g, c => (c ^ randomBytes() & 15 >> c / 4).toString(16) ; } for (var i = 0; i < 10; i++) console.log(uuidv4());

碰撞:

我们可以喜欢谷歌分析,并添加一个时间标签: uuidv4() + "。

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}

var uuid = function() {
    var buf = new Uint32Array(4);
    window.crypto.getRandomValues(buf);
    var idx = -1;
    return 'xxxxxxxx-xxxx-4xxx-yxxx-xxxxxxxxxxxx'.replace(/[xy]/g, function(c) {
        idx++;
        var r = (buf[idx>>3] >> ((idx%8)*4))&15;
        var v = c == 'x' ? r : (r&0x3|0x8);
        return v.toString(16);
    });
};

这个版本是基于Briguy37的答案和一些Bitwise运营商从泡沫中提取Nibble大小的窗户。

它应该遵守RFC类型4(随机)方案,因为我上次遇到麻烦与Java的UUID不符合的UUID。

  // RFC 4122
  //
  // A UUID is 128 bits long
  //
  // String representation is five fields of 4, 2, 2, 2, and 6 bytes.
  // Fields represented as lowercase, zero-filled, hexadecimal strings, and
  // are separated by dash characters
  //
  // A version 4 UUID is generated by setting all but six bits to randomly
  // chosen values
  var uuid = [
    Math.random().toString(16).slice(2, 10),
    Math.random().toString(16).slice(2, 6),

    // Set the four most significant bits (bits 12 through 15) of the
    // time_hi_and_version field to the 4-bit version number from Section
    // 4.1.3
    (Math.random() * .0625 /* 0x.1 */ + .25 /* 0x.4 */).toString(16).slice(2, 6),

    // Set the two most significant bits (bits 6 and 7) of the
    // clock_seq_hi_and_reserved to zero and one, respectively
    (Math.random() * .25 /* 0x.4 */ + .5 /* 0x.8 */).toString(16).slice(2, 6),

    Math.random().toString(16).slice(2, 14)].join('-');

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);
        }
    );
};

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;
        };