这里有一个测试片段供您测试它的独特性。 灵感来自@scalabl3的评论

有趣的是，你可以连续生成2个完全相同的结果，当然是在令人难以置信的巧合、运气和神的干预下，尽管有不可思议的几率，但这仍然是可能的!:是的，不会发生的。我这么说只是为了好玩，想想你创造了一个复制品的那一刻!视频截图!- scalab13 10月20日15日19:11

如果您觉得幸运，请选中复选框，它只检查当前生成的id。如果您希望进行历史记录检查，请不勾选。 请注意，如果您不勾选它，您可能会在某些时候耗尽ram。我试图使它对cpu友好，以便在需要时可以快速中止，只需再次点击运行代码片段按钮或离开页面。

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 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); }); }; function logit(item1, item2) { console.log("Do "+item1+" and "+item2+" equal? "+(item1 == item2 ? "OMG! take a screenshot and you'll be epic on the world of cryptography, buy a lottery ticket now!":"No they do not. shame. no fame")+ ", runs: "+window.numberofRuns); } numberofRuns = 0; function test() { window.numberofRuns++; var x = Math.guid(); var y = Math.guid(); var test = x == y || historyTest(x,y); logit(x,y); return test; } historyArr = []; historyCount = 0; function historyTest(item1, item2) { if(window.luckyDog) { return false; } for(var i = historyCount; i > -1; i--) { logit(item1,window.historyArr[i]); if(item1 == history[i]) { return true; } logit(item2,window.historyArr[i]); if(item2 == history[i]) { return true; } } window.historyArr.push(item1); window.historyArr.push(item2); window.historyCount+=2; return false; } luckyDog = false; document.body.onload = function() { document.getElementById('runit').onclick = function() { window.luckyDog = document.getElementById('lucky').checked; var val = document.getElementById('input').value if(val.trim() == '0') { var intervaltimer = window.setInterval(function() { var test = window.test(); if(test) { window.clearInterval(intervaltimer); } },0); } else { var num = parseInt(val); if(num > 0) { var intervaltimer = window.setInterval(function() { var test = window.test(); num--; if(num < 0 || test) { window.clearInterval(intervaltimer); } },0); } } }; }; Please input how often the calulation should run. set to 0 for forever. Check the checkbox if you feel lucky.<BR/> <input type="text" value="0" id="input"><input type="checkbox" id="lucky"><button id="runit">Run</button><BR/>

这个问题的答案很大程度上取决于UUID版本。

许多UUID生成器使用版本4的随机数。然而，其中许多使用伪随机数生成器来生成它们。

如果使用一个短周期的低种子PRNG来生成UUID，我认为这一点都不安全。一些随机数生成器的方差也很差。也就是说，更倾向于某些数字。这不会有好结果的。

因此，它的安全性取决于生成它的算法。

另一方面，如果您知道这些问题的答案，那么我认为使用版本4的uuid应该是非常安全的。事实上，我正在使用它来识别网络块文件系统上的块，到目前为止还没有发生冲突。

在我的情况下，我使用的PRNG是一个梅森龙卷风，我很小心，它的播种方式是来自多个来源，包括/dev/ urrandom。梅森龙卷风的周期为2^19937−1。在我看到一个重复的uuid之前，会有很长很长的时间。

因此，选择一个好的库或自己生成它，并确保使用合适的PRNG算法。

我同意其他的答案。uuid对于几乎所有的实际用途都是足够安全的，当然对你来说也是如此。

但假设(假设)它们不是。

是否有更好的系统或某种类型的模式来缓解这个问题?

这里有一些方法:

Use a bigger UUID. For instance, instead of a 128 random bits, use 256 or 512 or ... Each bit you add to a type-4 style UUID will reduce the probability of a collision by a half, assuming that you have a reliable source of entropy2. Build a centralized or distributed service that generates UUIDs and records each and every one it has ever issued. Each time it generates a new one, it checks that the UUID has never been issued before. Such a service would be technically straight-forward to implement (I think) if we assumed that the people running the service were absolutely trustworthy, incorruptible, etcetera. Unfortunately, they aren't ... especially when there is the possibility of governments' security organizations interfering. So, this approach is probably impractical, and may be3 impossible in the real world.

1 - If uniqueness of UUIDs determined whether nuclear missiles got launched at your country's capital city, a lot of your fellow citizens would not be convinced by "the probability is extremely low". Hence my "nearly all" qualification. 2 - And here's a philosophical question for you. Is anything ever truly random? How would we know if it wasn't? Is the universe as we know it a simulation? Is there a God who might conceivably "tweak" the laws of physics to alter an outcome? 3 - If anyone knows of any research papers on this problem, please comment.

对于UUID4，我认为在一个边长360000公里的立方体盒子中，id的数量大约与沙粒的数量相同。这是一个边长约为木星直径2.5倍的盒子。

如果我搞砸了单位，就会有人告诉我:

沙粒体积0.00947mm^3 (Guardian) UUID4有122个随机位-> 5.3e36可能的值(维基百科) 那么多沙粒的体积= 5.0191e34 mm^3或5.0191e+25m^3 体积= 3.69E8m或369,000km的立方箱的边长 木星直径:139,820公里(谷歌)

UUID类型不止一种，因此“安全程度”取决于您使用的类型(UUID规范称为“版本”)。

Version 1 is the time based plus MAC address UUID. The 128-bits contains 48-bits for the network card's MAC address (which is uniquely assigned by the manufacturer) and a 60-bit clock with a resolution of 100 nanoseconds. That clock wraps in 3603 A.D. so these UUIDs are safe at least until then (unless you need more than 10 million new UUIDs per second or someone clones your network card). I say "at least" because the clock starts at 15 October 1582, so you have about 400 years after the clock wraps before there is even a small possibility of duplications. Version 4 is the random number UUID. There's six fixed bits and the rest of the UUID is 122-bits of randomness. See Wikipedia or other analysis that describe how very unlikely a duplicate is. Version 3 is uses MD5 and Version 5 uses SHA-1 to create those 122-bits, instead of a random or pseudo-random number generator. So in terms of safety it is like Version 4 being a statistical issue (as long as you make sure what the digest algorithm is processing is always unique). Version 2 is similar to Version 1, but with a smaller clock so it is going to wrap around much sooner. But since Version 2 UUIDs are for DCE, you shouldn't be using these.

所以对于所有实际问题，它们都是安全的。如果你不喜欢把它留给概率(例如，你是那种担心地球在你的一生中被一颗大小行星摧毁的人)，只要确保你使用版本1的UUID，并且它保证是唯一的(在你的一生中，除非你计划活到公元3603年以后)。

那么，为什么不是每个人都使用版本1的uuid呢?这是因为版本1的uuid揭示了生成它的机器的MAC地址，并且它们是可以预测的——这两件事可能会对使用这些uuid的应用程序产生安全影响。

我已经做了很多年了。永远不要遇到问题。

我通常设置我的数据库有一个表，其中包含所有的键和修改的日期等。我从没遇到过钥匙重复的问题。

它的唯一缺点是，当您编写一些查询来快速查找一些信息时，您需要进行大量的复制和粘贴键。你不再有简单易记的id了。