UUID token = UUID.randomUUID(); user.setSessionId(token.toString()); Sessions.INSTANCE.sessions.put(user.getName(),user.getSessionId());
但是,我不知道如何防止重复的sessionTokens.
例如：当user1登录并获得令牌87955dc9-d2ca-4f79-b7c8-b0223a32532a和user2登录时,可能会出现一种情况,并且还会获得令牌87955dc9-d2ca-4f79-b7c8-b0223a32532a.
有更好的做法吗？
解决方法
维基百科：
Randomly generated UUIDs have 122 random bits. Out of a total of 128
bits,four bits are used for the version (‘Randomly generated UUID’),
and two bits for the variant (‘Leach-Salz’).With random UUIDs,the
chance of two having the same value can be calculated using
probability theory (Birthday paradox). Using the approximation
p(n)\approx 1-e^{-\tfrac{n^2}{{2x}}}these are the probabilities of an
accidental clash after calculating n UUIDs,with x=2122:n probability
68,719,476,736 = 236 0.0000000000000004 (4 × 10−16)
2,199,023,255,552 = 241 0.0000000000004 (4 × 10−13)
70,368,744,177,664 = 246 0.0000000004 (4 × 10−10)To put these numbers into perspective,the annual risk of someone being hit by a meteorite is estimated to be one chance in 17 billion,which means the probability is about 0.00000000006 (6 × 10−11),equivalent to the odds of creating a few tens of trillions of > UUIDs in a year and having one duplicate. In other words,only after generating 1 billion UUIDs every second for the next 100 years,the probability of creating just one duplicate would be about 50%. The probability of one duplicate would be about 50% if every person on earth owns 600 million UUIDs.