java 生成长类型的 UUID
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generate UUID of long type
提问by Mohamed Saligh
Please give me sample code to generate UUIDof longtype in java without using timestamp.
请给我示例代码生成UUID的long，而无需使用时间戳Java类型。
Thanks
谢谢
回答by Stephen C
A real UUID is 128 bits. A long is 64 bits.
真正的 UUID 是 128 位。long 是 64 位。
This is not just pedantry. UUID stands for UniversalUnique IDentifier.
这不仅仅是迂腐。UUID 代表通用唯一标识符。
The "universal uniqueness" of the established UUID schemesare based on:
已建立的 UUID 方案的“通用唯一性”基于：
- encoding a MAC address and a timestamp,
- encoding a hash of a DNS name and a timestamp, or
- using a 122 bit random number ... which is large enough that the probability of a collision is very very small.
- 对 MAC 地址和时间戳进行编码，
- 编码 DNS 名称和时间戳的散列，或
- 使用 122 位随机数......它足够大，碰撞的概率非常非常小。
With 64 bits, there are simply not enough bits for "universal uniqueness". For instance, the birthday paradox means that if we had a number of computers generating random 64 bit numbers, the probability of a potentially detectable collision would be large enough to be of concern.
对于 64 位，根本没有足够的位来实现“通用唯一性”。例如，生日悖论意味着，如果我们有许多计算机生成随机的 64 位数字，那么潜在可检测碰撞的概率将大到值得关注。
Now if you just want a UID (not a UUID), then any 64-bit sequence generator will do the job, provided that you take steps to guard against the sequence repeating. (If the sequence repeats, then the IDs are not unique in time; i.e. over time a given ID may denote different entities.)
现在，如果您只想要一个 UID（而不是 UUID），那么任何 64 位序列生成器都可以完成这项工作，前提是您采取措施防止序列重复。（如果序列重复，则 ID 在时间上不是唯一的；即随着时间的推移，给定的 ID 可能表示不同的实体。）
回答by Raghuram
Have you looked at java.util.UUID?
你看过java.util.UUID吗？
回答by Peter Lawrey
If you just want a simple unique long you can use AtomicLong.incrementAndGet(). This doesn't use a timestamp but does reset to 0 every time you start it and is not unique across JVMs.
如果你只想要一个简单的唯一 long 你可以使用 AtomicLong.incrementAndGet()。这不使用时间戳，但每次启动时都会重置为 0，并且在 JVM 中不是唯一的。
What is the requirement not to use timestamps all about? UUID uses a timestamp. (amoungst other things)
不使用时间戳的要求是什么？UUID 使用时间戳。（除其他外）
