从 PostgreSQL 中的 v1 UUID 投射或提取时间戳

问题描述：

我正在尝试从第 1 版 UUID，天真地希望这能奏效:

I'm trying to extract the timestamp from a Version 1 UUID, naively wished this worked:

SELECT '3efe0a20-f1b3-11e3-bb44-14109fec739e'::uuid::timestamp;

这里是一个简单的例子，展示了如何在 Go 中提取时间，但我希望 PostgreSQL 有内置的东西，而不是创建一次性的 PL/pgSql 函数:
http://play.golang.org/p/XRCooLgfaG

Here is a quick example showing how to extract the time in Go, but I'm hoping PostgreSQL has something built-in rather than creating a one-off PL/pgSql function:
http://play.golang.org/p/XRCooLgfaG

答

我已经用我的数据库中的 uuid 测试了这个，它似乎工作得很好，即使没有 unsigned bigints

I've tested this with uuid's from my database and it seems to work very well, even without the unsigned bigints

CREATE FUNCTION uuid_timestamp(id uuid) RETURNS timestamptz AS $$
  select TIMESTAMP WITH TIME ZONE 'epoch' +
      (((('x' || lpad(split_part(id::text, '-', 1), 16, '0'))::bit(64)::bigint) +
      (('x' || lpad(split_part(id::text, '-', 2), 16, '0'))::bit(64)::bigint << 32) +
      ((('x' || lpad(split_part(id::text, '-', 3), 16, '0'))::bit(64)::bigint&4095) << 48) - 122192928000000000) / 10000000 ) * INTERVAL '1 second';    
$$ LANGUAGE SQL
  IMMUTABLE
  RETURNS NULL ON NULL INPUT;

我在 2099 年未来创建的 V1 uuid！

a V1 uuid I created in the 2099 future!

select uuid_timestamp('6d248400-65b7-1243-a57a-14109fec739e');
uuid_timestamp     
------------------------
 2099-08-01 11:30:00-07
(1 row)

从 PostgreSQL 中的 v1 UUID 投射或提取时间戳

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