PostgreSQL-按UUID v1时间戳排序 [英] Postgresql - sort by UUID v1 timestamp

问题描述

我正在使用UUID v1作为主键。我想对UUID v1时间戳进行排序。现在，如果我执行

I am using UUID v1 as the primary key. I would like to sort on UUID v1 timestamp. Right now if I do something like

select id, title 
from table 
order by id desc

Postgresql不会按UUID时间戳对记录进行排序，而是按UUID字符串表示对字符串进行排序，最终导致意外的排序结果

Postgresql does not sort records by UUID timestamp, but by UUID string representation, which ends up with unexpected sorting result in my case.

我错过了什么吗，或者在Postgresql中没有内置的方法来做到这一点？

Am I missing something, or there is not a built in way to do this in Postgresql?

推荐答案

时间戳是v1 UUID的一部分。自 1582-10-15 00:00 起，它以十六进制格式存储为数百纳秒。此函数提取时间戳记：

The timestamp is one of the parts of a v1 UUID. It is stored in hex format as hundreds nanoseconds since 1582-10-15 00:00. This function extracts the timestamp:

create or replace function uuid_v1_timestamp (_uuid uuid)
returns timestamp with time zone as $$

    select
        to_timestamp(
            (
                ('x' || lpad(h, 16, '0'))::bit(64)::bigint::double precision -
                122192928000000000
            ) / 10000000
        )
    from (
        select
            substring (u from 16 for 3) ||
            substring (u from 10 for 4) ||
            substring (u from 1 for 8) as h
        from (values (_uuid::text)) s (u)
    ) s
    ;

$$ language sql immutable;

select uuid_v1_timestamp(uuid_generate_v1());
       uuid_v1_timestamp       
-------------------------------
 2016-06-16 12:17:39.261338+00

122192928000000000 是

在您的查询中：

select id, title
from t
order by uuid_v1_timestamp(id) desc

为提高性能，可以在该索引上创建索引：

To improve performance an index can be created on that:

create index uuid_timestamp_ndx on t (uuid_v1_timestamp(id));

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