sql - 从表中选择时间戳彼此相差不到一小时的项目

Question

我正在使用 postgres (postgis)，我想执行一个查询，返回时间戳在一个小时内的所有行。

SQL 语句会是什么样子？

表格如下所示:

• id { 整数}
• 名字{字符变化}

• 时间戳{带时区的时间戳}

  id   name       timestamp
  ---+----------+--------------------------------
  1    one        "2010-09-24 21:10:39.515+00"
  2    two        "2010-09-16 09:21:09.362+00"
  3    three      "2010-07-08 00:00:46.549+00"
  

编辑#1

这是一个更好的例子。 (在 Tometzky 的帮助下)所有结果彼此相差不到 1 小时。也就是说，对于每一行，给我 1 小时内的任何其他行:

select * from myTable order by t

 id |               t               
----+-------------------------------
  9 | 2011-07-15 18:20:20.05+02
 10 | 2011-07-15 19:05:00.05+02
 11 | 2011-07-15 19:40:20.05+02
 13 | 2011-07-15 20:31:01.05+02
 14 | 2011-07-15 20:35:11.05+02
(5 rows)

result of needed query:

 id |  matchid |  origTime                  | matchTime
----+----------+----------------------------+------------------------------
  9 |   10     | 2011-07-15 18:20:20.05+02  | 2011-07-15 19:05:00.05+02
 10 |    9     | 2011-07-15 19:05:00.05+02  | 2011-07-15 18:20:20.05+02
 10 |   11     | 2011-07-15 19:05:00.05+02  | 2011-07-15 19:40:20.05+02
 11 |   10     | 2011-07-15 19:40:20.05+02  | 2011-07-15 19:05:00.05+02
 11 |   13     | 2011-07-15 19:40:20.05+02  | 2011-07-15 20:31:01.05+02
 11 |   14     | 2011-07-15 19:40:20.05+02  | 2011-07-15 20:35:11.05+02
 13 |   11     | 2011-07-15 20:31:01.05+02  | 2011-07-15 19:40:20.05+02
 13 |   14     | 2011-07-15 20:31:01.05+02  | 2011-07-15 20:35:11.05+02
 14 |   11     | 2011-07-15 20:35:11.05+02  | 2011-07-15 19:40:20.05+02
 14 |   13     | 2011-07-15 20:35:11.05+02  | 2011-07-15 20:31:01.05+02
(10 rows)

Answer 1

select 
    t0.id, t1.id as matchid, 
    t0.ts as origTime, t1.ts as matchTime
from t t0
inner join t t1 on 
    t1.ts between 
        t0.ts - interval '1 hour'
        and 
        t0.ts + interval '1 hour'
    and t0.id != t1.id
;