sql - PostgreSQL:如何返回与找到的行相关的行(相对结果)？

Question

如果我的例子没有意义，请原谅我。我将尝试使用简化版来鼓励更多参与。

考虑如下表格:

•        dt     |    mnth    |  foo
  --------------+------------+--------
    2012-12-01  |  December  |
      ...
    2012-08-01  |  August    |
    2012-07-01  |  July      |
    2012-06-01  |  June      |
    2012-05-01  |  May       |
    2012-04-01  |  April     |
    2012-03-01  |  March     |
      ...
    1997-01-01  |  January   |  
  

如果您寻找 dt 最接近今天的记录，而没有过去，那么最好的方法是预先返回 3 条记录 和 7 条记录之后?

我决定试试窗口函数:

• WITH dates AS (
     select  row_number() over (order by dt desc)
           , dt
           , dt - now()::date as dt_diff
     from    foo
  )
  , closest_date AS (
     select * from dates
     where dt_diff = ( select max(dt_diff) from dates where dt_diff <= 0 )
  )
  
  SELECT * 
  FROM   dates
  WHERE  row_number - (select row_number from closest_date) >= -3
     AND row_number - (select row_number from closest_date) <=  7 ;
  

我觉得一定有更好的方法来使用窗口函数返回相关记录，但我已经有一段时间没看过它们了。

Answer 1

create table foo (dt date);
insert into foo values
('2012-12-01'),
('2012-08-01'),
('2012-07-01'),
('2012-06-01'),
('2012-05-01'),
('2012-04-01'),
('2012-03-01'),
('2012-02-01'),
('2012-01-01'),
('1997-01-01'),
('2012-09-01'),
('2012-10-01'),
('2012-11-01'),
('2013-01-01')
;

select dt
from (
(
    select dt
    from foo
    where dt <= current_date
    order by dt desc
    limit 4
)
union all
(
    select dt
    from foo
    where dt > current_date
    order by dt
    limit 7
)) s
order by dt
;
     dt     
------------
 2012-03-01
 2012-04-01
 2012-05-01
 2012-06-01
 2012-07-01
 2012-08-01
 2012-09-01
 2012-10-01
 2012-11-01
 2012-12-01
 2013-01-01
(11 rows)