查询以从每个组中找到第二大值

Question

我有三个表：

project: project_id, project_name
milestone: milestone_id, milestone_name
project_milestone: id, project_id, milestone_id, completed_date

我想从按 project_id 分组的 project_milestone 中获得第二高的 completed_date 和 milestone_id。那就是我想为每个项目获得第二高 completed_date 的 milestone_id。正确的查询是什么？

Answer 1

我想你可以用 project_milestone table 和 row_number() 做你想做的事:

select pm.*
from (select pm.*,
             row_number() over (partition by project_id order by completed_date desc) as seqnum
      from project_milestone pm
      where pm.completed_date is not null
     ) pm
where seqnum = 2;

如果您需要包括所有个项目，即使是那些没有两个里程碑的项目，您也可以使用 left join:

select p.project_id, pm.milestone_id, pm.completed_date
from projects p left join
     (select pm.*,
             row_number() over (partition by project_id order by completed_date desc) as seqnum
      from project_milestone pm
      where pm.completed_date is not null
     ) pm
     on p.project_id = pm.project_id and pm.seqnum = 2;

Answer 2

使用 LATERAL (PG 9.3+) 可以产生比 window 函数版本更好的性能。

SELECT * FROM project;
 project_id | project_name 
------------+--------------
          1 | Project A
          2 | Project B

SELECT * FROM project_milestone;
 id | project_id | milestone_id |     completed_date     
----+------------+--------------+------------------------
  1 |          1 |            1 | 2000-01-01 00:00:00+01
  2 |          1 |            2 | 2000-01-02 00:00:00+01
  3 |          1 |            5 | 2000-01-03 00:00:00+01
  4 |          1 |            6 | 2000-01-04 00:00:00+01
  5 |          2 |            3 | 2000-02-01 00:00:00+01
  6 |          2 |            4 | 2000-02-02 00:00:00+01
  7 |          2 |            7 | 2000-02-03 00:00:00+01
  8 |          2 |            8 | 2000-02-04 00:00:00+01


SELECT *
FROM project p
CROSS JOIN LATERAL (
    SELECT milestone_id, completed_date
    FROM project_milestone pm
    WHERE pm.project_id = p.project_id
    ORDER BY completed_date ASC
    LIMIT 1
    OFFSET 1
) second_highest;
 project_id | project_name | milestone_id |     completed_date     
------------+--------------+--------------+------------------------
          1 | Project A    |            2 | 2000-01-02 00:00:00+01
          2 | Project B    |            4 | 2000-02-02 00:00:00+01

Answer 3

实现此目的的最简单方法是使用 window 函数。

SELECT *, nth_value(completed_date,2)
OVER (
    PARTITION BY project_id ORDER BY completed_date DESC
    RANGE BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING
)
AS date2
FROM project_milestone;

查询以从每个组中找到第二大值

Query to find second largest value from every group

sql

postgresql

postgresql-9.3