选择每所学校的 N 个最新行，但跳过同一学生的重复行

Question

在 Postgres 数据库中，我有一个名为

school_id_cards

的视图。该视图捕获学校系统中每个学校和学生的身份证列表。身份证会定期重新发放给学生，因此每个学生可能有任意数量的身份证。每个

card_id

都是独一无二的。一个学生可能属于多个学校。记录样本如下所示：

卡_id	学校_id	学生id
1	1	1
2	1	1
3	1	2
4	1	2
5	1	3
6	1	3
7	1	4
8	1	4
9	2	5
10	2	5
11	2	6
12	3	7

给定一个

school_ids

列表，我想检索每个学校+学生最近创建的身份证列表，仅限于每个

school_id

选定的学生数量。

我有以下查询来获取我需要的东西，没有任何限制：

SELECT card_id FROM school_id_cards
WHERE card_id IN (
  SELECT MAX(card_id) FROM school_id_cards
  WHERE
    school_id in (1, 2, 3)
  GROUP BY
    school_id,
    student_id
);

...对于上述示例，返回

(2,4,6,8,10,11,12)

但是，在我的子查询中，我想限制为 where 子句中列出的每个

school_id

返回的学校+学生记录的数量。例如，限制为 2，以获取学校 1 的最多 2 个最近添加的学生、学校 2 的最多 2 个最近添加的学生以及学校 3 的最多 2 个最近添加的学生。在这种情况下，最终结果为

(6,8,10,11,12)

。

有一个查询可以完成这个任务吗？

Answer 1

您可以通过构建子查询的查询来做到这一点，并且

dense_rank

。

首先，获取每个学生/学校的最新卡片。 （注意：ID 不能很好地替代时间排序。添加日期时间列。）

select 
  *,
  dense_rank() over(partition by school_id, student_id order by card_id desc) as student_card_order
from school_id_cards

接下来，我们使用它作为子查询来获取学校最近发给学生的卡片的顺序。最近的卡片有

student_card_order = 1

。

with ordered_student_cards as (
  select 
    *,
    dense_rank() over(partition by school_id, student_id order by card_id desc) as student_card_order
  from school_id_cards
)
select
  *,
  dense_rank() over(partition by school_id order by card_id desc) as school_card_order
from ordered_student_cards
where student_card_order = 1

最后，我们只能获取每所学校的前两个。

school_card_order <= 2;

with ordered_student_cards as (
  select 
    *,
    dense_rank() over(partition by school_id, student_id order by card_id desc) as student_card_order
  from school_id_cards
), ordered_school_cards as (
  select
    *,
    dense_rank() over(partition by school_id order by card_id desc) as school_card_order
  from ordered_student_cards
  where student_card_order = 1
)
select card_id
from ordered_school_cards
where school_card_order <= 2;

示范.

可能有更紧凑或更高效的方法来做到这一点，但是窗口函数和子查询是分解复杂查询的一种方法。

Answer 2

如果您的表很大，您希望避免昂贵的全表顺序扫描。使用智能查询从匹配索引中选择具有索引（仅）扫描的合格行。速度快得多。

通常，您的数据库中应该存在某种“学校”表，每个相关学校只有一行。使查询更简单、更快：

WITH RECURSIVE latest_card AS (
   SELECT c.*
   FROM   school s
   CROSS  JOIN LATERAL (
      SELECT c.school_id, c.card_id, ARRAY[c.student_id] AS leading_ids
      FROM   school_id_cards c
      WHERE  c.school_id = s.school_id
      ORDER  BY c.card_id DESC
      LIMIT  1
      ) c

   UNION ALL
   SELECT c.*
   FROM   latest_card l
   JOIN   LATERAL (
      SELECT l.school_id, c.card_id, l.leading_ids || student_id
      FROM   school_id_cards c
      WHERE  c.school_id = l.school_id
      AND    c.card_id < l.card_id
      AND    c.student_id <> ALL (l.leading_ids)
      ORDER  BY c.card_id DESC
      LIMIT  1
      ) C ON cardinality(l.leading_ids) < 2  -- your limit per school here!
   )
SELECT card_id
FROM   latest_card
ORDER  BY card_id;

小提琴

正如您所演示的那样，这对于每所学校的小限制来说可以很好地扩展。对于较大的限制，我会切换到不同的查询。

关于递归CTE（rCTE）的使用：

• 优化 GROUP BY 查询以检索每个用户的最新行

确保有一个匹配的索引，例如

CREATE INDEX ON school_id_cards (school_id DESC, card_id DESC);

具有（默认）升序排序顺序的更简单的索引几乎没有任何糟糕。 Postgres 可以向后扫描 B 树索引。只有相反的排序顺序才不太理想。

如果没有

school

表：

WITH RECURSIVE latest_card AS (
   (
   SELECT DISTINCT ON (school_id)
          school_id, card_id, ARRAY[student_id] AS leading_ids
   FROM   school_id_cards c
   ORDER  BY school_id DESC, card_id DESC
   )

   UNION ALL
   SELECT c.*
   FROM   latest_card l
   JOIN   LATERAL (
      SELECT l.school_id, c.card_id, l.leading_ids || student_id
      FROM   school_id_cards c
      WHERE  c.school_id = l.school_id
      AND    c.card_id < l.card_id
      AND    c.student_id <> ALL (l.leading_ids)
      ORDER  BY c.card_id DESC
      LIMIT  1
      ) C ON cardinality(l.leading_ids) < 2  -- your limit per school here!
   )
SELECT card_id
FROM   latest_card
ORDER  BY card_id;

关于

DISTINCT ON

：

• 选择每个 GROUP BY 组中的第一行？

您可以用另一个嵌套的 rCTE 替换非递归项来生成学校列表（可能使用最新的卡片来启动）...
但确实应该有一张

school

桌子。如果没有，请创建它。