关于sql：PostgreSQL中的分组LIMIT：显示每组的前N行？

Grouped LIMIT in PostgreSQL: show the first N rows for each group?

我需要为每个组取前N行，按自定义列排序。

鉴于下表：

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db=# SELECT * FROM xxx;
id | section_id | name
----+------------+------
1 | 1 | A
2 | 1 | B
3 | 1 | C
4 | 1 | D
5 | 2 | E
6 | 2 | F
7 | 3 | G
8 | 2 | H
(8 ROWS)

我需要每个section_id的前两行(按名称排序)，即类似于的结果：

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id | section_id | name
----+------------+------
1 | 1 | A
2 | 1 | B
5 | 2 | E
6 | 2 | F
7 | 3 | G
(5 ROWS)

我正在使用PostgreSQL 8.3.5。

新解决方案(PostgreSQL 8.4)

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SELECT
*
FROM (
SELECT
ROW_NUMBER() OVER (PARTITION BY section_id ORDER BY name) AS r,
t.*
FROM
xxx t) x
WHERE
x.r <= 2;

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从v9.3开始，你可以进行横向连接

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SELECT DISTINCT t_outer.section_id, t_top.id, t_top.name FROM t t_outer
JOIN lateral (
SELECT * FROM t t_inner
WHERE t_inner.section_id = t_outer.section_id
ORDER BY t_inner.name
LIMIT 2
) t_top ON TRUE
ORDER BY t_outer.section_id;

它可能更快，但当然，您应该专门针对您的数据和用例测试性能。

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这是另一个解决方案(PostgreSQL <= 8.3)。

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SELECT
*
FROM
xxx a
WHERE (
SELECT
COUNT(*)
FROM
xxx
WHERE
section_id = a.section_id
AND
name <= a.name
) <= 2

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-- ranking without WINDOW functions
-- EXPLAIN ANALYZE
WITH rnk AS (
SELECT x1.id
, COUNT(x2.id) AS rnk
FROM xxx x1
LEFT JOIN xxx x2 ON x1.section_id = x2.section_id AND x2.name <= x1.name
GROUP BY x1.id
)
SELECT this.*
FROM xxx this
JOIN rnk ON rnk.id = this.id
WHERE rnk.rnk <=2
ORDER BY this.section_id, rnk.rnk
;

-- The same without using a CTE
-- EXPLAIN ANALYZE
SELECT this.*
FROM xxx this
JOIN ( SELECT x1.id
, COUNT(x2.id) AS rnk
FROM xxx x1
LEFT JOIN xxx x2 ON x1.section_id = x2.section_id AND x2.name <= x1.name
GROUP BY x1.id
) rnk
ON rnk.id = this.id
WHERE rnk.rnk <=2
ORDER BY this.section_id, rnk.rnk
;

相关讨论

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SELECT x.*
FROM (
SELECT section_id,
COALESCE
(
(
SELECT xi
FROM xxx xi
WHERE xi.section_id = xo.section_id
ORDER BY
name, id
OFFSET 1 LIMIT 1
),
(
SELECT xi
FROM xxx xi
WHERE xi.section_id = xo.section_id
ORDER BY
name DESC, id DESC
LIMIT 1
)
) AS mlast
FROM (
SELECT DISTINCT section_id
FROM xxx
) xo
) xoo
JOIN xxx x
ON x.section_id = xoo.section_id
AND (x.name, x.id) <= ((mlast).name, (mlast).id)

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