Dav*_*d E 2 sql postgresql group-by greatest-n-per-group
这可能是一个标准问题,我已经排除了其他一些每组最佳答案,但到目前为止无法解决我当前的问题。
A B C
+----+-------+ +----+------+ +----+------+-------+
| id | start | | id | a_id | | id | b_id | name |
+----+-------+ +----+------+ +----+------+-------+
| 1 | 1 | | 1 | 1 | | 1 | 1 | aname |
| 2 | 2 | | 2 | 1 | | 2 | 2 | aname |
+----+-------+ | 3 | 2 | | 3 | 3 | aname |
+----+------+ | 4 | 3 | bname |
+----+------+-------+
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用英语我想完成的是:
我尝试过的SQL如下:
SELECT a.id, a.start, c.id, c.name
FROM a
INNER JOIN (
SELECT id, MAX(start) as start
FROM a
GROUP BY id
) a2 ON a.id = a2.id AND a.start = a2.start
JOIN b
ON a.id = b.a_id
JOIN c
on b.id = c.b_id
GROUP BY c.name;
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它失败并出现以下错误:
ERROR: column "a.id" must appear in the GROUP BY clause or be used in an aggregate function Position: 8
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为了有用,我确实需要查询中的 id,但无法对它们进行分组,因为它们是唯一的。这是我想要的上述第一种情况的输出示例:
+------+---------+------+--------+
| a.id | a.start | c.id | c.name |
+------+---------+------+--------+
| 2 | 2 | 3 | aname |
| 2 | 2 | 4 | bname |
+------+---------+------+--------+
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这是一个Sqlfiddle
编辑-删除了第二种情况
情况1
select distinct on (c.name)
a.id, a.start, c.id, c.name
from
a
inner join
b on a.id = b.a_id
inner join
c on b.id = c.b_id
order by c.name, a.start desc
;
id | start | id | name
----+-------+----+-------
2 | 2 | 3 | aname
2 | 2 | 4 | bname
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案例2
select distinct on (c.name)
a.id, a.start, c.id, c.name
from
a
inner join
b on a.id = b.a_id
inner join
c on b.id = c.b_id
where
b.a_id in (
select a_id
from b
group by a_id
having count(*) > 1
)
order by c.name, a.start desc
;
id | start | id | name
----+-------+----+-------
1 | 1 | 1 | aname
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