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I have a foo table which has a one-to-many relationship with the BAR table. FOO also has a many-to-many relationship with QUX.

To illustrate:

FOO
+--------+
| fid    | 
+--------+
|    1   |
|    2   |
|    3   |
|    4   |
+--------+


BAR
+------+-------+-----+
| bid  | cost  | fid |
+------+-------+-----+
|    1 |     2 |  1  |  
|    2 |     4 |  1  |
|    3 |     8 |  2  |
|    4 |    42 |  3  |
|    5 |    21 |  3  |
|    6 |     5 |  4  |
|    7 |    98 |  2  |
|    8 |    30 |  4  |
+------+-------+-----+

FOOQUX
+------+------+
| fid  | qid  |
+------+------+
|    1 |    2 |
|    2 |    1 |
|    2 |    3 |
|    4 |    1 |
+------+------+

QUX
+------+
| qid  |
+------+
|    1 |
|    2 | 
|    3 |
|    4 |
+------+

Now I want to calculate the sum of BAR.cost for every FOO entry, but I want to filter by QUX in terms of "FOO has an association with any element of QUX"

I tried using this sentence:

SELECT fid, sum(cost)
FROM FOO
INNER JOIN BAR on BAR.fid = FOO.fid
LEFT JOIN FOOQUX ON FOOQUX.fid = FOO.fid

But this returns an entry for each instance of FOO in FOOQUX, so in the case of fid=2 (which has two QUX associated to it) we'll have double the amount of cost than in reality.

Maybe this is a simple problem but I'm struggling on how to solve it.

4

1 回答 1

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我想这会对你有所帮助。Group by foo.fid 将保证每条 foo 记录只会出现一次，并加上其 bar-costs 的总和。Having过滤聚合记录：仅当 FOOQUX 表包含其fid.

SELECT fid, SUM(cost)
FROM foo JOIN bar on bar.fid=foo.fid
GROUP BY foo.fid
HAVING foo.fid IN (SELECT fid FROM FOOQUX)
于 2013-06-20T14:08:01.287 回答