[英]get calculation from two tables related with a third table
我有这张桌子
table1
| id | name |
| 1 | axe |
| 2 | bow |
| 3 | car |
| 4 | dart |
这两个表
table2 table3
| t1_id | number | | t1_id | letter |
| 1 | 5 | | 1 | a |
| 1 | 6 | | 1 | b |
| 1 | 2 | | 1 | c |
| 2 | 2 | | 2 | a |
| 2 | 2 | | 2 | c |
| 2 | 3 | | 2 | r |
| 3 | 8 | | 3 | y |
| 3 | 3 | | 3 | i |
| 3 | 1 | | 3 | a |
| 4 | 8 | | 4 | a |
| 4 | 9 | | 4 | b |
| 4 | 10 | | 4 | c |
其中t1_id是table1 id
我想做的事是让具有表3字母ABC和他们的人数像这样为了通过letter_count DESC AVG的所有table1的记录先然后由avg_numbers DESC
| id | name | letter_count | avg_number |
| 4 | dart | 3 | 9 |
| 1 | axe | 3 | 4.3333333333 |
| 2 | bow | 2 | 2.3333333333 |
| 3 | car | 1 | 4 |
我预期可以正常工作的查询是http://www.sqlfiddle.com/#!9/69086b/3/0
SELECT
t1.id,
t1.name,
COUNT(t3.letter) AS letter_count,
AVG(t2.number) AS avg_number
FROM
table1 t1
INNER JOIN
table2 t2
ON t2.t1_id = t1.id
LEFT JOIN
table3 t3
ON t3.t1_id = t1.id
AND t3.letter IN ('a', 'b', 'c')
GROUP BY
t1.id
ORDER BY
letter_count DESC,
avg_number DESC
但数字完全不同且准确,但顺序正确
我不想获取letter_count和avg_number值,但我只想按它们排序,但它们的值让我担心查询性能
我不会注意到这个怪异的值,因为我的实际查询是
SELECT
t1.id,
t1.name
FROM
table1 t1
INNER JOIN
table2 t2
ON t2.t1_id = t1.id
LEFT JOIN
table3 t3
ON t3.t1_id = t1.id
AND t3.letter IN ('a', 'b', 'c')
GROUP BY
t1.id
ORDER BY
COUNT(t3.letter) DESC,
AVG(t2.number) DESC
只给我适当的秩序
| id | name |
| 4 | dart |
| 1 | axe |
| 2 | bow |
| 3 | car |
但是在检查了值之后,我是否对letter_count感到惊讶, 是否只是忽略了这些值,这不会影响我的大表的性能?
您正在汇总两个不同的维度。 这导致笛卡尔积。 解决此问题的一种方法是在加入之前进行汇总:
SELECT t1.id, t1.name, t2.letter_count, t2.avg_number
FROM table1 t1 INNER JOIN
(SELECT t2.t1_id, AVG(t2.number) as avg_number
FROM table2 t2
GROUP BY t2.t1_id
) t2
ON t2.t1_id = t1.id LEFT JOIN
(SELECT t3.t2_id, COUNT(t3.letter) as letter_count
FROM table3 t3
WHERE t3.letter IN ('a', 'b', 'c')
GROUP BY t3.t2_id
) t3
ON t3.t1_id = t1.id
ORDER BY t3.letter_count DESC, t2.avg_number DESC;
在第三个表上,IN-SELECT子查询。
SELECT
t1.id,
t1.name,
(
SELECT count(letter)
FROM t3
where t3.t1_id = t1.id
) as lettercount,
AVG(t2.number) AS avg_number
FROM
table1 t1
INNER JOIN
table2 t2
ON t2.t1_id = t1.id
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