Question

    SELECT AVG(`col5`)
    FROM `table1`
    WHERE `id` NOT IN (
        SELECT `id` FROM `table2`
        WHERE `col4` = 5
    )
    group by `col2` having sum(`col3`) > 0
UNION
    SELECT MAX(`col5`)
    FROM `table1`
    WHERE `id` NOT IN (
        SELECT `id` FROM `table2`
        WHERE `col4` = 5
    )
    group by `col2` having sum(`col3`) = 0

出于可读性和性能原因，我认为这段代码可以重构。但是如何？

EDITIONS

删除了外部选择
使第一个选择返回一个总和而第二个选择返回另一个值
将SUM替换为AVG

Answer 1

SELECT * 
FROM table1 t1
left outer join table2 t2 on t1.id = t2.id and t2.col4 = 5
where t2.id is null
group by t1.col2 
having sum(col3) >= 0

外部选择缺少FROM子句，并没有添加任何内容，因此我删除了它。与LEFT OUTER JOIN方法相比，NOT IN效率低，所以我替换了它。使用UNION可以很容易地将两个>=组合成一个。

<强>更新请注意使用UNION ALL而不是UNION。我不想想你想删除重复项，它会以这种方式执行得更快。

SELECT AVG(t1.col5) 
FROM table1 t1 
left outer join table2 t2 on t1.id = t2.id and t2.col4 = 5 
where t2.id is null 
group by t1.col2  
having sum(t1.col3) > 0  
UNION ALL
SELECT MAX(t1.col5) 
FROM table1 t1 
left outer join table2 t2 on t1.id = t2.id and t2.col4 = 5 
where t2.id is null 
group by t1.col2  
having sum(t1.col3) = 0

Answer 2

SELECT t1.* FROM table1 t1 LEFT OUTER JOIN table2 t2 ON t1.id = t2.id 
WHERE t2.col4 <> 5 AND SUM(t1.col3) > 0 GROUP BY t1.col2

Answer 3

我猜你想要这个：

SELECT * FROM `table`
WHERE col2 IN
(SELECT col2
    FROM `table1` 
    WHERE `id` NOT IN ( 
        SELECT `id` FROM `table2` 
        WHERE `col4` = 5 
    ) 
    group by `col2` having sum(`col3`) >= 0
)

使用GROUP BY时，只应返回GROUP BY子句中指定的列或包含聚合函数的列。因此，此处的内部SELECT获取总和大于或等于零的col2值，然后外部SELECT获取这些值的整个行。

Answer 4

SELECT
    IF(SUM(`table1`.`col3`) > 0, AVG(`table1`.`col5`), MAX(`table1`.`col5`))
FROM `table1`
    LEFT JOIN `table2` ON `table2`.`id` = `table1`.`id` AND `table2`.`col4` = 5
WHERE `table2`.`id` IS NULL
GROUP BY `table1`.`col2`
HAVING SUM(`table1`.`col3`) >= 0

同样*被认为是有害的。如果要使查询向前兼容以后可能对数据库模型进行更改，请按名称指定列。

如何重构这个MySQL代码？

4 个答案: