从两个表连接两列的MySQL SUM（）

Question

我有两个表，我需要连接并获得每个表中一列的SUM（）。

mysql> select * from day_sales2; 
    +--------+------------+ 
    | id  | ds_oversht | 
    +--------+------------+ 
    | 119263 |  -0.17 | 
    | 119336 |  0.55 | 
    | 119409 |  3.08 | 
    | 119482 |  -33.25 | 
    | 119555 |  1.27 | 
    | 119628 |  0.32 | 
    | 119701 |  -0.15 | 
    | 119774 |  6.57 | 
    | 119847 |  -0.06 | 
    | 119920 |  -12.04 | 
    | 119993 |  -12.40 | 
    | 120066 |  -38.90 | 
    | 120139 |  20.56 | 
    | 120212 |  -17.11 | 
    | 120285 |  -0.51 | 
    | 120358 |  0.04 | 
    | 120431 |  3.23 | 
    | 120504 |  -0.89 | 
    | 120577 |  2.13 | 
    | 120650 |  0.57 | 
    | 120723 |  -1.15 | 
    | 120796 |  1.99 | 
    | 120869 |  9.29 | 
    | 120942 |  6.95 | 
    | 121015 |  3.22 | 
    | 121088 |  -0.59 | 
    | 121161 |  -0.89 | 
    | 121234 |  -1.00 | 
    +--------+------------+ 

    mysql> select * from over_short_adj2; 
    +-------+---------+-----------+ 
    | id | main_id | cd_amount | 
    +-------+---------+-----------+ 
    | 45881 | 119920 |  0.66 | 
    | 45882 | 119920 |  1.19 | 
    | 45907 | 119920 |  4.81 | 
    | 46017 | 120212 |  3.35 | 
    | 46018 | 120066 |  16.85 | 
    +-------+---------+-----------+ 

我使用这个查询来弄完：

SELECT SUM(t1.ds_oversht) sum1, SUM(t2.cd_amount) sum2 
    FROM day_sales2 t1 
    LEFT JOIN over_short_adj2 t2 ON t1.id = t2.main_id 

但SUM1不正确，因为联接是一对多。我怎么能得到这个查询工作？

Answer 1

你应该得到的第一个day_sales2的SUM做LEFT JOIN前：

SELECT 
    t1.id, 
    t1.sum1, 
    SUM(t2.cd_amount) AS sum2 
FROM (
    SELECT id, SUM(ds_oversht) AS sum1 
    FROM day_sales2 
    GROUP BY id 
) t1 
LEFT JOIN over_short_adj2 t2 
    ON t1.id = t2.main_id 
GROUP BY t1.id, t1.sum1 

编辑：

如果你只想要一个行，我觉得不需要JOIN：

SELECT 
    (SELECT SUM(ds_oversht) FROM day_sales2), 
    (SELECT SUM(cd_amount) FROM over_short_adj2) 

Answer 2

如果你只想从两个表中获得一行总和，你可以使用UNION

SELECT SUM(ds_oversht) from day_sales2 
UNION 
SELECT SUM(cd_amount) from over_short_adj2; 

Answer 3

SELECT SUM(sum1) sum1, SUM(sum2) sum2 FROM (
    SELECT 
     t1.id, 
     t1.sum1, 
     SUM(t2.cd_amount) sum2 
    FROM (
     SELECT id, SUM(ds_oversht) sum1 
     FROM day_sales2 
     GROUP BY id 
    ) t1 
    LEFT JOIN over_short_adj2 t2 
     ON t1.id = t2.main_id 
    GROUP BY t1.id, t1.sum1 
    ) t3 

这确实它，谢谢费利克斯的答案。