根据匹配条件连接表，并从Table1的总和中减去Table2.Seantity的数量。匹配行的数量

Question

我使用下面的代码：table1-add，table2-remove

SELECT add.description, add.size, (sum(add.qty) - sum(remove.qty)) AS OnHand
`FROM add LEFT JOIN remove ON (add.description = remove.description) AND `(add.size = remove.size) 
group by add.description, add.size
OnHand FROM add RIGHT JOIN remove ON (add.description = remove.description) AND 
(add.size = remove.size) WHERE add.description IS NULL and add.size is null 
group by add.description, add.size;

我的目标在标题中表达。

我尝试了一些其他的方法，但它没有在减法之前对列进行求和。 它仅减去第一个匹配的条目。

其他方法：

SELECT add.description, add.size, sum(add.qty) - sum(remove.qty) AS OnHand
FROM add INNER JOIN remove 
ON add.description = remove.description
WHERE add.size = remove.size
GROUP BY add.description, add.size;

Answer 1

看起来你需要先做组，计算每个表的总和 - 然后进行减法;

SELECT t1.description, t1.size, (sum_add - sum_rem) AS OnHand
FROM 
( SELECT add.description, add.size, sum(add.qty) as sum_add from add GROUP BY add.description, add.size )t1 
INNER JOIN 
( SELECT remove.description, remove.size, sum(remove.qty) as sum_rem from remove GROUP BY remove.description, remove.size )t2
ON t1.description = t2.description and t1.size = t2.size

代码未经测试，可能以更好的方式完成 - 但这应该有效。