获取 4 个不同的总和结果 table

Question

我有 4 个 table 具有相同的列（id & banka_id & miktar）。

2 table 的收入（drink_earn、eat_earn）和

2 table 秒支出 (drink_spending, eat_spending )

|id|banka_id|miktar|
---|--------|------|
|1 |2       |20    |
|2 |1       |20    |
|3 |1       |30    |

我的问题。

如何一次性计算所有行？

drink_earn : table SUM(miktar) = 50$

eat_earn : table SUM(miktar) = 50$

drink_spending : table SUM(miktar) = 15$

eat_spending : table SUM(miktar) = 5$

收入：drink_earn + eat_earn = 100$

消费：drink_spending + eat_spending 20$

总计：收入 - 支出 = 80 美元

最后 GROUP BY banka_id

能否请您编写代码示例。问候。

Answer 1

使用 UNION ALL 将所有 table 合并为一个：

select banka_id, sum(earn) - sum(spend) as total
from (
    select banka_id, miktar as earn, 0 as spend
    from drink_earn
    union all
    select banka_id, miktar as earn, 0 as spend
    from eat_earn
    union all
    select banka_id, 0 as earn, miktar as spend
    from drink_spending
    union all
    select banka_id, 0 as earn, miktar as spend
    from eat_spending
) t
group by banka_id

但是，如果您将所有数据都放在一个 table 中，再加上一个额外的类型列，那就更好了。

获取 4 个不同的总和结果 table

Get Total SUM Result of 4 differents table

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mysqli

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