Question

我试图在两个日期之间的E1_CHECK为真时对行进行求和。尽管我最好解释这里是我的代码

select e.e1_fecha2,a.a1_nombre,a.A1_NUMCUENTA,f.CF_KIT,e.E1_CC,e.E1_MONTo,f.CF_NUM,f.CF_MONTO
from
e1 as e inner join cf as f on e.CF_num=f.CF_num
inner join a1 as a on f.a1_ID=a.a1_id
where
e.E1_CHECK=true and E1_FECHA2 between '2016-02-01' and '2016-02-29'

它显示了这个

 e1_fecha2 | a1_nombre | A1_NUMCUENTA | CF_KIT    | E1_CC | E1_Monto | CF_NUM        |CF_Monto
-------------------------------------------------------------------------------------------
 2016-02-29|khaldun    |123456        | 55        |1/5    |1320000   |001-001-0000002|6600000
 2016-02-29|khaldun    |123456        |8 cameras  |1/5    |1100000   |001-001-0000003|5500000 
 2016-02-29|khaldun    |123456        |8 cameras  |2/5    |1100000   |001-001-0000003|5500000
 2016-02-29|oiuiouio   |64            | fff       |1/1    |1375000   |001-001-0000004|1375000 
 2016-02-29|khaldun    |123456        | 44444     |1/1    |5500000   |444-444-4422222|5500000
 2016-02-29|khaldun    |123456        |rrrr       |1/1    |1125000   |001-001-0000006|1125000

和E1_Monto是付款的一部分，CF_MONTO是我想要的总账单减去CF_Monto - E1_MONTO以及需要支付多少费用

这就是我想要的样子

 e1_fecha2 | a1_nombre | A1_NUMCUENTA | CF_KIT    | E1_CC | E1_Monto | CF_NUM        | CF_Monto  | Payed     | Left_to_Pay 
---------------------------------------------------------------------------------------------------------------------------
 2016-02-29|khaldun    |123456        | 55        |1/5    |1320000   |001-001-0000002| 6600000   |1320000    |5280000
 2016-02-29|khaldun    |123456        |8 cameras  |1/5    |1100000   |001-001-0000003| 5500000   |1100000    |4400000
 2016-02-29|khaldun    |123456        |8 cameras  |2/5    |1100000   |001-001-0000003| 5500000   |2200000    |3300000
 2016-02-29|oiuiouio   |64            | fff       |1/1    |1375000   |001-001-0000004| 1375000   |1375000    |0
 2016-02-29|khaldun    |123456        | 44444     |1/1    |5500000   |444-444-4422222| 5500000   |5500000    |0
 2016-02-29|khaldun    |123456        |rrrr       |1/1    |1125000   |001-001-0000006| 1125000   |1125000    |0

我尝试了其他方法，但它只显示了1行并且总结了所有东西。

Answer 1

只需选择列并执行操作（减去）。我改变了

SELECT *, E1_MONTO AS payed,CF_Monto - E1_MONTO AS left_to_pay这一行。希望它有所帮助。

SELECT *, E1_MONTO  AS payed,CF_Monto - E1_MONTO AS left_to_pay
FROM e1 AS e INNER JOIN cf AS f ON e.CF_num=f.CF_num
INNER JOIN a1 AS a ON f.a1_ID=a.a1_id
WHERE
e.E1_CHECK=true AND E1_FECHA2 BETWEEN '2016-02-01' AND '2016-02-29'

Answer 2

我想出了怎么做这就是我做的事情

select
e.e1_fecha2,e.e1_check, a.a1_nombre,a.A1_NUMCUENTA,f.CF_KIT,e.E1_CC,e.E1_MONTo, 
(select sum(IF(t.E1_CHECK = true, t.e1_monto , 0)) from e1 as t where e.CF_NUM=t.cf_num ) as payed,
CF_Monto - (select sum(IF(t.E1_CHECK = true, t.e1_monto , 0)) from e1 as t where e.CF_NUM=t.cf_num )AS left_to_pay
from e1 as e
inner join cf as f on e.CF_num=f.CF_num
inner join a1 as a on f.a1_ID=a.a1_id
where 
e.E1_CHECK=true and E1_FECHA2 between '2016-02-01' and '2016-02-29'

感谢大家的帮助：）

MySQL查询Sum和Substract

2 个答案: