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mysql - 数学运算后得到mysql SUM

转载 作者:行者123 更新时间:2023-11-29 13:41:11 24 4
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这是我的查询的一部分

(SUM(dr.drv)/100)*
(st.suppliers1_value+st.suppliers2_value+st.suppliers3_value+st.suppliers4_value)

这将生成多行。我想得到所有这些的总和。我试过了

SUM(SUM(dr.drv)/100)*
(st.suppliers1_value+st.suppliers2_value+st.suppliers3_value+st.suppliers4_value)
as total drv_amount

但是它抛出了一个错误。有人可以引导我走向正确的方向吗?

这是整个查询:

SELECT SUM(dr.tickets_sold) as tickets_total_amount,
SUM(dr.drv) as drv_total_amount,
FORMAT(((SUM(dr.drv)/100) * (st.suppliers1_value+st.suppliers2_value+st.suppliers3_value+st.suppliers4_value) ),2 ) as supplier_commission,
FORMAT(((SUM(dr.drv)/100) * st.exhibitor_value),2) as exhibitor_commission,
FORMAT(((SUM(dr.drv)/100) * st.circuit_value),2) as distributer_commission,
FORMAT(((SUM(dr.drv)/100) * (st.suppliers1_value+st.suppliers2_value+st.suppliers3_value+st.suppliers4_value)+((SUM(dr.drv)/100) * st.circuit_value )),2) as film_hire_total
FROM com_pro_dr as dr
INNER JOIN com_pro_ratecard_rates AS rt
ON (dr.movie_id=rt.movie_id and
dr.theater_id=rt.theater_id and
dr.showtime_id=rt.showtime_id and
dr.category_id=rt.category_id and
dr.applied_date = rt.date_apply)
INNER JOIN com_pro_ratecard as rc ON (rc.id=rt.ratecard_id)
INNER JOIN com_pro_movie as m ON (m.id=dr.movie_id)
INNER JOIN com_pro_theater as t ON (t.id=dr.theater_id)
INNER JOIN com_pro_share as st ON (st.id=rc.share_id)
INNER JOIN com_pro_theater_dr as td ON (td.theater_id=dr.theater_id)
INNER JOIN com_pro_circuit as c ON (c.id =td.circuit_id)
WHERE 1

最佳答案

你总是可以做这样的事情:

SELECT SUM(val)
FROM
(
SELECT (SUM(dr.drv)/100)*
(st.suppliers1_value+st.suppliers2_value+st.suppliers3_value+st.suppliers4_value)
AS val
/* the rest of your current query */
) AS a

关于mysql - 数学运算后得到mysql SUM,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/18057617/

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