MySQL 来自 1 个以上的 sum() table

Question

SELECT fee +
    sum(((hourEnd-hourStart)*4) as sumAct3 
    from reserve join activity
    on reserve.idactivity=activity.idactivity
    join customer
    on customer.idcustomer=reserve.idcustomer
    where activity.typeAct=1 and customer.idcustomer='S1')
    +
    sum(((hourEnd-hourStart)*2) as sumAct2
    from reserve join activity
    on reserve.idactivity=activity.idactivity
    join customer
    on customer.idcustomer=reserve.idcustomer
    where activity.typeAct=2 and customer.idcustomer='S1')
FROM customer 
WHERE idcustomer='S1';

好的，所以每个 sum() returns 都是想要的值，我已经检查过了，但是我在尝试将所有内容加在一起然后得到总值时遇到了问题。

基本上我必须这样做：

初始费用（是"customer"table上定义的设定值）+activity1成本总和+activity2成本总和==值

但我不知道如何让它工作所以我得到了所有的总和，我尝试了一段时间但我总是得到 "SELECT is not valid at this position"

Answer 1

您只需 JOIN 将 table 彼此 table 并使用条件聚合来获得 activity 总和，即可更轻松地获得小时总和：

SELECT c.idcustomer, 
       SUM(CASE WHEN a.typeAct = 1 THEN hourEnd - hourStart ELSE 0 END) * 4 AS sumAct1,
       SUM(CASE WHEN a.typeAct = 2 THEN hourEnd - hourStart ELSE 0 END) * 2 AS sumAct2
FROM customer c
JOIN reserve r ON r.idcustomer = c.idcustomer
JOIN activity a ON a.idactivity = r.idactivity
WHERE a.typeAct IN (1, 2) AND c.idcustomer = 'S1'
GROUP BY c.idcustomer

然后使用这个结果作为派生的 table 进行最后的加法：

SELECT c.idcustomer, c.fee + f.sumAct1 + f.sumAct2
FROM customer c
JOIN (
    SELECT c.idcustomer, 
           SUM(CASE WHEN a.typeAct = 1 THEN hourEnd - hourStart ELSE 0 END) * 4 AS sumAct1,
           SUM(CASE WHEN a.typeAct = 2 THEN hourEnd - hourStart ELSE 0 END) * 2 AS sumAct2
    FROM customer c
    JOIN reserve r ON r.idcustomer = c.idcustomer
    JOIN activity a ON a.idactivity = r.idactivity
    WHERE a.typeAct IN (1, 2) AND c.idcustomer = 'S1'
    GROUP BY c.idcustomer
) f ON f.idcustomer = c.idcustomer

MySQL 来自 1 个以上的 sum() table

MySQL sum() from more than 1 table

mysql

sql

database

sum

multi-table