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，所以每个sum（）返回所需的值，我已经检查过了，但是在尝试将所有内容加在一起然后获得总值时遇到了问题。

基本上我必须这样做：

初始费用（这是在“客户”表上定义的设置值）+活动1费用之和+活动2费用之和==值

但是我不知道如何使它工作，所以我得到了所有东西的总和，我尝试了一段时间，但总是得到“ SELECT在此位置无效”

Answer 1

通过将表相互JOIN并使用条件聚合来获得活动总和，您可以更容易地获得小时总和：

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

然后使用此结果作为派生表进行最终添加：

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

来自1个以上表的MySQL sum（）

1 个答案: