Question

我有两张表，如下面

书

id TID reserveddate fee1 fee2 fee3 Noofplayers total
-------------------------------------
1   11  2012-11-25 100  200  300      3        800
2   12  2012-11-25 100  200  300      3        800
3   13  2012-11-28 200  100  200      1        500

球员

 TID  fee1 fee2 
-------------------------------------
  11  100  200  
  11  100  200  
  11  100  200 
  12  100  200  
  12  100  200
  12  100  200
  13  200  100

现在我想通过Reserveddate从两个表中检索数据。

我有一个查询，

select sum(b.fee1)+sum(b.fee2)+sum(a.fee3) as total 
from Players b,book a where a.TID = b.TID  and a.ReservedDate ='25-nov-2012' 
group by a.ReservedDate

以上查询总和（fee3）添加3（noofplayers）次我想添加一本时间表表。

请按照客户要求更新查询...

Answer 1

我相信这会奏效，总共会给出2400：

select  sum(b.fee1_fee2_totalPerPlayer) + sum(a.fee3) as total 
from    book a
join    (
            select  TID,
                    sum(fee1) + sum(fee2) 
                        as fee1_fee2_totalPerPlayer
            from    Players
            group   by TID
        ) b on
        a.TID = b.TID
where   a.ReservedDate ='2012-11-25' 
group by a.ReservedDate

以下是计算总数的方法：

对于TID = 11：300（Players.fee1的总和）+ 600（Players.fee2的总和）= 900

对于TID = 12：300（Players.fee1之和）+ 600（Players.fee2之和）= 900

900 + 900 = 1800

1800 + 600（日期'2012-11-25'的Book.fee3总和）= 2400

以下是一个示例： SQL Fiddle

Answer 2

所以你想每天SUM？您可以使用带SUM(...)OVER的CTE：

WITH CTE AS
(
   SELECT b.id, b.TID AS bTID, b.reserveddate, b.fee1, b.fee2, b.fee3, b.Noofplayers
     ,    p.TID AS pTID,  p.fee1 AS pFee1, p.fee2 AS pFee2
     ,    SUM(b.fee1 + b.fee2 + a.fee3) OVER (PARTITION BY DATEADD(day, DATEDIFF(dd, 0, ReservedDate), 0)) As total
     ,    DATEADD(day, DATEDIFF(dd, 0, ReservedDate), 0) AS ReservedDateDay
   FROM Book b INNER JOIN Players p ON b.TID=p.TID
)
SELECT total FROM CTE WHERE ReservedDateDay='20121125';

Answer 3

如果您确定 noofplayers 是播放器表中的条目数，您可以简单地将 fee3 的总和除以 noofplayer

select sum(b.fee1)+sum(b.fee2)+(sum(a.fee3)/noofplayer) as total 
from Players b,book a where a.TID = b.TID  and a.ReservedDate ='25-nov-2012' 
group by a.ReservedDate

如果你不确定 noofplayer 是否正确，你可以在同一条记录上过滤玩家表，如

select sum(b.fee1)+sum(b.fee2)+sum(a.fee3) as total 
from book a 
join (select distinct TID, fee1, fee3 from players) as b
on a.TID = b.TID
where  a.ReservedDate ='25-nov-2012' 
group by a.ReservedDate

如何从带有条件的两个表中检索数据总和

3 个答案: