Question

我创建了一个表格：

Create table rooms
(
roomNo int not null,
hotelNo int not null,
hotelName varchar (40) not null,
roomType varchar (40)
default 'standard' check (roomType in ('single', 'double', 'master', 
'standard', 'president')),
price decimal (8,2) check (price>=0 and price<=99999),
guestNo int null
)

roomNo和hotelNo是主键，guestNo是外键。

该表填充了以下值：

(101, 123456, 'Home','standard',300, 54321)
(101, 654321, 'Ibis','standard',300, Null)
(101, 987654, 'Hiton','president',15000, null)
(222, 987654, 'Hiton','standard',300, 12345)
(321, 987654, 'Hiton','master',1000, 11111)
(333, 123456, 'Home','single',100, null)
(333, 987654, 'Hiton','standard',300, 12345)
(567, 654321, 'Ibis','standard',300, null)
(888, 654321, 'Ibis','single' ,100, null)
(999, 654321, 'Ibis','double' ,200, null)

如何回答“选择同一人预订的房间的总价”这个问题？

我希望总价为600，但根据此查询，我得到1900：

select sum(price)
from rooms r inner join guests g
on r.guestNo = g.guestNo and g.guestName = g.guestName

Answer 1

如果您想要每个人的总价，则需要按该值分组：

select g.guestNo, sum(r.price)
from rooms r inner join guests g
on r.guestNo = g.guestNo
group by g.guestNo
order by g.guestNo;

   GUESTNO SUM(R.PRICE)
---------- ------------
     11111         1000
     12345          600
     54321          300

如果您只对进入一个以上房间的客人感兴趣，则可以添加一个having子句以检查有多少个房间：

select g.guestNo, sum(r.price)
from rooms r inner join guests g
on r.guestNo = g.guestNo
group by g.guestNo
having count(r.roomNo) > 1
order by g.guestNo;

   GUESTNO SUM(R.PRICE)
---------- ------------
     12345          600

如何选择同一人预订的房间的总价

1 个答案: