Question

我有4张桌子：

create table Hotel (
numHotel int primary key,
nomHotel varchar(30),
ville varchar(30),
etoiles tinyint
);

create table Chambre (
numChambre int identity primary key,
numHotel int foreign key references Hotel(numHotel),
etage tinyint,
prixnuit smallmoney not null check (prixnuit>=100)
);

create table Client (
cinClient varchar(10) primary key,
nom varchar(30),
prenom varchar(30),
adresse varchar(255) default 'non renseignée',
telephone varchar(10) check (telephone like '0%' and len(telephone)=10)
);

create table reservation (
numReservation int identity primary key,
numChambre int foreign key references Chambre(numChambre),
numCl varchar(10) foreign key references Client(cinClient),
dateArrivee date,
dateDepart date,
constraint ck_dates check ((Datediff(day,dateArrivee,dateDepart)>0))
);

列名是法语，但我认为它们是可以理解的。

所以问题是：我需要选择总价高于某个价格（例如10000）的酒店名称（nomHotel）

这就是我在纸上所做的：

select nomHotel from Hotel h join Chambre ch on h.numHotel=ch.numChambre join reservation r on ch.numChambre=r.numChambre
group by nomHotel
having COUNT(r.numChambre)*ch.prixnuit>10000

而且（当然）我弄错了。感谢任何帮助，谢谢。

翻译：＆＃34; prixnuit＆＃34;是每晚的房间（chambre）价格。

Answer 1

总收入（我认为）基于预订数量。我想你需要计算每次预订的晚数，不是吗？所以，也许：

select nomHotel,
    sum(Datediff(day,r.dateArrivee,r.dateDepart)*ch.prixnuit)
from Hotel h join Chambre ch on h.numHotel=ch.numChambre join reservation r       
on ch.numChambre=r.numChambre
group by nomHotel
having sum(Datediff(day,r.dateArrivee,r.dateDepart)*ch.prixnuit)>10000

Answer 2

select nomHotel, sum(Datediff(day,dateArrivee,dateDepart) * prixnuit)
from Hotel h join Chambre ch on h.numHotel=ch.numHotel 
           join reservation r on ch.numChambre=r.numChambre
group by nomHotel
having sum(Datediff(day,dateArrivee,dateDepart) * prixnuit)>1000

加入多个表+ HAVING子句

2 个答案: