Question

我创建了三个表

CREATE TABLE guest(

name varchar(100),
ranking int,
PRIMARY KEY (name)
);

CREATE TABLE room(

 roomname varchar(100),
 wallcolor varchar(100),
 rating int,
 PRIMARY KEY(roomnane)
 );

 CREATE TABLE reservation(

 name varchar(100),
 roomname varchar(100),
 day varchar(100),
 moveinday int,
 moveoutday int,
 PRIMARY KEY(roomname, day, start, finish),
  FOREIGN KEY(roomname) REFERENCES room(roomname),
  FOREIGN KEY(name) REFERENCES guest(name)
 );

我正在尝试创建两个查询

对于每个预订的房间，列出客人的平均排名保留那个房间。我无法列出每个房间的平均值。这是我试过的

select reservation.roomname, AVG(guest.ranking)
from reservation, room, guest
 where reservation.roomname == room.roomname AND reservation.name = guest.name

这不能提供所需的结果。我该如何修复我的查询？

查询以确认房间没有冲突的预订。（同一个房间不能为多人搬家和搬家日保留。我不知道如何开始这个

Answer 1

别忘了GROUP BY：

SELECT reservation.roomname, AVG(guest.ranking)
FROM reservation, room, guest
WHERE reservation.roomname = room.roomname 
AND reservation.name = guest.name
GROUP BY reservation.roomname

Answer 2

您应该学会使用正确的join语法和表别名。

select r.roomname, AVG(g.ranking)
from room r join
     guest g
     on r.name = g.name
group by r.roomname;

此外，实际上并不需要room表。

Answer 3

关于你的第二个问题，基本应该是这样一个联接：

select room.roomname 
from room
join reservation r1 on room.roomname = r1.roomname 
join reservation r2 on room.roomname = r2.roomname

然后where子句是复杂的部分：

where (r1.moveinday < r2.moveinday and r1.moveoutday > r2.moveoutday)
or (r1.moveinday < r2.moveinday and r1.moveoutday > r2.moveinday)
or (r1.moveinday < r2.moveoutday and r1.moveoutday > r2.moveoutday)

第一种情况：r1完全包含r2。

第二种情况：r1与r2中的移动重叠

第三种情况：r1与r2的移出日重叠

你可能想要使用＆lt; =，＆gt; =，取决于你如何考虑这种情况。如果你想使用＆lt; =，＆gt; =（而不是＆lt;。＆gt;），我建议你为每个预订添加一个唯一的ID，因为你必须添加到where子句中，例如 r1.ID＆lt;＆gt; r2.ID

SQL查询查找平均值

3 个答案: