这只是一次大学练习,所以我并不着急。但是我想得到答案。考虑这些表和行:
create table course (
numco integer,
nameco varchar(20) not null,
primary key (numco)
);
create table prereq (
numco integer,
numcoprereq integer,
primary key (numco, numcoprereq),
foreign key (numco) references course (numco),
foreign key (numcoprereq) references course (numco)
);
insert into course values (1, 'course 1');
insert into course values (2, 'course 2');
insert into course values (3, 'course 3');
insert into course values (4, 'course 4');
insert into course values (1, 'course 5');
insert into course values (2, 'course 6');
insert into prereq values (4, 2);
insert into prereq values (2, 1);
我没有做到这一点。我只是翻译它并删除了一些不相关的位。我知道有两个course
行具有相同的PK,但我不得不问我的老师。
他让我们写一个查询来获取课程的名称和课程的先决条件的名称。如果它只是课程的编号,那么左连接就可以了。但我无法使用这些名称。它应输出:
course 1 (null)
course 2 course 1
course 3 (null)
course 4 course 2
course 5 (null)
course 6 course 1
我经常搜索并写了两次尝试:
select C1.nameco, C2.nameco as namecoprereq
from course C2 left join
(course C1 join prereq P on C1.numco = P.numco)
on C2.numco = P.numcoprereq;
select C1.nameco, C2.nameco as namecoprereq
from (course C1 join prereq P on C1.numco = P.numco)
left join course C2 on C2.numco = P.numcoprereq;
他们分别输出:
course 2 course 1
course 6 course 1
course 4 course 2
(null) course 3
(null) course 4
course 2 course 5
course 6 course 5
course 4 course 6
和
course 2 course 1
course 2 course 5
course 4 course 2
course 4 course 6
course 6 course 1
course 6 course 5
我知道(我想,我还没试过)我可以用联盟和减号来做,但我想知道是否可以在单个选择上进行。如果没有kludging怎么做呢?
答案 0 :(得分:1)
您的查询几乎可以得到正确的答案,但我认为您的搜索有点远。只需用“英语”来想一想:
FROM course
)LEFT JOIN prereq
)LEFT JOIN course
)这应该有效:
SELECT
c1.nameco,
c2.nameco AS nameprereq
FROM course AS c1
LEFT JOIN prereq AS p ON p.numcoprereq = c1.numco
LEFT JOIN course AS c2 ON c2.numco = p.numco