一套题学会多表查询,请认真做这一套!
练习2:
数据表介绍:
--1.学生表
Student(SId,Sname,Sage,Ssex)
--SId 学生编号,Sname 学生姓名,Sage 出生年月,Ssex 学生性别
--2.课程表
Course(CId,Cname,TId)
--CId 课程编号,Cname 课程名称,TId 教师编号
--3.教师表
Teacher(TId,Tname)
--TId 教师编号,Tname 教师姓名
--4.成绩表
SC(SId,CId,score)
--SId 学生编号,CId 课程编号,score 分数
学生表 Student
create table Student(SId varchar(10),Sname varchar(10),Sage datetime,Ssex varchar(10));
insert into Student values('01' , '赵雷' , '1990-01-01' , '男');
insert into Student values('02' , '钱电' , '1990-12-21' , '男');
insert into Student values('03' , '孙风' , '1990-12-20' , '男');
insert into Student values('04' , '李云' , '1990-12-06' , '男');
insert into Student values('05' , '周梅' , '1991-12-01' , '女');
insert into Student values('06' , '吴兰' , '1992-01-01' , '女');
insert into Student values('07' , '郑竹' , '1989-01-01' , '女');
insert into Student values('09' , '张三' , '2017-12-20' , '女');
insert into Student values('10' , '李四' , '2017-12-25' , '女');
insert into Student values('11' , '李四' , '2012-06-06' , '女');
insert into Student values('12' , '赵六' , '2013-06-13' , '女');
insert into Student values('13' , '孙七' , '2014-06-01' , '女');
科目表 Course
create table Course(CId varchar(10),Cname nvarchar(10),TId varchar(10)); insert into Course values('01' , '语文' , '02');
insert into Course values('02' , '数学' , '01');
insert into Course values('03' , '英语' , '03');
教师表 Teacher
create table Teacher(TId varchar(10),Tname varchar(10));
insert into Teacher values('01' , '张三');
insert into Teacher values('02' , '李四');
insert into Teacher values('03' , '王五');
成绩表 SC
create table SC(SId varchar(10),CId varchar(10),score decimal(18,1)); insert into SC values('01' , '01' , 80);
insert into SC values('01' , '02' , 90);
insert into SC values('01' , '03' , 99);
insert into SC values('02' , '01' , 70);
insert into SC values('02' , '02' , 60);
insert into SC values('02' , '03' , 80);
insert into SC values('03' , '01' , 80);
insert into SC values('03' , '02' , 80);
insert into SC values('03' , '03' , 80);
insert into SC values('04' , '01' , 50);
insert into SC values('04' , '02' , 30);
insert into SC values('04' , '03' , 20);
insert into SC values('05' , '01' , 76);
insert into SC values('05' , '02' , 87);
insert into SC values('06' , '01' , 31);
insert into SC values('06' , '03' , 34);
insert into SC values('07' , '02' , 89);
insert into SC values('07' , '03' , 98);
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1.查询" 01 "课程比" 02 "课程成绩高的学生的信息及课程分数
select * from Student RIGHT JOIN
( select t1.SId, class1, class2 from
(select SId, score as class1 from sc where sc.CId = '01')as t1,
(select SId, score as class2 from sc where sc.CId = '02')as t2
where t1.SId = t2.SId AND t1.class1 > t2.class2
)r
on Student.SId = r.SId;
因为需要全部的学生信息,则需要在sc表中得到符合条件的SId后与student表进行join,可以left join 也可以 right join。
2、查询存在" 01 "课程但可能不存在" 02 "课程的情况(不存在时显示为 null )
select * from
(select * from sc where sc.CId = '01') as t1
left join
(select * from sc where sc.CId = '02') as t2
on t1.SId = t2.SId;
这一道就是明显需要使用join的情况了,02可能不存在,即为left join的右侧或right join 的左侧即可。
3、查询平均成绩大于等于 60 分的同学的学生编号和学生姓名和平均成绩这里只用根据学生ID把成绩分组,对分组中的score求平均值,最后在选取结果中AVG大于60的即可. 注意,这里必须要给计算得到的AVG结果一个alias.(AS ss)
得到学生信息的时候既可以用join也可以用一般的联合搜索。
select student.SId,sname,ss from student,(
select SId, AVG(score) as ss from sc
GROUP BY SId
HAVING AVG(score)> 60
)r
where student.sid = r.sid;
4、查询至少有一门课与学号为" 01 "的同学所学相同的同学的信息。 这个用联合查询也可以,但是逻辑不清楚,我觉得较为清楚的逻辑是这样的:从sc表查询01同学的所有选课cid--从sc表查询所有同学的sid如果其cid在前面的结果中--从student表查询所有学生信息如果sid在前面的结果中。
SELECT * FROM student WHERE sid IN
(SELECT sc.`SId` FROM sc WHERE sc.`CId` IN (SELECT sc.`CId` FROM sc WHERE sc.`SId`='01'));
5、查询没有学全所有课程的同学的信息。
因为有学生什么课都没有选,反向思考,先查询选了所有课的学生,再选择这些人之外的学生。
select * from student where student.sid not in ( select sc.sid from sc group by sc.sid having count(sc.cid)= (select count(cid) from course) );
6、查询学过「张三」老师授课的同学的信息。
多表联合查询
select student.* from student,teacher,course,sc
where
student.sid = sc.sid
and course.cid=sc.cid
and course.tid = teacher.tid
and tname = '张三';
其他习题请参考以下地址,做完这套题基本已经能够熟练掌握多表查询的题目了!