SQL练习50题(包含题目和答案及运行结果)

SQL经典练习50题

  • 数据准备
-- 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
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' , '1990-05-20' , '男');
insert into Student values('04' , '李云' , '1990-08-06' , '男');
insert into Student values('05' , '周梅' , '1991-12-01' , '女');
insert into Student values('06' , '吴兰' , '1992-03-01' , '女');
insert into Student values('07' , '郑竹' , '1989-07-01' , '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' , '2017-12-30' , '女');
insert into Student values('12' , '赵六' , '2017-01-01' , '女');
insert into Student values('13' , '孙七' , '2018-01-01' , '女');

-- 课程表 Course(CId,Cname,TId)
-- CId 课程编号,Cname 课程名称,TId 教师编号
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(TId,Tname)
-- TId 教师编号,Tname 教师姓名
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(SId,CId,score)
-- SId 学生编号,CId 课程编号,score 分数
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);
  • 练习

1、查询"01"课程比"02"课程成绩高的学生的信息及课程分数

方法一:

select 
    stu.*,sc1.cid,sc1.score 
from 
    student as stu 
join(
    select sid,cid,score from sc 
) sc1 on stu.sid = sc1.sid
where stu.sid in(
    select 
        s1.sid
    from
        (select sid,score from sc where cid = 01) as s1 -- 课程01的成绩和学生id
        inner join
        (select sid,score from sc where cid = 02) as s2 -- 课程02的成绩和学生id
        on s1.sid = s2.sid
    where 
        s1.score > s2.score
)
方式一运行结果

方法二:

select stu.*,c1_sc,c2_sc from student stu 
join(
    select s1.sid,s1.score as c1_sc,s2.score as c2_sc from 
    (select sid,score from sc where cid = 01) as s1 -- 课程01的成绩和学生id
    inner join
    (select sid,score from sc where cid = 02) as s2 -- 课程02的成绩和学生id
    on s1.sid = s2.sid
    where s1.score > s2.score
) sc1 on stu.sid = sc1.sid
result

2、查询同时存在" 01 "课程和" 02 "课程的情况 (查看同时学习01和02课程的同学信息)

select s1.sid,s1.score as c01_sc,s2.score as c02_sc
from
(select sid,cid,score from sc where cid = 01) as s1
join
(select sid,cid,score from sc where cid = 02) as s2
on s1.sid = s2.sid 
result

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

select stu.sid,stu.sname,sc1.avg_sc 
from student stu
join(
    select sid,avg(score) as avg_sc
    from score
    group by sid
) sc1 on stu.sid = sc1.sid 
where avg_sc > 60
result

4、查询平均成绩小于60分的同学的学生编号和学生姓名和平均成绩 (包括有成绩的和无成绩的)

SELECT * from student stu1
LEFT JOIN(
    SELECT sid,avg(score)
    from sc
    group by sid
) sc2 on stu1.sid = sc2.sid
where stu1.sid not in (
    select stu.sid
    from student stu
    join(
            select sid,avg(score) as avg_sc
            from sc
            group by sid
    ) sc1 on stu.sid = sc1.sid 
    where avg_sc > 60
)
result

5、查询所有同学的学生编号、学生姓名、选课总数、所有课程的总成绩

select stu.sid,stu.sname,t.cnt,t.sum_sc 
from student stu
left join(
    select sid, count(sid) as cnt,sum(score) as sum_sc 
    from sc group by sid
) t on stu.sid = t.sid;
result

6、查询"李"姓老师的数量

SELECT count(1) from teacher where tname like "li%"
result

7、查询学过"张三"老师授课的同学的信息

SELECT stu.* from student stu where sid in (
    SELECT sid from sc where cid in (
        select cid from course where tid in (
            SELECT tid from teacher where tname = "zhangsan"
        )
    )
)
result

8、查询没学过"张三"老师授课的同学的信息

select * from student where sid not in (
    select sid 
    from sc 
    where cid in (
        SELECT cid from course where tid in (
            select tid from teacher where tname = "zhangsan"
        )
    )
    group by sid
)
result

你可能感兴趣的:(SQL练习50题(包含题目和答案及运行结果))