MySQL练习

原创已于 2024-01-30 11:00:59 修改 · 277 阅读

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#mysql #数据库

于 2024-01-30 10:26:02 首次发布

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学生表：

老师表：

课程表及定位：

成绩表：

1.查询"01"课程比"02"课程成绩高的学生的信息及课程分数

查询代码：

select st.*,sc.s_Score as '语文' ,sc2.s_Score '数学'

from Student st

left join Score sc on sc.s_id=st.s_id and sc.c_id='01'

left join Score sc2 on sc2.s_id=st.s_id and sc2.c_id='02'

where sc.s_Score>sc2.s_Score;

结果

2、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

查询代码：

select st.s_id,st.s_name,ROUND(AVG(sc.s_Score),2) cjScore from Student st

left join Score sc on sc.s_id=st.s_id

group by st.s_id having AVG(sc.s_Score)>=60;

3、查询学过编号为"01"并且也学过编号为"02"的课程的同学的信息

查询代码：

select st.* from Student st

inner join Score sc on sc.s_id = st.s_id

inner join Course c on c.c_id=sc.c_id and c.c_id="01"

where st.s_id in (

select st2.s_id from Student st2

inner join Score sc2 on sc2.s_id = st2.s_id

inner join Course c2 on c2.c_id=sc2.c_id and c2.c_id="02"

)；

4、查询没有学全所有课程的同学的信息

查询代码

select * from Student where s_id not in (

select st.s_id from Student st

inner join Score sc on sc.s_id = st.s_id and sc.c_id="01"

where st.s_id in (

select st2.s_id from Student st2

inner join Score sc2 on sc2.s_id = st2.s_id and sc2.c_id="02"

) and st.s_id in (

select st2.s_id from Student st2

inner join Score sc2 on sc2.s_id = st2.s_id and sc2.c_id="03"

));

5、查询两门及其以上不及格课程的同学的学号，姓名及其平均成绩

查询代码

select st.s_id,st.s_name,avg(sc.s_Score) from Student st

left join Score sc on sc.s_id=st.s_id

where sc.s_id in (

select sc.s_id from Score sc

where sc.s_Score<60 or sc.s_Score is NULL

group by sc.s_id having COUNT(sc.s_id)>=2

)

group by st.s_id;