Linux 第五次练习(MySQL-DQL语句练习)

导入hellodb.sql生成数据库

 

1. 在students表中，查询年龄大于25岁，且为男性的同学的名字和年龄

select name,age from students where age>25 and Gender="M";

 2. 以ClassID为分组依据，显示每组的平均年龄

select avg(age) from students group by classid;

3. 显示第2题中平均年龄大于30的分组及平均年龄

select avg(age),classid from students group by classid having (avg(age)>30);

4. 显示以L开头的名字的同学的信息

select * from students where name like 'l%';

5. 显示TeacherID非空的同学的相关信息

select * from students where TeacherID is not null;

6. 以年龄排序后，显示年龄最大的前10位同学的信息

select * from students order by age DESC limit 10;

7. 查询年龄大于等于20岁，小于等于25岁的同学的信息

select * from students where age>=20 and age<=25;

8. 以ClassID分组，显示每班的同学的人数

select ClassID,count(*) 学生人数 from students group by ClassID;

9. 以Gender分组，显示其年龄之和

select Gender,sum(age) from students group by Gender;

10. 以ClassID分组，显示其平均年龄大于25的班级

select ClassID,avg(age) from students group by ClassID having (avg(age)>25);

11. 以Gender分组，显示各组中年龄大于25的学员的年龄之和

select Gender,sum(age) from students where age>25 group by Gender;

12. 显示前5位同学的姓名、课程及成绩

select st.name,co.Course,sc.Score from students st inner join scores sc on st.StuID=sc.StuID inner join courses co on sc.CourseID=co.CourseID where st.StuID<=5;

13. 显示其成绩高于80的同学的名称及课程

select st.name,co.Course,sc.Score from students st inner join scores sc on st.StuID=sc.StuID inner join courses co on sc.CourseID=co.CourseID where sc.Score>80;

14. 取每位同学各门课的平均成绩，显示成绩前三名的同学的姓名和平均成绩

select st.name,avg(sc.Score) 平均成绩 from students st inner join courses co on st.ClassID=co.CourseID inner join scores sc on st.StuID=sc.StuID group by st.name order by avg(sc.Score) DESC limit 3;

15. 显示每门课程课程名称及学习了这门课的同学的个数

select co.Course,count(st.name) from students st inner join courses co on st.ClassID=co.CourseID group by co.Course;

16. 显示其年龄大于平均年龄的同学的名字

select name from students where age>(select avg(age)from students);

17. 显示其学习的课程为第1、2，4或第7门课的同学的名字

select name from students where ClassID in (1,2,4,7);

18. 显示其成员数最少为3个的班级的同学中年龄大于同班同学平均年龄的同学

没搞懂

select st.name,st.age,avg.平均年龄 from students st,(select classid,count(*),avg(age) 平均年龄 from students group by classid having count(*)>=3) avg where st.classid=avg.classid and st.age > avg.平均年龄;

 

19. 统计各班级中年龄大于全校同学平均年龄的同学

select name,ClassID 班级 from students where age>(select avg(age) from students) order by 班级