mysql练习题

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
MariaDB [hellodb]> 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
MariaDB [hellodb]> select ClassID,avg(age) as 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
MariaDB [hellodb]> select ClassID,avg(age) as avg_age 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
MariaDB [hellodb]> select * from students where Name like ‘L%’;
MariaDB [hellodb]> select * from students where Name rlike ‘^L’;
(5) 显示TeacherID非空的同学的相关信息
MariaDB [hellodb]> 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
MariaDB [hellodb]> 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
MariaDB [hellodb]> select * from students where age>=20 and age<=25;
MariaDB [hellodb]> select * from students where age between 20 and 25;

 

导入hellodb.sql,以下操作在students表上执行
1、以ClassID分组,显示每班的同学的人数
MariaDB [hellodb]> select classid,count(name) from students group by classid;
2、以Gender分组,显示其年龄之和
MariaDB [hellodb]> select gender,sum(age) from students group by gender;
3、以ClassID分组,显示其平均年龄大于25的班级
select classid from students group by classid having avg(age) > 25;
4、以Gender分组,显示各组中年龄大于25的学员的年龄之和
MariaDB [hellodb]> select gender,sum(age) from (select age,gender from students where age>25) as t group by gender;
5、显示前5位同学的姓名、课程及成绩
MariaDB [hellodb]> select name,course,score from (select * from students limit 5) as t,courses,scores where t.stuid=scores.stuid and scores.courseid=courses.courseid;
6、显示其成绩高于80的同学的名称及课程;
select name,course,score from students,(select * from scores where score > 80) as s,courses where students.stuid=s.stuid and s.courseid=courses.courseid;
7、求前8位同学每位同学自己两门课的平均成绩,并按降序排列
select name,a from (select * from students limit 8) as s,(select stuid,avg(score) as a from scores group by stuid) as c where s.stuid=c.stuid order by a desc;
8、显示每门课程课程名称及学习了这门课的同学的个数
select course,count(name) from (select name,course from students,courses,scores where students.stuid=scores.stuid and scores.courseid=courses.courseid) as a group by course;
select courses.course,count(stuid) from scores left join courses on scores.courseid=courses.courseid group by scores.courseid;
9、显示其年龄大于平均年龄的同学的名字
MariaDB [hellodb]> select name from students where age > (select avg(age) from students);
10、显示其学习的课程为第1、2,4或第7门课的同学的名字
MariaDB [hellodb]> select name from students,(select distinct stuid from (select * from scores where courseid in (1,2,4,7)) as a) as b where b.stuid=students.stuid;
select students.name from students,(select distinct stuid from scores where courseid in (1,2,4,7))as s where s.stuid=students.stuid;
11、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
MariaDB [hellodb]> select student.name,student.age,student.classid,second.avg_age from (select students.name as name ,students.age as age,students.classid as classid from students left join (select count(name) as num,classid as classid from students group by classid having num>=3) as first on first.classid=students.classid) as student,(select avg(age) as avg_age,classid as classid from students group by classid) as second where student.age>second.avg_age and student.classid=second.classid;
12、统计各班级中年龄大于全校同学平均年龄的同学
select name,age,classid from students where age > (select avg(age) as a from students);

本文来自投稿,不代表Linux运维部落立场,如若转载,请注明出处:http://www.178linux.com/100759

发表评论

登录后才能评论

联系我们

400-080-6560

在线咨询:点击这里给我发消息

邮件:1823388528@qq.com

工作时间:周一至周五,9:30-18:30,节假日同时也值班