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2020-05-02 阅读量: 898
SQL面试进阶题1

##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 stu.*,sc.c_id,sc.score from (select * from sc where c_id='01') t1

left join (select * from sc where c_id='02') t2 on t1.s_id=t2.s_id left join sc on sc.s_id=t1.s_id

left join stu on stu.s_id=t1.s_id where t1.score>t2.score;

##2、查询"01"课程比"02"课程成绩低的学生的信息及课程分数

select stu.*,sc.c_id,sc.score from (select * from sc where c_id='01') t1

left join (select * from sc where c_id='02') t2 on t1.s_id=t2.s_id left join sc on sc.s_id=t1.s_id

left join stu on stu.s_id=t1.s_id where t1.score<t2.score;

##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.s_id,s_name,avg(score) 平均成绩 from stu join sc on stu.s_id=sc.s_id group by stu.s_id having avg(score)>=60;

##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

select stu.s_id,s_name,avg(score) 平均成绩 from stu join sc on stu.s_id=sc.s_id group by stu.s_id having avg(score)<=60;

##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.s_id,s_name,count(c_id) 选课总数,sum(score) 总成绩 from stu join sc on stu.s_id=sc.s_id group by stu.s_id order by 总成绩 desc;

##6、查询"李"姓老师的数量

select count(t_id) from te where t_name like '李%';

##7、查询学过"张三"老师授课的同学的信息

select stu.* from co left join te on co.t_id=te.t_id left join sc on sc.c_id=co.c_id left join stu on stu.s_id=sc.s_id

where t_name='张三';

##8、查询没学过"张三"老师授课的同学的信息

select *from stu where stu.s_id not in (select stu.s_id from co left join te on co.t_id=te.t_id left join sc on sc.c_id=co.c_id left join stu on stu.s_id=sc.s_id

where t_name='张三');

##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

select stu.* from stu left join sc on stu.s_id=sc.s_id where c_id in ('01','02')

group by stu.s_id having count(sc.c_id)=2;

##10、查询学过编号为"01"但是没有学过编号为"02"的课程的同学的信息

select stu.*

from stu left join sc on stu.s_id=sc.s_id

where c_id in ('01','02')

group by sc.s_id

having group_concat(c_id)='01';

##11、查询没有学全所有课程的同学的信息

select stu.* from stu join sc on stu.s_id=sc.s_id join co on sc.c_id=co.c_id group by stu.s_id

having count(co.c_id)<(select count(co.c_id) from co);

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