• [转载]SQL语句练习


    2、查询“生物”课程比“物理”课程成绩高的所有学生的学号;
    
    思路:
    
        获取所有有生物课程的人(学号,成绩) - 临时表
    
        获取所有有物理课程的人(学号,成绩) - 临时表
    
        根据【学号】连接两个临时表:
    
            学号  物理成绩   生物成绩
    
     
    
        然后再进行筛选
    
     
    
            select A.student_id,sw,ty from
    
     
    
            (select student_id,num as sw from score left join course on score.course_id = course.cid where course.cname = '生物') as A
    
     
    
            left join
    
     
    
            (select student_id,num  as ty from score left join course on score.course_id = course.cid where course.cname = '体育') as B
    
     
    
            on A.student_id = B.student_id where sw > if(isnull(ty),0,ty);
    
     
    
    3、查询平均成绩大于60分的同学的学号和平均成绩; 
    
        思路:
    
            根据学生分组,使用avg获取平均值,通过having对avg进行筛选
    
     
    
            select student_id,avg(num) from score group by student_id having avg(num) > 60
    
     
    
    4、查询所有同学的学号、姓名、选课数、总成绩;
    
     
    
        select score.student_id,sum(score.num),count(score.student_id),student.sname 
    
        from
    
            score left join student on score.student_id = student.sid   
    
        group by score.student_id
    
     
    
    5、查询姓“李”的老师的个数;
    
        select count(tid) from teacher where tname like '李%'
    
     
    
        select count(1) from (select tid from teacher where tname like '李%') as B
    
     
    
    6、查询没学过“叶平”老师课的同学的学号、姓名;
    
        思路:
    
            先查到“李平老师”老师教的所有课ID
    
            获取选过课的所有学生ID
    
            学生表中筛选
    
        select * from student where sid not in (
    
            select DISTINCT student_id from score where score.course_id in (
    
                select cid from course left join teacher on course.teacher_id = teacher.tid where tname = '李平老师'
    
            )
    
        )
    
     
    
    7、查询学过“001”并且也学过编号“002”课程的同学的学号、姓名;
    
        思路:
    
            先查到既选择001又选择002课程的所有同学
    
            根据学生进行分组,如果学生数量等于2表示,两门均已选择
    
     
    
        select student_id,sname from
    
     
    
        (select student_id,course_id from score where course_id = 1 or course_id = 2) as B
    
          
    
        left join student on B.student_id = student.sid group by student_id HAVING count(student_id) > 1
    
     
    
     
    
    8、查询学过“叶平”老师所教的所有课的同学的学号、姓名;
    
     
    
        同上,只不过将001和002变成 in (叶平老师的所有课)
    
     
    
    9、查询课程编号“002”的成绩比课程编号“001”课程低的所有同学的学号、姓名;
    
        同第1题
    
     
    
     
    
    10、查询有课程成绩小于60分的同学的学号、姓名;
    
             
    
        select sid,sname from student where sid in (
    
            select distinct student_id from score where num < 60
    
        )
    
     
    
    11、查询没有学全所有课的同学的学号、姓名;
    
        思路:
    
            在分数表中根据学生进行分组,获取每一个学生选课数量
    
            如果数量 == 总课程数量,表示已经选择了所有课程
    
     
    
            select student_id,sname 
    
            from score left join student on score.student_id = student.sid 
    
            group by student_id HAVING count(course_id) = (select count(1) from course)
    
     
    
     
    
    12、查询至少有一门课与学号为“001”的同学所学相同的同学的学号和姓名;
    
        思路:
    
            获取 001 同学选择的所有课程
    
            获取课程在其中的所有人以及所有课程
    
            根据学生筛选,获取所有学生信息
    
            再与学生表连接,获取姓名
    
     
    
            select student_id,sname, count(course_id) 
    
            from score left join student on score.student_id = student.sid
    
            where student_id != 1 and course_id in (select course_id from score where student_id = 1) group by student_id
    
     
    
    13、查询至少学过学号为“001”同学所有课的其他同学学号和姓名;
    
            先找到和001的学过的所有人
    
            然后个数 = 001所有学科     ==》 其他人可能选择的更多
    
     
    
            select student_id,sname, count(course_id) 
    
            from score left join student on score.student_id = student.sid
    
            where student_id != 1 and course_id in (select course_id from score where student_id = 1) group by student_id having count(course_id) = (select count(course_id) from score where student_id = 1)
    
     
    
    14、查询和“002”号的同学学习的课程完全相同的其他同学学号和姓名;
    
             
    
            个数相同
    
            002学过的也学过
    
     
    
            select student_id,sname from score left join student on score.student_id = student.sid where student_id in (
    
                select student_id from score  where student_id != 1 group by student_id HAVING count(course_id) = (select count(1) from score where student_id = 1)
    
            ) and course_id in (select course_id from score where student_id = 1) group by student_id HAVING count(course_id) = (select count(1) from score where student_id = 1)
    
     
    
     
    
    15、删除学习“叶平”老师课的score表记录;
    
     
    
        delete from score where course_id in (
    
            select cid from course left join teacher on course.teacher_id = teacher.tid where teacher.name = '叶平'
    
        )
    
     
    
    16、向SC表中插入一些记录,这些记录要求符合以下条件:①没有上过编号“002”课程的同学学号;②插入“002”号课程的平均成绩; 
    
        思路:
    
            由于insert 支持  
    
                    inset into tb1(xx,xx) select x1,x2 from tb2;
    
            所有,获取所有没上过002课的所有人,获取002的平均成绩
    
     
    
        insert into score(student_id, course_id, num) select sid,2,(select avg(num) from score where course_id = 2) 
    
        from student where sid not in (
    
            select student_id from score where course_id = 2
    
        )
    
         
    
    17、按平均成绩从低到高 显示所有学生的“语文”、“数学”、“英语”三门的课程成绩,按如下形式显示: 学生ID,语文,数学,英语,有效课程数,有效平均分;
    
        select sc.student_id,
    
            (select num from score left join course on score.course_id = course.cid where course.cname = "生物" and score.student_id=sc.student_id) as sy,
    
            (select num from score left join course on score.course_id = course.cid where course.cname = "物理" and score.student_id=sc.student_id) as wl,
    
            (select num from score left join course on score.course_id = course.cid where course.cname = "体育" and score.student_id=sc.student_id) as ty,
    
            count(sc.course_id),
    
            avg(sc.num)
    
        from score as sc
    
        group by student_id desc        
    
     
    
    18、查询各科成绩最高和最低的分:以如下形式显示:课程ID,最高分,最低分;
    
         
    
        select course_id, max(num) as max_num, min(num) as min_num from score group by course_id;
    
     
    
    19、按各科平均成绩从低到高和及格率的百分数从高到低顺序;
    
        思路:case when .. then
    
        select course_id, avg(num) as avgnum,sum(case when score.num > 60 then 1 else 0 END)/count(1)*100 as percent from score group by course_id order by avgnum asc,percent desc;
    
     
    
    20、课程平均分从高到低显示(现实任课老师);
    
     
    
        select avg(if(isnull(score.num),0,score.num)),teacher.tname from course 
    
        left join score on course.cid = score.course_id 
    
        left join teacher on course.teacher_id = teacher.tid
    
     
    
        group by score.course_id
    
     
    
     
    
    21、查询各科成绩前三名的记录:(不考虑成绩并列情况) 
    
        select score.sid,score.course_id,score.num,T.first_num,T.second_num from score left join
    
        (
    
        select
    
            sid,
    
            (select num from score as s2 where s2.course_id = s1.course_id order by num desc limit 0,1) as first_num,
    
            (select num from score as s2 where s2.course_id = s1.course_id order by num desc limit 3,1) as second_num
    
        from
    
            score as s1
    
        ) as T
    
        on score.sid =T.sid
    
        where score.num <= T.first_num and score.num >= T.second_num
    
     
    
    22、查询每门课程被选修的学生数;
    
         
    
        select course_id, count(1) from score group by course_id;
    
     
    
    23、查询出只选修了一门课程的全部学生的学号和姓名;
    
        select student.sid, student.sname, count(1) from score
    
     
    
        left join student on score.student_id  = student.sid
    
     
    
         group by course_id having count(1) = 1
    
     
    
     
    
    24、查询男生、女生的人数;
    
        select * from
    
        (select count(1) as man from student where gender='') as A ,
    
        (select count(1) as feman from student where gender='') as B 
    
     
    
    25、查询姓“张”的学生名单;
    
        select sname from student where sname like '张%';
    
     
    
    26、查询同名同姓学生名单,并统计同名人数;
    
     
    
        select sname,count(1) as count from student group by sname;
    
     
    
    27、查询每门课程的平均成绩,结果按平均成绩升序排列,平均成绩相同时,按课程号降序排列;
    
        select course_id,avg(if(isnull(num), 0 ,num)) as avg from score group by course_id order by avg     asc,course_id desc;
    
     
    
    28、查询平均成绩大于85的所有学生的学号、姓名和平均成绩;
    
     
    
        select student_id,sname, avg(if(isnull(num), 0 ,num)) from score left join student on score.student_id = student.sid group by student_id;
    
     
    
    29、查询课程名称为“数学”,且分数低于60的学生姓名和分数;
    
     
    
        select student.sname,score.num from score 
    
        left join course on score.course_id = course.cid
    
        left join student on score.student_id = student.sid
    
        where score.num < 60 and course.cname = '生物'
    
     
    
    30、查询课程编号为003且课程成绩在80分以上的学生的学号和姓名; 
    
        select * from score where score.student_id = 3 and score.num > 80
    
     
    
    31、求选了课程的学生人数
    
     
    
        select count(distinct student_id) from score
    
     
    
        select count(c) from (
    
            select count(student_id) as c from score group by student_id) as A
    
     
    
    32、查询选修“杨艳”老师所授课程的学生中,成绩最高的学生姓名及其成绩;
    
         
    
        select sname,num from score 
    
        left join student on score.student_id = student.sid
    
        where score.course_id in (select course.cid from course left join teacher on course.teacher_id = teacher.tid where tname='张磊老师') order by num desc limit 1;
    
     
    
    33、查询各个课程及相应的选修人数;
    
        select course.cname,count(1) from score
    
        left join course on score.course_id = course.cid
    
        group by course_id;
    
     
    
     
    
    34、查询不同课程但成绩相同的学生的学号、课程号、学生成绩;
    
        select DISTINCT s1.course_id,s2.course_id,s1.num,s2.num from score as s1, score as s2 where s1.num = s2.num and s1.course_id != s2.course_id;
    
     
    
    35、查询每门课程成绩最好的前两名;
    
     
    
        select score.sid,score.course_id,score.num,T.first_num,T.second_num from score left join
    
        (
    
        select
    
            sid,
    
            (select num from score as s2 where s2.course_id = s1.course_id order by num desc limit 0,1) as first_num,
    
            (select num from score as s2 where s2.course_id = s1.course_id order by num desc limit 1,1) as second_num
    
        from
    
            score as s1
    
        ) as T
    
        on score.sid =T.sid
    
        where score.num <= T.first_num and score.num >= T.second_num
    
     
    
    36、检索至少选修两门课程的学生学号;
    
        select student_id from score group by student_id having count(student_id) > 1
    
     
    
    37、查询全部学生都选修的课程的课程号和课程名;
    
        select course_id,count(1) from score group by course_id having count(1) = (select count(1) from student);
    
     
    
    38、查询没学过“叶平”老师讲授的任一门课程的学生姓名;
    
        select student_id,student.sname from score 
    
        left join student on score.student_id = student.sid
    
        where score.course_id not in (
    
            select cid from course left join teacher on course.teacher_id = teacher.tid where tname = '张磊老师'
    
        ) 
    
        group by student_id
    
     
    
    39、查询两门以上不及格课程的同学的学号及其平均成绩;
    
     
    
        select student_id,count(1) from score where num < 60 group by student_id having count(1) > 2
    
     
    
    40、检索“004”课程分数小于60,按分数降序排列的同学学号;
    
        select student_id from score where num< 60 and course_id = 4 order by num desc;
    
     
    
    41、删除“002”同学的“001”课程的成绩;
    
        delete from score where course_id = 1 and student_id = 2
  • 相关阅读:
    第1章 数据类型 第1节 JavaScript中的几个重要概念
    第1章 数据类型 第3节 JavaScript数值类型(2)
    第1章 数据类型 第4节 JavaScript字符类型
    GDI编程框架代码
    linux清空日志文件内容 比如log日志
    快速点对点批量向员工推送企业微信消息的方法
    在EasySQLMAIL中实现表格的行列转置
    快速信息分发技巧:按员工手机号通过企业微信点对点分发信息
    网络
    【odoo】【相识篇】一、Odoo VS SAP?
  • 原文地址:https://www.cnblogs.com/xu-xiaofeng/p/7562467.html
Copyright © 2020-2023  润新知