时间:2021-07-01 10:21:17 帮助过:7人阅读
练习题答案:
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=‘男‘) as A , (select count(1) as feman from student where gender=‘女‘) 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
mysql练习题
标签:最好 tin 排列 同名 创建 inf cond const 要求