练习题答案：spa
二、查询“生物”课程比“物理”课程成绩高的全部学生的学号；
思路：
获取全部有生物课程的人（学号，成绩） - 临时表
获取全部有物理课程的人（学号，成绩） - 临时表
根据【学号】链接两个临时表：
学号 物理成绩 生物成绩
而后再进行筛选
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 = '物理') as B
on A.student_id = B.student_id where sw > if(isnull(ty),0,ty);
注释：if(isnull(ty),0,ty); 至关于三元运算
三、查询平均成绩大于60分的同窗的学号和平均成绩； 思路： 根据学生分组，使用avg获取平均值，经过having对avg进行筛选 select student_id,avg(num) from score group by student_id having avg(num) > 60 四、查询全部同窗的学号、姓名、选课数、总成绩；
思路：先从score表中对学号分组，做为一个临时表 A ：【学号】，【选课数】，【总成绩】,
而后将A表与学生表进行链接，查出结果集。
selectcode
sname,blog
student_id,utf-8
course_count,ci
total
from
(SELECT
student_id,
count(course_id) as course_count,
SUM( num ) as total
FROM
score
GROUP BY
student_id)
AS A left JOIN student on A.student_id = student.sid
五、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
六、查询没学过“李平”老师课的同窗的学号、姓名；
思路：
先查到“李平老师”老师教的全部课ID
获取选过课的全部学生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'
)
)
七、查询学过“001”而且也学过编号“002”课程的同窗的学号、姓名；
思路：
先查到既选择001又选择002课程的全部同窗
根据学生进行分组，若是学生数量等于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
八、查询学过“李平”老师所教的全部课的同窗的学号、姓名；
同上，只不过将001和002变成 in (叶平老师的全部课)
九、查询课程编号“002”的成绩比课程编号“001”课程低的全部同窗的学号、姓名；
同第1题
十、查询有课程成绩小于60分的同窗的学号、姓名；
select sid,sname from student where sid in (
select distinct student_id from score where num < 60
)
十一、查询没有学全全部课的同窗的学号、姓名；
思路：
在分数表中根据学生进行分组，获取每个学生选课数量
若是数量 == 总课程数量，表示已经选择了全部课程
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)
十二、查询至少有一门课与学号为“001”的同窗所学相同的同窗的学号和姓名；
思路：
获取 001 同窗选择的全部课程
获取课程在其中的全部人以及全部课程
根据学生筛选，获取全部学生信息
再与学生表链接，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
1三、查询至少学过学号为“001”同窗全部课的其余同窗学号和姓名；
先找到和001的学过的全部人
而后个数 ＝ 001全部学科 ＝＝》 其余人可能选择的更多
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)
1四、查询和“002”号的同窗学习的课程彻底相同的其余同窗学号和姓名；
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)
1五、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'
)
1六、向SC表中插入一些记录，这些记录要求符合如下条件：①没有上过编号“002”课程的同窗学号；②插入“002”号课程的平均成绩；
思路：
因为insert 支持
inset into tb1(xx,xx) select x1,x2 from tb2;
全部，获取全部没上过002课的全部人，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
)
1七、按平均成绩从低到高 显示全部学生的“语文”、“数学”、“英语”三门的课程成绩，按以下形式显示： 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
1八、查询各科成绩最高和最低的分：以以下形式显示：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;
1九、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
2一、查询各科成绩前三名的记录:(不考虑成绩并列状况)
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
2二、查询每门课程被选修的学生数；
select course_id, count(1) from score group by course_id;
2三、查询出只选修了一门课程的所有学生的学号和姓名；
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
2四、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
2五、查询姓“张”的学生名单；
select sname from student where sname like '张%';
2六、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;
2七、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;
2八、查询平均成绩大于85的全部学生的学号、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;
2九、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
3一、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
3二、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;
3三、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;
3四、查询不一样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;
3五、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
3六、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
3七、查询所有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);
3八、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
3九、查询两门以上不及格课程的同窗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窗学号；
select student_id from score where num< 60 and course_id = 4 order by num desc;
4一、删除“002”同窗的“001”课程的成绩；
delete from score where course_id = 1 and student_id = 2