能独立完成这50道sql的我叫他爸爸
建表
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–1.学生表
Student(s_id,s_name,s_birth,s_sex) –学生编号,学生姓名, 出生年月,学生性别
–2.课程表
Course(c_id,c_name,t_id) – –课程编号, 课程名称, 教师编号
–3.教师表
Teacher(t_id,t_name) –教师编号,教师姓名
–4.成绩表
Score(s_id,c_id,s_score) –学生编号,课程编号,分数
题目
查询”01”课程比”02”课程成绩高的学生的信息及课程分数
select st.*,sc.s_score as ‘语文’ ,sc2.s_score ‘数学’
from student st
left join score sc on sc.s_id=st.s_id and sc.c_id=’01’
left join score sc2 on sc2.s_id=st.s_id and sc2.c_id=’02’
where sc.s_score>sc2.s_score
查询”01”课程比”02”课程成绩低的学生的信息及课程分数
select st.*,sc.s_score ‘语文’,sc2.s_score ‘数学’ from student st
left join score sc on sc.s_id=st.s_id and sc.c_id=’01’
left join score sc2 on sc2.s_id=st.s_id and sc2.c_id=’02’
where sc.s_score<sc2.s_score
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 st.s_id,st.s_name,ROUND(AVG(sc.s_score),2) “平均成绩” from student st
left join score sc on sc.s_id=st.s_id
group by st.s_id having AVG(sc.s_score)>=60
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
## (包括有成绩的和无成绩的)
select st.s_id,st.s_name,(case when ROUND(AVG(sc.s_score),2) is null then 0 else ROUND(AVG(sc.s_score),2) end ) “平均成绩” from student st
left join score sc on sc.s_id=st.s_id
group by st.s_id having AVG(sc.s_score)<60 or AVG(sc.s_score) is NULL
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 st.s_id,st.s_name,count(sc.c_id) “选课总数”,sum(case when sc.s_score is null then 0 else sc.s_score end) “总成绩”
from student st
left join score sc on st.s_id = sc.s_id
group by st.s_id
查询”李”姓老师的数量
select t.t_name,count(t.t_id) from teacher t
group by t.t_id having t.t_name like “李%”;
查询学过”张三”老师授课的同学的信息
select st.* from student st
left join score sc on sc.s_id=st.s_id
left join course c on c.c_id=sc.c_id
left join teacher t on t.t_id=c.t_id
where t.t_name=”张三”
查询没学过”张三”老师授课的同学的信息
张三老师教的课
select c.* from course c left join teacher t on t.t_id=c.t_id where t.t_name=”张三”
有张三老师课成绩的st.s_id
select sc.s_id from score sc where sc.c_id in (select c.c_id from course c left join teacher t on t.t_id=c.t_id where t.t_name=”张三”)
不在上面查到的st.s_id的学生信息,即没学过张三老师授课的同学信息
select st.* from student st where st.s_id not in(
select sc.s_id from score sc where sc.c_id in (select c.c_id from course c left join teacher t on t.t_id=c.t_id where t.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 st.* from student st
inner join score sc on sc.s_id = st.s_id
inner join course c on c.c_id=sc.c_id and c.c_id=”01”
where st.s_id in (
select st2.s_id from student st2
inner join score sc2 on sc2.s_id = st2.s_id
inner join course c2 on c2.c_id=sc2.c_id and c2.c_id=”02”
)
select a.*
from
student a,
score b,
score c
where
a.s_id = b.s_id
and a.s_id = c.s_id
and b.c_id = ‘01’
and c.c_id = ‘02’;
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 st.* from student st
inner join score sc on sc.s_id = st.s_id
inner join course c on c.c_id=sc.c_id and c.c_id=”01”
where st.s_id not in (
select st2.s_id from student st2
inner join score sc2 on sc2.s_id = st2.s_id
inner join course c2 on c2.c_id=sc2.c_id and c2.c_id=”02”
)
查询没有学全所有课程的同学的信息
select * from student where s_id not in (
select st.s_id from student st
inner join score sc on sc.s_id = st.s_id and sc.c_id=”01”
where st.s_id in (
select st1.s_id from student st1
inner join score sc2 on sc2.s_id = st1.s_id and sc2.c_id=”02”
) and st.s_id in (
select st2.s_id from student st2
inner join score sc2 on sc2.s_id = st2.s_id and sc2.c_id=”03”
))
select a.*
from student a
left join score b on a.s_id = b.s_id
group by
a.s_id
having
count(b.c_id) != ‘3’;
查询至少有一门课与学号为”01”的同学所学相同的同学的信息
select distinct st.* from student st
left join score sc on sc.s_id=st.s_id
where sc.c_id in (
select sc2.c_id from student st2
left join score sc2 on sc2.s_id=st2.s_id
where st2.s_id =’01’
)
查询和”01”号的同学学习的课程完全相同的其他同学的信息
select st.* from student st
left join score sc on sc.s_id=st.s_id
group by st.s_id
having group_concat(sc.c_id) =
(
select group_concat(sc2.c_id) from student st2
left join score sc2 on sc2.s_id=st2.s_id
where st2.s_id =’01’
)
查询没学过”张三”老师讲授的任一门课程的学生姓名
select st.s_name from student st
where st.s_id not in (
select sc.s_id from score sc
inner join course c on c.c_id=sc.c_id
inner join teacher t on t.t_id=c.t_id and t.t_name=”张三”
)
查询两门及其以上不及格课程的同学的学号,姓名及其平均成绩
select st.s_id,st.s_name,avg(sc.s_score) from student st
left join score sc on sc.s_id=st.s_id
where sc.s_id in (
select sc.s_id from score sc
where sc.s_score<60 or sc.s_score is NULL
group by sc.s_id having COUNT(1)>=2
)
group by st.s_id
检索”01”课程分数小于60,按分数降序排列的学生信息
select st.*,sc.s_score from student st
inner join score sc on sc.s_id=st.s_id and sc.c_id=”01” and sc.s_score<60
order by sc.s_score desc
select st.*,sc.s_score from student st
left join score sc on sc.s_id=st.s_id
where sc.c_id=”01” and sc.s_score<60
order by sc.s_score desc
按平均成绩从高到低显示所有学生的所有课程的成绩以及平均成绩
select st.s_id,st.s_name,avg(sc4.s_score) “平均分”,sc.s_score “语文”,sc2.s_score “数学”,sc3.s_score “英语” from student st
left join score sc on sc.s_id=st.s_id and sc.c_id=”01”
left join score sc2 on sc2.s_id=st.s_id and sc2.c_id=”02”
left join score sc3 on sc3.s_id=st.s_id and sc3.c_id=”03”
left join score sc4 on sc4.s_id=st.s_id
group by st.s_id
order by avg(sc4.s_score) desc
select st.s_id,st.s_name,
(case when avg(sc4.s_score) is null then 0 else avg(sc4.s_score) end) “平均分”,
(case when sc.s_score is null then 0 else sc.s_score end) “语文”,
(case when sc2.s_score is null then 0 else sc2.s_score end) “数学”,
(case when sc3.s_score is null then 0 else sc3.s_score end) “英语”
from student st
left join score sc on sc.s_id=st.s_id and sc.c_id=”01”
left join score sc2 on sc2.s_id=st.s_id and sc2.c_id=”02”
left join score sc3 on sc3.s_id=st.s_id and sc3.c_id=”03”
left join score sc4 on sc4.s_id=st.s_id
group by st.s_id
order by avg(sc4.s_score) desc
查询各科成绩最高分、最低分和平均分:以如下形式显示:课程ID,课程name,最高分,最低分,平均分,及格率,中等率,优良率,优秀率
及格为>=60,中等为:70-80,优良为:80-90,优秀为:>=90
select c.c_id,c.c_name,max(sc.s_score) “最高分”,MIN(sc2.s_score) “最低分”,avg(sc3.s_score) “平均分”
,((select count(s_id) from score where s_score>=60 and c_id=c.c_id )/(select count(s_id) from score where c_id=c.c_id)) “及格率”
,((select count(s_id) from score where s_score>=70 and s_score<80 and c_id=c.c_id )/(select count(s_id) from score where c_id=c.c_id)) “中等率”
,((select count(s_id) from score where s_score>=80 and s_score<90 and c_id=c.c_id )/(select count(s_id) from score where c_id=c.c_id)) “优良率”
,((select count(s_id) from score where s_score>=90 and c_id=c.c_id )/(select count(s_id) from score where c_id=c.c_id)) “优秀率”
from course c
left join score sc on sc.c_id=c.c_id
left join score sc2 on sc2.c_id=c.c_id
left join score sc3 on sc3.c_id=c.c_id
group by c.c_id
19、按各科成绩进行排序,并显示排名(实现不完全)
mysql没有rank函数
加@score是为了防止用union all 后打乱了顺序
select c1.s_id,c1.c_id,c1.c_name,@score:=c1.s_score,@i:=@i+1 from (select c.c_name,sc.* from course c
left join score sc on sc.c_id=c.c_id
where c.c_id=”01” order by sc.s_score desc) c1 ,
(select @i:=0) a
union all
select c2.s_id,c2.c_id,c2.c_name,c2.s_score,@ii:=@ii+1 from (select c.c_name,sc.* from course c
left join score sc on sc.c_id=c.c_id
where c.c_id=”02” order by sc.s_score desc) c2 ,
(select @ii:=0) aa
union all
select c3.s_id,c3.c_id,c3.c_name,c3.s_score,@iii:=@iii+1 from (select c.c_name,sc.* from course c
left join score sc on sc.c_id=c.c_id
where c.c_id=”03” order by sc.s_score desc) c3;
set @iii=0;
查询学生的总成绩并进行排名
select st.s_id,st.s_name
,(case when sum(sc.s_score) is null then 0 else sum(sc.s_score) end)
from student st
left join score sc on sc.s_id=st.s_id
group by st.s_id order by sum(sc.s_score) desc
查询不同老师所教不同课程平均分从高到低显示
select t.t_id,t.t_name,c.c_name,avg(sc.s_score) from teacher t
left join course c on c.t_id=t.t_id
left join score sc on sc.c_id =c.c_id
group by t.t_id
order by avg(sc.s_score) desc
查询所有课程的成绩第2名到第3名的学生信息及该课程成绩
select a.* from (
select st.,c.c_id,c.c_name,sc.s_score from student st
left join score sc on sc.s_id=st.s_id
inner join course c on c.c_id =sc.c_id and c.c_id=”01”
order by sc.s_score desc LIMIT 1,2 ) a
union all
select b. from (
select st.,c.c_id,c.c_name,sc.s_score from student st
left join score sc on sc.s_id=st.s_id
inner join course c on c.c_id =sc.c_id and c.c_id=”02”
order by sc.s_score desc LIMIT 1,2) b
union all
select c. from (
select st.*,c.c_id,c.c_name,sc.s_score from student st
left join score sc on sc.s_id=st.s_id
inner join course c on c.c_id =sc.c_id and c.c_id=”03”
order by sc.s_score desc LIMIT 1,2) c
统计各科成绩各分数段人数:课程编号,课程名称,[100-85],[85-70],[70-60],[0-60]及所占百分比
select c.c_id,c.c_name
,((select count(1) from score sc where sc.c_id=c.c_id and sc.s_score<=100 and sc.s_score>80)/(select count(1) from score sc where sc.c_id=c.c_id )) “100-85”
,((select count(1) from score sc where sc.c_id=c.c_id and sc.s_score<=85 and sc.s_score>70)/(select count(1) from score sc where sc.c_id=c.c_id )) “85-70”
,((select count(1) from score sc where sc.c_id=c.c_id and sc.s_score<=70 and sc.s_score>60)/(select count(1) from score sc where sc.c_id=c.c_id )) “70-60”
,((select count(1) from score sc where sc.c_id=c.c_id and sc.s_score<=60 and sc.s_score>=0)/(select count(1) from score sc where sc.c_id=c.c_id )) “60-0”
from course c order by c.c_id
查询学生平均成绩及其名次
set @i=0;
select a.*,@i:=@i+1 from (
select st.s_id,st.s_name,round((case when avg(sc.s_score) is null then 0 else avg(sc.s_score) end),2) “平均分” from student st
left join score sc on sc.s_id=st.s_id
group by st.s_id order by sc.s_score desc) a
查询各科成绩前三名的记录
select a.* from (
select st.s_id,st.s_name,c.c_id,c.c_name,sc.s_score from student st
left join score sc on sc.s_id=st.s_id
inner join course c on c.c_id=sc.c_id and c.c_id=’01’
order by sc.s_score desc LIMIT 0,3) a
union all
select b.* from (
select st.s_id,st.s_name,c.c_id,c.c_name,sc.s_score from student st
left join score sc on sc.s_id=st.s_id
inner join course c on c.c_id=sc.c_id and c.c_id=’02’
order by sc.s_score desc LIMIT 0,3) b
union all
select c.* from (
select st.s_id,st.s_name,c.c_id,c.c_name,sc.s_score from student st
left join score sc on sc.s_id=st.s_id
inner join course c on c.c_id=sc.c_id and c.c_id=’03’
order by sc.s_score desc LIMIT 0,3) c
查询每门课程被选修的学生数
select c.c_id,c.c_name,count(1) from course c
left join score sc on sc.c_id=c.c_id
inner join student st on st.s_id=c.c_id
group by st.s_id
查询出只有两门课程的全部学生的学号和姓名
select st.s_id,st.s_name from student st
left join score sc on sc.s_id=st.s_id
inner join course c on c.c_id=sc.c_id
group by st.s_id having count(1)=2
查询男生、女生人数
select st.s_sex,count(1) from student st group by st.s_sex
查询名字中含有”风”字的学生信息
select st.* from student st where st.s_name like “%风%”;
查询同名同性学生名单,并统计同名人数
select st.*,count(1) from student st group by st.s_name,st.s_sex having count(1)>1
查询1990年出生的学生名单
select st.* from student st where st.s_birth like “1990%”;
查询每门课程的平均成绩,结果按平均成绩降序排列,平均成绩相同时,按课程编号升序排列
select c.c_id,c.c_name,avg(sc.s_score) from course c
inner join score sc on sc.c_id=c.c_id
group by c.c_id order by avg(sc.s_score) desc,c.c_id asc
查询平均成绩大于等于85的所有学生的学号、姓名和平均成绩
select st.s_id,st.s_name,avg(sc.s_score) from student st
left join score sc on sc.s_id=st.s_id
group by st.s_id having avg(sc.s_score)>=85
查询课程名称为”数学”,且分数低于60的学生姓名和分数
select st.s_id,st.s_name,sc.s_score from student st
inner join score sc on sc.s_id=st.s_id and sc.s_score<60
inner join course c on c.c_id=sc.c_id and c.c_name =”数学”
查询所有学生的课程及分数情况;
select st.s_id,st.s_name,c.c_name,sc.s_score from student st
left join score sc on sc.s_id=st.s_id
left join course c on c.c_id =sc.c_id
order by st.s_id,c.c_name
查询任何一门课程成绩在70分以上的姓名、课程名称和分数
select st2.s_id,st2.s_name,c2.c_name,sc2.s_score from student st2
left join score sc2 on sc2.s_id=st2.s_id
left join course c2 on c2.c_id=sc2.c_id
where st2.s_id in(
select st.s_id from student st
left join score sc on sc.s_id=st.s_id
group by st.s_id having min(sc.s_score)>=70)
order by s_id
查询不及格的课程
select st.s_id,c.c_name,st.s_name,sc.s_score from student st
inner join score sc on sc.s_id=st.s_id and sc.s_score<60
inner join course c on c.c_id=sc.c_id
查询课程编号为01且课程成绩在80分以上的学生的学号和姓名
select st.s_id,st.s_name,sc.s_score from student st
inner join score sc on sc.s_id=st.s_id and sc.c_id=”01” and sc.s_score>=80
求每门课程的学生人数
select c.c_id,c.c_name,count(1) from course c
inner join score sc on sc.c_id=c.c_id
group by c.c_id
查询选修”张三”老师所授课程的学生中,成绩最高的学生信息及其成绩
select st.*,c.c_name,sc.s_score,t.t_name from student st
inner join score sc on sc.s_id=st.s_id
inner join course c on c.c_id=sc.c_id
inner join teacher t on t.t_id=c.t_id and t.t_name=”张三”
order by sc.s_score desc
limit 0,1
查询不同课程成绩相同的学生的学生编号、课程编号、学生成绩
select st.s_id,st.s_name,sc.c_id,sc.s_score from student st
left join score sc on sc.s_id=st.s_id
left join course c on c.c_id=sc.c_id
where (
select count(1) from student st2
left join score sc2 on sc2.s_id=st2.s_id
left join course c2 on c2.c_id=sc2.c_id
where sc.s_score=sc2.s_score and c.c_id!=c2.c_id
)>1
查询每门功成绩最好的前两名
select a.* from (select st.s_id,st.s_name,c.c_name,sc.s_score from student st
left join score sc on sc.s_id=st.s_id
inner join course c on c.c_id=sc.c_id and c.c_id=”01”
order by sc.s_score desc limit 0,2) a
union all
select b.* from (select st.s_id,st.s_name,c.c_name,sc.s_score from student st
left join score sc on sc.s_id=st.s_id
inner join course c on c.c_id=sc.c_id and c.c_id=”02”
order by sc.s_score desc limit 0,2) b
union all
select c.* from (select st.s_id,st.s_name,c.c_name,sc.s_score from student st
left join score sc on sc.s_id=st.s_id
inner join course c on c.c_id=sc.c_id and c.c_id=”03”
order by sc.s_score desc limit 0,2) c
借鉴(更准确,漂亮):
select a.s_id,a.c_id,a.s_score from score a
where (select COUNT(1) from score b where b.c_id=a.c_id and b.s_score>=a.s_score)<=2 order by a.c_id
43、统计每门课程的学生选修人数(超过5人的课程才统计)。要求输出课程号和选修人数,查询结果按人数降序排列,若人数相同,按课程号升序排列
select sc.c_id,count(1) from score sc
left join course c on c.c_id=sc.c_id
group by c.c_id having count(1)>5
order by count(1) desc,sc.c_id asc
检索至少选修两门课程的学生学号
select st.s_id from student st
left join score sc on sc.s_id=st.s_id
group by st.s_id having count(1)>=2
查询选修了全部课程的学生信息
select st.* from student st
left join score sc on sc.s_id=st.s_id
group by st.s_id having count(1)=(select count(1) from course)
查询各学生的年龄
select st.*,timestampdiff(year,st.s_birth,now()) from student st
查询本周过生日的学生
此处可能有问题,week函数取的为当前年的第几周,2017-12-12是第50周而2018-12-12是第49周,可以取月份,day,星期几(%w),
再判断本周是否会持续到下一个月进行判断,太麻烦,不会写
select st.* from student st
where week(now())=week(date_format(st.s_birth,’%Y%m%d’))
查询下周过生日的学生
select st.* from student st
where week(now())+1=week(date_format(st.s_birth,’%Y%m%d’))
查询本月过生日的学生
select st.* from student st
where month(now())=month(date_format(st.s_birth,’%Y%m%d’))
查询下月过生日的学生
注意:当 当前月为12时,用month(now())+1为13而不是1,可用timestampadd()函数或mod取模
select st.* from student st
where month(timestampadd(month,1,now()))=month(date_format(st.s_birth,’%Y%m%d’))
或
select st.* from student st where (month(now()) + 1) mod 12 = month(date_format(st.s_birth,’%Y%m%d’))