MySQL经典50题-2019更新版

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--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 a.SID , a.Sname , cast(avg(b.score) as decimal(18,2)) avg_score
from Student a , sc b
where a.SID = b.SID
group by a.SID , a.Sname
having cast(avg(b.score) as decimal(18,2)) >= 60
order by a.SID;

CAST函数用于将某种数据类型的表达式显式转换为另一种数据类型。CAST()函数的参数是一个表达式，它包括用AS关键字分隔的源值和目标数据类型。
CAST (expression AS data_type)；
decimal(a,b)
a指定指定小数点左边和右边可以存储的十进制数字的最大个数，最大精度38。
b指定小数点右边可以存储的十进制数字的最大个数。小数位数必须是从 0 到 a之间的值。默认小数位数是 0。

--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
--4.1、查询在sc表存在成绩的学生信息的SQL语句。
select a.SID , a.Sname , cast(avg(b.score) as decimal(18,2)) avg_score
from Student a , sc b
where a.SID = b.SID
group by a.SID , a.Sname
having cast(avg(b.score) as decimal(18,2)) < 60
order by a.SID

67--4.2、查询在sc表中不存在成绩的学生信息的SQL语句(有成绩不等于学生信息在sc表里)。
select a.SID , a.Sname , cast(avg(b.score) as decimal(18,2)) avg_score
from Student a left join sc b
                         on a.SID = b.SID
group by a.SID , a.Sname
having cast(avg(b.score) as decimal(18,2)) < 60
order by a.SID;

--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
--5.1、查询所有有成绩的SQL。
select a.SID  学生编号 , a.Sname  学生姓名 , count(b.CID) 选课总数, sum(score)  所有课程的总成绩 
from Student a , SC b
where a.SID = b.SID
group by a.SID,a.Sname
order by a.SID

--5.2、查询所有(包括有成绩和无成绩)的SQL。
select a.SID  学生编号 , a.Sname  学生姓名 , count(b.CID) 选课总数, sum(score)  所有课程的总成绩 
from Student a left join SC b
on a.SID = b.SID
group by a.SID,a.Sname
order by a.SID

--6、查询"李"姓老师的数量
--方法1
select count(Tname)  李姓老师的数量  from Teacher where Tname like ‘李%‘
--方法2
select count(Tname)  李姓老师的数量  from Teacher where left(Tname,1) = ‘李‘

--7、查询学过"张三"老师授课的同学的信息
select distinct Student.* from Student , SC , Course , Teacher
where Student.SID = SC.SID and SC.CID = Course.CID and Course.TID = Teacher.TID and Teacher.Tname = ‘张三‘
order by Student.SID

--8、查询没学过"张三"老师授课的同学的信息
select m.* from Student m where SID not in (select distinct SC.SID from SC , Course , Teacher where SC.CID = Course.CID and Course.TID = Teacher.TID and Teacher.Tname = ‘张三‘) order by m.SID

--9、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
--方法1
select Student.* from Student , SC where Student.SID = SC.SID and SC.CID = ‘01‘ and exists (Select 1 from SC SC_2 where SC_2.SID = SC.SID and SC_2.CID = ‘02‘) order by Student.SID

--方法2
select Student.* from Student , SC where Student.SID = SC.SID and SC.CID = ‘02‘ and exists (Select 1 from SC SC_2 where SC_2.SID = SC.SID and SC_2.CID = ‘01‘) order by Student.SID
--方法3
select m.* from Student m where SID in
(
  select SID from
  (
    select distinct SID from SC where CID = ‘01‘
    union all
    select distinct SID from SC where CID = ‘02‘
  ) t group by SID having count(1) = 2
)
order by m.SID

--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
--方法1
select Student.* from Student , SC where Student.SID = SC.SID and SC.CID = ‘01‘ and not exists (Select 1 from SC SC_2 where SC_2.SID = SC.SID and SC_2.CID = ‘02‘) order by Student.SID

--方法2
select Student.* from Student , SC where Student.SID = SC.SID and SC.CID = ‘01‘ and Student.SID not in (Select SC_2.SID from SC SC_2 where SC_2.SID = SC.SID and SC_2.CID = ‘02‘) order by Student.SID

--11、查询没有学全所有课程的同学的信息
--11.1、
select Student.*
from Student , SC
where Student.SID = SC.SID
group by Student.SID , Student.Sname , Student.Sage , Student.Ssex having count(CID) < (select count(CID) from Course)

--11.2
select Student.*
from Student left join SC
on Student.SID = SC.SID
group by Student.SID , Student.Sname , Student.Sage , Student.Ssex having count(CID) < (select count(CID) from Course)

--12、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 Student.* from Student , SC where Student.SID = SC.SID and SC.CID in (select CID from SC where SID = ‘01‘) and Student.SID <> ‘01‘

--13、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 Student.* from Student where SID in
(select distinct SC.SID from SC where SID <> ‘01‘ and SC.CID in (select distinct CID from SC where SID = ‘01‘)
group by SC.SID having count(1) = (select count(1) from SC where SID=‘01‘))

--14、查询没学过"张三"老师讲授的任一门课程的学生姓名
select student.* from student where student.SID not in
(select distinct sc.SID from sc , course , teacher where sc.CID = course.CID and course.TID = teacher.TID and teacher.tname = ‘张三‘)
order by student.SID

--15、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 student.SID , student.sname , cast(avg(score) as decimal(18,2)) avg_score from student , sc
where student.SID = SC.SID and student.SID in (select SID from SC where score < 60 group by SID having count(1) >= 2)
group by student.SID , student.sname

--16、检索"01"课程分数小于60，按分数降序排列的学生信息
select student.* , sc.CID , sc.score from student , sc
where student.SID = SC.SID and sc.score < 60 and sc.CID = ‘01‘
order by sc.score desc

--17、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
--17.1 SQL 2000 静态
select a.SID 学生编号 , a.Sname 学生姓名 ,
       max(case c.Cname when ‘语文‘ then b.score else null end)  语文 ,
       max(case c.Cname when ‘数学‘ then b.score else null end)  数学 ,
       max(case c.Cname when ‘英语‘ then b.score else null end)  英语 ,
       cast(avg(b.score) as decimal(18,2)) 平均分
from Student a
left join SC b on a.SID = b.SID
left join Course c on b.CID = c.CID
group by a.SID , a.Sname
order by 平均分 desc

--17.2 SQL 2000 动态
declare @sql nvarchar(4000)
set @sql = ‘select a.SID ‘ + ‘学生编号‘ + ‘ , a.Sname ‘ + ‘学生姓名‘
select @sql = @sql + ‘,max(case c.Cname when ‘‘‘+Cname+‘‘‘ then b.score else null end)  ‘+Cname+‘ ‘
from (select distinct Cname from Course) as t
set @sql = @sql + ‘ , cast(avg(b.score) as decimal(18,2)) ‘ + ‘平均分‘ + ‘ from Student a left join SC b on a.SID = b.SID left join Course c on b.CID = c.CID
group by a.SID , a.Sname order by ‘ + ‘平均分‘ + ‘ desc‘
exec(@sql)

--18、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
--方法1
select m.CID  课程编号 , m.Cname  课程名称 ,
  max(n.score)  最高分 ,
  min(n.score)  最低分 ,
  cast(avg(n.score) as decimal(18,2))  平均分 ,
  cast((select count(1) from SC where CID = m.CID and score >= 60)*100.0 / (select count(1) from SC where CID = m.CID) as decimal(18,2))  及格率 ,
  cast((select count(1) from SC where CID = m.CID and score >= 70 and score < 80 )*100.0 / (select count(1) from SC where CID = m.CID) as decimal(18,2))  中等率 ,
  cast((select count(1) from SC where CID = m.CID and score >= 80 and score < 90 )*100.0 / (select count(1) from SC where CID = m.CID) as decimal(18,2))  优良率 ,
  cast((select count(1) from SC where CID = m.CID and score >= 90)*100.0 / (select count(1) from SC where CID = m.CID) as decimal(18,2))  优秀率
from Course m , SC n
where m.CID = n.CID
group by m.CID , m.Cname
order by m.CID;

--方法2
select m.CID  课程编号 , m.Cname  课程名称 ,
  (select max(score) from SC where CID = m.CID)  最高分 ,
  (select min(score) from SC where CID = m.CID)  最低分 ,
  (select cast(avg(score) as decimal(18,2)) from SC where CID = m.CID)  平均分 ,
  cast((select count(1) from SC where CID = m.CID and score >= 60)*100.0 / (select count(1) from SC where CID = m.CID) as decimal(18,2))  及格率,
  cast((select count(1) from SC where CID = m.CID and score >= 70 and score < 80 )*100.0 / (select count(1) from SC where CID = m.CID) as decimal(18,2))  中等率 ,
  cast((select count(1) from SC where CID = m.CID and score >= 80 and score < 90 )*100.0 / (select count(1) from SC where CID = m.CID) as decimal(18,2))  优良率 ,
  cast((select count(1) from SC where CID = m.CID and score >= 90)*100.0 / (select count(1) from SC where CID = m.CID) as decimal(18,2))  优秀率
from Course m
order by m.CID

--19、按各科成绩进行排序，并显示排名
--19.1 sql 2000用子查询完成
--Score重复时保留名次空缺
select t.* , (select count(1) from SC where CID = t.CID and score > t.score) + 1 px from sc t order by t.cid , px;

方法二（推荐）：

select t.* ,  (select count( score) from SC where CID = t.CID and score > t.score) +1 px from sc t order by t.cid , px;

--Score重复时合并名次
select t.* ,  (select count(distinct score) from SC where CID = t.CID and score >= t.score) px from sc t order by t.cid , px;

方法二（推荐）：

select t.* ,  (select count(distinct score) from SC where CID = t.CID and score > t.score) +1 px from sc t order by t.cid , px;

--19.2 sql 2005用rank,DENSE_RANK完成
--Score重复时保留名次空缺(rank完成)
select t.* , px = rank() over(partition by cid order by score desc) from sc t order by t.CID , px
--Score重复时合并名次(DENSE_RANK完成)
select t.* , px = DENSE_RANK() over(partition by cid order by score desc) from sc t order by t.CID , px

--20、查询学生的总成绩并进行排名
--20.1 查询学生的总成绩
select m.SID  学生编号  ,
       m.Sname  学生姓名  ,
       sum(score) 总成绩
from Student m left join SC n on m.SID = n.SID
group by m.SID , m.Sname
order by  总成绩  desc;

--20.2 查询学生的总成绩并进行排名，sql 2000用子查询完成，分总分重复时保留名次空缺和不保留名次空缺两种。
select t1.* , px = (select count(1) from 
(
  select m.SID  学生编号  ,
         m.Sname  学生姓名  ,
         isnull(sum(score),0)  总成绩
  from Student m left join SC n on m.SID = n.SID
  group by m.SID , m.Sname
) t2 where 总成绩 > t1.总成绩) + 1 from 
(
  select m.SID  学生编号  ,
         m.Sname  学生姓名  ,
         isnull(sum(score),0)  总成绩
  from Student m left join SC n on m.SID = n.SID
  group by m.SID , m.Sname
) t1
order by px

select t1.* , px = (select count(distinct 总成绩) from 
(
  select m.SID  学生编号  ,
         m.Sname  学生姓名  ,
         isnull(sum(score),0)  总成绩
  from Student m left join SC n on m.SID = n.SID
  group by m.SID , m.Sname
) t2 where 总成绩 >= t1.总成绩) from 
(
  select m.SID  学生编号  ,
         m.Sname  学生姓名  ,
         isnull(sum(score),0)  总成绩
  from Student m left join SC n on m.SID = n.SID
  group by m.SID , m.Sname
) t1
order by px
--20.3 查询学生的总成绩并进行排名，sql 2005用rank,DENSE_RANK完成，分总分重复时保留名次空缺和不保留名次空缺两种。
select t.* , px = rank() over(order by  总成绩  desc) from
(
  select m.SID  学生编号  ,
         m.Sname  学生姓名  ,
         isnull(sum(score),0)  总成绩
  from Student m left join SC n on m.SID = n.SID
  group by m.SID , m.Sname
) t
order by px

select t.* , px = DENSE_RANK() over(order by  总成绩  desc) from
(
  select m.SID  学生编号  ,
         m.Sname  学生姓名  ,
         isnull(sum(score),0)  总成绩
  from Student m left join SC n on m.SID = n.SID
  group by m.SID , m.Sname
) t
order by px

--21、查询不同老师所教不同课程平均分从高到低显示
select m.TID ,n.Cname, m.Tname , cast(avg(o.score) as decimal(18,2)) avg_score
from Teacher m , Course n , SC o
where m.TID = n.TID and n.CID = o.CID
group by m.TID, n.Cname , m.Tname
order by avg_score desc;

--22、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
--22.1 sql 2000用子查询完成
--Score重复时保留名次空缺
方法一：select * from (select t.* , (select count(1) from SC where CID = t.CID and score > t.score) + 1 px from sc t) m where px between 2 and 3 order by m.cid;

方法二（推荐）：select  t.*,(select count(score) from sc where cid=t.cid and score>t.score) +1 px from sc t having px>=2 and px<=3 order by t.cid,px;

--Score重复时合并名次
方法一：select * from (select t.* , (select count(distinct score) from SC where CID = t.CID and score >= t.score) px from sc t) m where px between 2 and 3 order by m.cid;

方法二（推荐）：select t.* ,  (select count(distinct score) from SC where CID = t.CID and score > t.score) + 1 px from sc t having px>=2 and px<=3 order by t.cid , px;

--22.2 sql 2005用rank,DENSE_RANK完成
--Score重复时保留名次空缺(rank完成)
select * from (select t.* , px = rank() over(partition by cid order by score desc) from sc t) m where px between 2 and 3 order by m.CID , m.px
--Score重复时合并名次(DENSE_RANK完成)
select * from (select t.* , px = DENSE_RANK() over(partition by cid order by score desc) from sc t) m where px between 2 and 3 order by m.CID , m.px

--23、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
--23.1 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
--横向显示
select Course.CID  课程编号  , Cname as  课程名称  ,
       sum(case when score >= 85 then 1 else 0 end)  ‘85-100‘ ,
       sum(case when score >= 70 and score < 85 then 1 else 0 end)  ‘70-85‘ ,
       sum(case when score >= 60 and score < 70 then 1 else 0 end)  ‘60-70‘ ,
       sum(case when score < 60 then 1 else 0 end)  ‘0-60‘
from sc , Course
where SC.CID = Course.CID
group by Course.CID , Course.Cname
order by Course.CID;

--纵向显示1(显示存在的分数段)
select m.CID  课程编号  , m.Cname  课程名称  ,  (
    case when n.score >= 85 then ‘85-100‘
         when n.score >= 70 and n.score < 85 then ‘70-85‘
         when n.score >= 60 and n.score < 70 then ‘60-70‘
         else ‘0-60‘
        end) 分数段 ,
       count(1) 数量
from Course m , sc n
where m.CID = n.CID
group by m.CID , m.Cname , (
    case when n.score >= 85 then ‘85-100‘
         when n.score >= 70 and n.score < 85 then ‘70-85‘
         when n.score >= 60 and n.score < 70 then ‘60-70‘
         else ‘0-60‘
        end)
order by m.CID , m.Cname , 分数段;

--纵向显示2(显示存在的分数段，不存在的分数段用0显示)
select m.CID  课程编号  , m.Cname  课程名称  ,  (
    case when n.score >= 85 then ‘85-100‘
         when n.score >= 70 and n.score < 85 then ‘70-85‘
         when n.score >= 60 and n.score < 70 then ‘60-70‘
         else ‘0-60‘
        end) 分数段,
       count(1) 数量
from Course m , sc n
where m.CID = n.CID
group by m.CID , m.Cname , (
    case when n.score >= 85 then ‘85-100‘
    when n.score >= 70 and n.score < 85 then ‘70-85‘
    when n.score >= 60 and n.score < 70 then ‘60-70‘
    else ‘0-60‘
    end)
order by m.CID , m.Cname , 分数段;

--23.2 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 , <60 及所占百分比
--横向显示
select m.CID 课程编号, m.Cname 课程名称,
       (select count(1) from SC where CID = m.CID and score < 60)  ‘0-60‘ ,
       cast((select count(1) from SC where CID = m.CID and score < 60)*100.0 / (select count(1) from SC where CID = m.CID) as decimal(18,2))  百分比 ,
       (select count(1) from SC where CID = m.CID and score >= 60 and score < 70)  ‘60-70‘ ,
       cast((select count(1) from SC where CID = m.CID and score >= 60 and score < 70)*100.0 / (select count(1) from SC where CID = m.CID) as decimal(18,2))  百分比 ,
       (select count(1) from SC where CID = m.CID and score >= 70 and score < 85)  ‘70-85‘ ,
       cast((select count(1) from SC where CID = m.CID and score >= 70 and score < 85)*100.0 / (select count(1) from SC where CID = m.CID) as decimal(18,2))  百分比 ,
       (select count(1) from SC where CID = m.CID and score >= 85)  ‘85-100‘ ,
       cast((select count(1) from SC where CID = m.CID and score >= 85)*100.0 / (select count(1) from SC where CID = m.CID) as decimal(18,2))  百分比
from Course m
order by m.CID;

--纵向显示1(显示存在的分数段)
select m.CID  课程编号  , m.Cname  课程名称  , (
  case when n.score >= 85 then ‘85-100‘
       when n.score >= 70 and n.score < 85 then ‘70-85‘
       when n.score >= 60 and n.score < 70 then ‘60-70‘
       else ‘0-60‘
  end) 分数段,
  count(1) 数量 ,  
  cast(count(1) * 100.0 / (select count(1) from sc where CID = m.CID) as decimal(18,2))  百分比
from Course m , sc n
where m.CID = n.CID
group by m.CID , m.Cname , (
  case when n.score >= 85 then ‘85-100‘
       when n.score >= 70 and n.score < 85 then ‘70-85‘
       when n.score >= 60 and n.score < 70 then ‘60-70‘
       else ‘0-60‘
  end)
order by m.CID , m.Cname , 分数段;

--纵向显示2(显示存在的分数段，不存在的分数段用0显示)
select m.CID  课程编号  , m.Cname  课程名称  ,(
    case when n.score >= 85 then ‘85-100‘
         when n.score >= 70 and n.score < 85 then ‘70-85‘
         when n.score >= 60 and n.score < 70 then ‘60-70‘
         else ‘0-60‘
        end) 分数段,
       count(1) 数量 ,
       cast(count(1) * 100.0 / (select count(1) from sc where CID = m.CID) as decimal(18,2))  百分比
from Course m , sc n
where m.CID = n.CID
group by m.CID , m.Cname , (
    case when n.score >= 85 then ‘85-100‘
    when n.score >= 70 and n.score < 85 then ‘70-85‘
    when n.score >= 60 and n.score < 70 then ‘60-70‘
    else ‘0-60‘
    end)
order by m.CID , m.Cname , 分数段;

--24、查询学生平均成绩及其名次
--24.1 查询学生的平均成绩并进行排名，sql 2000用子查询完成，分平均成绩重复时保留名次空缺和不保留名次空缺两种。

方法一（推荐）：

select a.sid 学生编号,a.sname 学生姓名,cast(avg(b.score) as decimal(18,2)) 平均成绩,
       (select count(*) from sc where sc.cid=b.cid and sc.score>=b.score)+1 px
from student a left join sc b on a.sid=b.sid group by a.sid,a.sname order by px;

select t1.* , (select count(1) from
    (
        select m.SID  学生编号  ,
               m.Sname  学生姓名  ,
               cast(avg(score) as decimal(18,2))  平均成绩
        from Student m left join SC n on m.SID = n.SID
        group by m.SID , m.Sname
    ) t2 where 平均成绩 > t1.平均成绩) + 1 px from
    (
        select m.SID  学生编号  ,
               m.Sname  学生姓名  ,
               cast(avg(score) as decimal(18,2))  平均成绩
        from Student m left join SC n on m.SID = n.SID
        group by m.SID , m.Sname
    ) t1
order by px;

--24.2 查询学生的平均成绩并进行排名，sql 2005用rank,DENSE_RANK完成，分平均成绩重复时保留名次空缺和不保留名次空缺两种。
select t.* , px = rank() over(order by  平均成绩  desc) from
(
  select m.SID  学生编号  ,
         m.Sname  学生姓名  ,
         isnull(cast(avg(score) as decimal(18,2)),0)  平均成绩 
  from Student m left join SC n on m.SID = n.SID
  group by m.SID , m.Sname
) t
order by px

select t.* , px = DENSE_RANK() over(order by  平均成绩  desc) from
(
  select m.SID  学生编号  ,
         m.Sname  学生姓名  ,
         isnull(cast(avg(score) as decimal(18,2)),0)  平均成绩 
  from Student m left join SC n on m.SID = n.SID
  group by m.SID , m.Sname
) t
order by px
 
--25、查询各科成绩前三名的记录
--25.1 分数重复时保留名次空缺
select m.* , n.CID , n.score from Student m, SC n where m.SID = n.SID and n.score in
(select top 3 score from sc where CID = n.CID order by score desc) order by n.CID , n.score desc

SELECT a.*,COUNT(B.score) +1 as ranking
FROM SC a LEFT JOIN SC b
                    ON a.CId = b.CId AND a.score<b.score
GROUP BY a.CId,a.SId,a.score
HAVING ranking <= 3
ORDER BY a.CId,ranking;

--25.2 分数重复时不保留名次空缺，合并名次
--sql 2000用子查询实现
select * from (select t.* , (select count(distinct score) from SC where CID = t.CID and score >= t.score) px from sc t) m where px between 1 and 3 order by m.Cid , m.px;

--sql 2005用DENSE_RANK实现
select * from (select t.* , px = DENSE_RANK() over(partition by Cid order by score desc) from sc t) m where px between 1 and 3 order by m.CID , m.px

--26、查询每门课程被选修的学生数
select sc.Cid,course.Cname , count(SID) 学生数  from course,sc where course.cid=sc.cid group by CID;

--27、查询出只有两门课程的全部学生的学号和姓名
select Student.SID , Student.Sname
from Student , SC
where Student.SID = SC.SID
group by Student.SID , Student.Sname
having count(SC.CID) = 2
order by Student.SID;

--28、查询男生、女生人数
select count(Ssex) as 男生人数 from Student where Ssex = N‘男‘
select count(Ssex) as 女生人数 from Student where Ssex = N‘女‘
select sum(case when Ssex = N‘男‘ then 1 else 0 end)  男生人数 ,sum(case when Ssex = N‘女‘ then 1 else 0 end)  女生人数  from student
select case when Ssex = N‘男‘ then N‘男生人数‘ else N‘女生人数‘ end  男女情况  , count(1)  人数  from student group by case when Ssex = N‘男‘ then N‘男生人数‘ else N‘女生人数‘ end;

--29、查询名字中含有"风"字的学生信息
select * from student where sname like N‘%风%‘;

--30、查询同名同性学生名单，并统计同名人数
select Sname  学生姓名 , count(*)  人数  from Student group by Sname having count(*) > 1;

--31、查询1990年出生的学生名单(注：Student表中Sage列的类型是datetime)

select * from student where Sage like ‘1990%‘;
select * from Student where year(sage) = 1990;

--32、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 m.CID , m.Cname , cast(avg(n.score) as decimal(18,2)) avg_score
from Course m, SC n
where m.CID = n.CID   
group by m.CID , m.Cname
order by avg_score desc, m.CID asc;

--33、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 a.SID , a.Sname , cast(avg(b.score) as decimal(18,2)) avg_score
from Student a , sc b
where a.SID = b.SID
group by a.SID , a.Sname
having cast(avg(b.score) as decimal(18,2)) >= 85
order by a.SID;

--34、查询课程名称为"数学"，且分数低于60的学生姓名和分数
select sname , score
from Student , SC , Course
where SC.SID = Student.SID and SC.CID = Course.CID and Course.Cname = N‘数学‘ and score < 60;

--35、查询所有学生的课程及分数情况；
select Student.* , Course.Cname , SC.CID , SC.score 
from Student, SC , Course
where Student.SID = SC.SID and SC.CID = Course.CID
order by Student.SID , SC.CID

--36、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 Student.* , Course.Cname , SC.CID , SC.score 
from Student, SC , Course
where Student.SID = SC.SID and SC.CID = Course.CID and SC.score >= 70
order by Student.SID , SC.CID

--37、查询不及格的课程
select Student.* , Course.Cname , SC.CID , SC.score 
from Student, SC , Course
where Student.SID = SC.SID and SC.CID = Course.CID and SC.score < 60
order by Student.SID , SC.CID

--38、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 Student.* , Course.Cname , SC.CID , SC.score 
from Student, SC , Course
where Student.SID = SC.SID and SC.CID = Course.CID and SC.CID = ‘01‘ and SC.score >= 80
order by Student.SID , SC.CID；

--39、求每门课程的学生人数
select Course.CID , Course.Cname,count(1) 学生人数
from course,sc
where course.cid=sc.CID
group by Course.CID , Course.Cname
order by Course.CID , Course.Cname;

方法二：

select CId,count(SId)  

from sc group by sc.CId

--40、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
--40.1 当最高分只有一个时
sSELECT Student.*,SC.CId,score
FROM Student JOIN SC ON Student.SId = SC.SId
             JOIN Course ON SC.CId = Course.CId
             JOIN Teacher ON Course.TId = Teacher.TId
WHERE Tname = ‘张三‘
ORDER BY score DESC LIMIT 1;

--40.2 当最高分出现多个时
select Student.* , Course.Cname , SC.CID , SC.score 
from Student, SC , Course , Teacher
where Student.SID = SC.SID and SC.CID = Course.CID and Course.TID = Teacher.TID and Teacher.Tname = N‘张三‘ and
SC.score = (select max(SC.score) from SC , Course , Teacher where SC.CID = Course.CID and Course.TID = Teacher.TID and Teacher.Tname = N‘张三‘)；

SELECT Student.*,SC.CId,score
FROM Student JOIN SC ON Student.SId = SC.SId
             JOIN Course ON SC.CId = Course.CId
             JOIN Teacher ON Course.TId = Teacher.TId
WHERE Tname = ‘张三‘ AND score IN
                       (SELECT MAX(score) FROM
                           SC JOIN Course ON SC.CId = Course.CId
                              JOIN Teacher ON Course.TId = Teacher.TId
                        WHERE Tname = ‘张三‘);

--41、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
--方法1
select m.* from SC m ,(select CID , score from SC group by CID , score having count(1) > 1) n
where m.CID= n.CID and m.score = n.score order by m.CID , m.score , m.SID;
--方法2
select m.* from SC m where exists (select 1 from (select CID , score from SC group by CID , score having count(1) > 1) n
where m.CID= n.CID and m.score = n.score) order by m.CID , m.score , m.SID

--42、查询每门功成绩最好的前两名
select t.* from sc t where score in (select top 2 score from sc where CID = T.CID order by score desc) order by t.CID , t.score desc

--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 Course.CID , Course.Cname , count(*)  学生人数 
from Course , SC
where Course.CID = SC.CID
group by  Course.CID , Course.Cname
having count(*) >= 5
order by  学生人数  desc , Course.CID;

--44、检索至少选修两门课程的学生学号
select student.SID , student.Sname
from student , SC
where student.SID = SC.SID
group by student.SID , student.Sname
having count(1) >= 2
order by student.SID

--45、查询选修了全部课程的学生信息
--方法1 根据数量来完成
select student.* from student where SID in
(select SID from sc group by SID having count(1) = (select count(1) from course))

--方法2 使用双重否定来完成
select t.* from student t where t.SID not in
(
  select distinct m.SID from
  (
    select SID , CID from student , course
  ) m where not exists (select 1 from sc n where n.SID = m.SID and n.CID = m.CID)
)
--方法3 使用双重否定来完成
select t.* from student t where not exists(select 1 from
(
  select distinct m.SID from
  (
    select SID , CID from student , course
  ) m where not exists (select 1 from sc n where n.SID = m.SID and n.CID = m.CID)
) k where k.SID = t.SID
)

--46、查询各学生的年龄
--46.1 只按照年份来算
select * , datediff(yy , sage , getdate())  年龄  from student

select Sname,Sage,
       year(now())-year(Sage) as 年龄
from student;

--46.2 按照出生日期来算，当前月日 < 出生年月的月日则，年龄减一
SELECT SId,Sname,TIMESTAMPDIFF(YEAR,Sage,NOW()) AS 年龄
FROM Student;

--47、查询本周过生日的学生
SELECT * FROM student where
        week(Sage)=week(now());

--48、查询下周过生日的学生
SELECT * FROM student where
        week(Sage)=week(now())+1;

--49、查询本月过生日的学生
SELECT * FROM student
where month(Sage)=month(now());

--50、查询下月过生日的学生
SELECT * FROM student
where month(Sage)=month(now())+1;

---恢复内容结束---

MySQL经典50题-2019更新版

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