标签：检索 方法 group res 语句 并且 恢复 总数 art
---恢复内容开始---
--1.学生表
Student(SID,Sname,Sage,Ssex) --SID 学生编号,Sname 学生姓名,Sage 出生年月,Ssex 学生性别
--2.课程表
Course(CID,Cname,TID) --CID --课程编号,Cname 课程名称,TID 教师编号
--3.教师表
Teacher(TID,Tname) --TID 教师编号,Tname 教师姓名
--4.成绩表
SC(SID,CID,score) --SID 学生编号,CID 课程编号,score 分数
*/
--创建测试数据
create table Student(SID varchar(10),Sname nvarchar(10),Sage datetime,Ssex nvarchar(10));
insert into Student values(‘01‘ , ‘赵雷‘ , ‘1990-01-01‘ , ‘男‘);
insert into Student values(‘02‘ , ‘钱电‘ , ‘1990-12-21‘ , ‘男‘);
insert into Student values(‘03‘ , ‘孙风‘ , ‘1990-05-20‘ , ‘男‘);
insert into Student values(‘04‘ , ‘李云‘ , ‘1990-08-06‘ , ‘男‘);
insert into Student values(‘05‘ , ‘周梅‘ , ‘1991-12-01‘ , ‘女‘);
insert into Student values(‘06‘ , ‘吴兰‘ , ‘1992-03-01‘ , ‘女‘);
insert into Student values(‘07‘ , ‘郑竹‘ , ‘1989-07-01‘ , ‘女‘);
insert into Student values(‘08‘ , ‘王菊‘ , ‘1990-01-20‘ , ‘女‘);
create table Course(CID varchar(10),Cname nvarchar(10),TID varchar(10));
insert into Course values(‘01‘ , ‘语文‘ , ‘02‘);
insert into Course values(‘02‘ , ‘数学‘ , ‘01‘);
insert into Course values(‘03‘ , ‘英语‘ , ‘03‘);
create table Teacher(TID varchar(10),Tname nvarchar(10));
insert into Teacher values(‘01‘ , ‘张三‘);
insert into Teacher values(‘02‘ , ‘李四‘);
insert into Teacher values(‘03‘ , ‘王五‘);
create table SC(SID varchar(10),CID varchar(10),score decimal(18,1));
insert into SC values(‘01‘ , ‘01‘ , 80);
insert into SC values(‘01‘ , ‘02‘ , 90);
insert into SC values(‘01‘ , ‘03‘ , 99);
insert into SC values(‘02‘ , ‘01‘ , 70);
insert into SC values(‘02‘ , ‘02‘ , 60);
insert into SC values(‘02‘ , ‘03‘ , 80);
insert into SC values(‘03‘ , ‘01‘ , 80);
insert into SC values(‘03‘ , ‘02‘ , 80);
insert into SC values(‘03‘ , ‘03‘ , 80);
insert into SC values(‘04‘ , ‘01‘ , 50);
insert into SC values(‘04‘ , ‘02‘ , 30);
insert into SC values(‘04‘ , ‘03‘ , 20);
insert into SC values(‘05‘ , ‘01‘ , 76);
insert into SC values(‘05‘ , ‘02‘ , 87);
insert into SC values(‘06‘ , ‘01‘ , 31);
insert into SC values(‘06‘ , ‘03‘ , 34);
insert into SC values(‘07‘ , ‘02‘ , 89);
insert into SC values(‘07‘ , ‘03‘ , 98);
--1、查询"01"课程比"02"课程成绩高的学生的信息及课程分数
--1.1、查询同时存在"01"课程和"02"课程的情况
select a.* , b.score 课程01的分数,c.score 课程02的分数 from Student a , SC b , SC c
where a.SID = b.SID and a.SID = c.SID and b.CID = ‘01‘ and c.CID = ‘02‘ and b.score > c.score;
--1.2、查询同时存在"01"课程和"02"课程的情况和存在"01"课程但可能不存在"02"课程的情况(不存在时显示为null)(以下存在相同内容时不再解释)
select a.* , b.score 课程01的分数,c.score 课程02的分数 from Student a
left join SC b on a.SID = b.SID and b.CID = ‘01‘
left join SC c on a.SID = c.SID and c.CID = ‘02‘
where b.score > isnull(c.score);
--2、查询"01"课程比"02"课程成绩低的学生的信息及课程分数
--2.1、查询同时存在"01"课程和"02"课程的情况
select a.* , b.score 课程01的分数 ,c.score 课程02的分数 from Student a , SC b , SC c
where a.SID = b.SID and a.SID = c.SID and b.CID = ‘01‘ and c.CID = ‘02‘ and b.score < c.score;
--2.2、查询同时存在"01"课程和"02"课程的情况和不存在"01"课程但存在"02"课程的情况
select a.* , b.score 课程01的分数 ,c.score 课程02的分数 from Student a
left join SC b on a.SID = b.SID and b.CID = ‘01‘
left join SC c on a.SID = c.SID and c.CID = ‘02‘
where isnull(b.score) < c.score;
如要显示没选课的学生(显示为NULL)，需要使用join:
--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---
标签：检索 方法 group res 语句 并且 恢复 总数 art
原文地址：https://www.cnblogs.com/szjblog/p/11258314.html
