sql练习题

sql练习题
问题及描述：

–1.学生表
Student(s_id,s_name,s_birth,s_sex) --s_id 学生编号,s_name 学生姓名,s_birth 出生年月,s_sex 学生性别
–2.课程表
Course(c_id,c_name,t_id) --c_id --课程编号,c_name 课程名称,t_id 教师编号
–3.教师表
Teacher(t_id,t_name) --t_id 教师编号,t_name 教师姓名
–4.成绩表
Score(s_id,c_id,s_score) --s_id 学生编号,c_id 课程编号,s_score 分数
*/

创建测试数据
```
create table Student(s_id varchar(10),s_name nvarchar(10),s_brith datetime,s_sex 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(c_id varchar(10),c_name nvarchar(10),t_id 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(t_id varchar(10),t_name 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 Score(s_id varchar(10),c_id varchar(10),s_score decimal(18,1));

insert into Score values('01' ,'01' , 80);
insert into Score values('01' ,'02' , 90);
insert into Score values('01' ,'03' , 99);
insert into Score values('02' ,'01' , 70);
insert into Score values('02' ,'02' , 60);
insert into Score values('02' ,'03' , 80);
insert into Score values('03' ,'01' , 80);
insert into Score values('03' ,'02' , 80);
insert into Score values('03' ,'03' , 80);
insert into Score values('04' ,'01' , 50);
insert into Score values('04' ,'02' , 30);
insert into Score values('04' ,'03' , 20);
insert into Score values('05' ,'01' , 76);
insert into Score values('05' ,'02' , 87);
insert into Score values('06' ,'01' , 31);
insert into Score values('06' ,'03' , 34);
insert into Score values('07' ,'02' , 89);
insert into Score values('07' ,'03' , 98);
```
５０道练习题

查询"01"课程比"02"课程成绩高的学生的信息及课程分数

解题思路
1. 需要三张表
  1. 学生信息 stu
  2. 01课程 01sco
  3. 02课程 02sco
2. 需要展现的信息
  1. stu的所有信息
  2. 01与02课程的分数
3. 关联关系
  1. stu.学生id = 01sco.学生id = 02sco.学生id
4. 逻辑关系
  1. 01sco.课程编号 = 01
  2. 01sco.课程编号 = 02
  3. 01sco.成绩 > 02sco.成绩
```
select stu.*, 01sco.s_score '01', 02sco.s_score '02'
from student stu,
     score 01sco,
     score 02sco
where stu.s_id = 01sco.s_id
  and stu.s_id = 02sco.s_id
  and 01sco.c_id = 01
  and 02sco.c_id = 02
  and 01sco.s_score > 02sco.s_score;
```
```
select a.*, b.s_score '01', c.s_score '02'
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
  and b.s_score > c.s_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.s_score '01', c.s_score '02'
from Student a
         right join Score b on a.s_id = b.s_id and b.c_id = '01'
         right join Score c on a.s_id = c.s_id and c.c_id = '02'
where b.s_score > c.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

2.1、查询同时存在"01"课程和"02"课程的情况
```
select a.*, b.s_score '01', c.s_score '02'
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'
  and b.s_score < c.s_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.s_score '01', c.s_score '02'
from student a
         left join score b on a.s_id = b.s_id and b.c_id = '01'
         left join score c on a.s_id = c.s_id and c.c_id = '02'
where b.s_score < c.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 a.s_id, a.s_name, cast(avg(b.s_score) as decimal(18, 2)) avg_s_score
from Student a,
     score b
where a.s_id = b.s_id
group by a.s_id, a.s_name
having cast(avg(b.s_score) as decimal(18, 2)) >= 60
order by a.s_id;
```
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.s_id,a.s_name,cast(avg(b.s_score) as decimal(18,2)) avf_s_score
from student a,
     score b
where a.s_id = b.s_id
group by a.s_id, a.s_name
having cast(avg(b.s_score) as decimal(8,1)) <=60
order by a.s_id;
```
4.2、查询在sc表中不存在成绩的学生信息的SQL语句。
```
select a.s_id, a.s_name, ifnull(cast(avg(b.s_score) as decimal(18, 2)), 0) avg_s_score
from Student a
         left join score b
                   on a.s_id = b.s_id
group by a.s_id, a.s_name
having ifnull(cast(avg(b.s_score) as decimal(18, 2)), 0) < 60
order by a.s_id
```
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.s_id 编号, a.s_name 姓名, count(b.c_id) 选课总数, sum(s_score) 所有课程总成绩from Student a,     Score bwhere a.s_id = b.s_idgroup by a.s_id, a.s_nameorder by a.s_id;
```
5.2、查询所有(包括有成绩和无成绩)的SQL。
```
select a.s_id 编号,a.s_name 学生姓名,count(b.c_id) 选课总数,sum(s_score) 所有课程总成绩from student a         left join score b on a.s_id = b.s_idgroup by a.s_id, a.s_nameorder by a.s_id;
```
查询"李"姓老师的数量

方法1
```
select count(t.t_name) 李姓老师的数量from teacherwhere t_name like '李%';
```
方法2
```
select count(t_name) 李姓老师的数量from Teacherwhere left(t_name, 1) = '李';
```
查询学过"张三"老师授课的同学的信息
```
select distinct a.*from student a         left join score s on a.s_id = s.s_id         left join course c on s.c_id = c.c_id         left join teacher t on c.t_id = t.t_idwhere t.t_name = '张三'order by a.s_id;
```
查询没学过"张三"老师授课的同学的信息
```
select a.*from student awhere s_id not in (    #先查询出上过张三老师课同学的编号    select distinct s.s_id    from score s             left join course c on s.c_id = c.c_id             left join teacher t on t.t_id = c.t_id    where t.t_name = '张三')order by a.s_id;
```
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,     Scorewhere Student.s_id = Score.s_id  and Score.c_id = '01'  and exists(Select 1 from Score Score_2 where Score_2.s_id = Score.s_id and Score_2.c_id = '02')order by Student.s_id;
```
–方法2
```
select Student.*from Student,     Scorewhere Student.s_id = Score.s_id  and Score.c_id = '02'  and exists(Select 1 from Score Score_2 where Score_2.s_id = Score.s_id and Score_2.c_id = '01')order by Student.s_id;
```
–方法3
```
select m.*from Student mwhere s_id in      (          select s_id          from (                   select distinct s_id                   from Score                   where c_id = '01'                   union all                   select distinct s_id                   from Score                   where c_id = '02'               ) t          group by s_id          having count(1) = 2      )order by m.s_id;
```
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,     Scorewhere Student.s_id = Score.s_id  and Score.c_id = '01'  and not exists(Select 1 from Score Score_2 where Score_2.s_id = Score.s_id and Score_2.c_id = '02')order by Student.s_id;
```
–方法2
```
select Student.*from Student,     Scorewhere Student.s_id = Score.s_id  and Score.c_id = '01'  and Student.s_id not in      (Select Score_2.s_id from Score Score_2 where Score_2.s_id = Score.s_id and Score_2.c_id = '02')order by Student.s_id;
```
查询没有学全所有课程的同学的信息

–11.1、
```
select Student.*from Student,     Scorewhere Student.s_id = Score.s_idgroup by Student.s_id, Student.s_name, student.s_brith, Student.s_sexhaving count(c_id) < (select count(c_id) from Course);
```
–11.2
```
select Student.*from Student         left join Score                   on Student.s_id = Score.s_idgroup by Student.s_id, Student.s_name, Student.s_brith, Student.s_sexhaving count(c_id) < (select count(c_id) 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 , Score where Student.s_id = Score.s_id and Score.c_id in (select c_id from Score where s_id ='01') and Student.s_id <>'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 s_id in
(select distinct Score.s_id from Score where s_id <>'01' and Score.c_id in (select distinct c_id from Score where s_id ='01')
group by Score.s_id having count(1) = (select count(1) from Score where s_id='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.s_id not in
(select distinct sc.s_id from sc , course , teacher where sc.c_id = course.c_id and course.t_id = teacher.t_id and teacher.t_name ='张三')
order by student.s_id

–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.s_id , student.s_name , cast(avg(s_score) as decimal(18,2)) avg_s_score from student , sc
where student.s_id = Score.s_id and student.s_id in (select s_id from Score where s_score < 60 group by s_id having count(1) >= 2)
group by student.s_id , student.s_name

–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.c_id , sc.s_score from student , sc
where student.s_id = Score.s_id and sc.s_score < 60 and sc.c_id ='01'
order by sc.s_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.s_id 学生编号 , a.s_name 学生姓名 ,
max(case c.c_name when'语文' then b.s_score else null end) 语文 ,
max(case c.c_name when'数学' then b.s_score else null end) 数学 ,
max(case c.c_name when'英语' then b.s_score else null end) 英语 ,
cast(avg(b.s_score) as decimal(18,2)) 平均分
from Student a
left join Score b on a.s_id = b.s_id
left join Course c on b.c_id = c.c_id
group by a.s_id , a.s_name
order by 平均分 desc
–17.2 SQL 2000 动态
declare @sql nvarchar(4000)
set @sql ='select a.s_id ' +'学生编号' + ' , a.s_name ' +'学生姓名'
select @sql = @sql +',max(case c.c_name when'''+c_name+''' then b.s_score else null end)'+c_name+' '
from (select distinct c_name from Course) as t
set @sql = @sql + ' , cast(avg(b.s_score) as decimal(18,2)) ' +'平均分' + ' from Student a left join Score b on a.s_id = b.s_id left join Course c on b.c_id = c.c_id
group by a.s_id , a.s_name 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.c_id 课程编号 , m.c_name 课程名称 ,
max(n.s_score) 最高分 ,
min(n.s_score) 最低分 ,
cast(avg(n.s_score) as decimal(18,2)) 平均分 ,
cast((select count(1) from Score where c_id = m.c_id and s_score >= 60)100.0 / (select count(1) from Score where c_id = m.c_id) as decimal(18,2)) 及格率 ,
cast((select count(1) from Score where c_id = m.c_id and s_score >= 70 and s_score < 80 )100.0 / (select count(1) from Score where c_id = m.c_id) as decimal(18,2)) 中等率 ,
cast((select count(1) from Score where c_id = m.c_id and s_score >= 80 and s_score < 90 )100.0 / (select count(1) from Score where c_id = m.c_id) as decimal(18,2)) 优良率 ,
cast((select count(1) from Score where c_id = m.c_id and s_score >= 90)100.0 / (select count(1) from Score where c_id = m.c_id) as decimal(18,2)) 优秀率
from Course m , Score n
where m.c_id = n.c_id
group by m.c_id , m.c_name
order by m.c_id
–方法2
select m.c_id 课程编号 , m.c_name 课程名称 ,
(select max(s_score) from Score where c_id = m.c_id) 最高分 ,
(select min(s_score) from Score where c_id = m.c_id) 最低分 ,
(select cast(avg(s_score) as decimal(18,2)) from Score where c_id = m.c_id) 平均分 ,
cast((select count(1) from Score where c_id = m.c_id and s_score >= 60)100.0 / (select count(1) from Score where c_id = m.c_id) as decimal(18,2)) 及格率,
cast((select count(1) from Score where c_id = m.c_id and s_score >= 70 and s_score < 80 )100.0 / (select count(1) from Score where c_id = m.c_id) as decimal(18,2)) 中等率 ,
cast((select count(1) from Score where c_id = m.c_id and s_score >= 80 and s_score < 90 )100.0 / (select count(1) from Score where c_id = m.c_id) as decimal(18,2)) 优良率 ,
cast((select count(1) from Score where c_id = m.c_id and s_score >= 90)100.0 / (select count(1) from Score where c_id = m.c_id) as decimal(18,2)) 优秀率
from Course m
order by m.c_id

–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
–s_score重复时保留名次空缺
select t.* , px = (select count(1) from Score where c_id = t.c_id and s_score > t.s_score) + 1 from sc t order by t.c_id , px
–s_score重复时合并名次
select t.* , px = (select count(distinct s_score) from Score where c_id = t.c_id and s_score >= t.s_score) from sc t order by t.c_id , px
–19.2 sql 2005用rank,DENSE_RANK完成
–s_score重复时保留名次空缺(rank完成)
select t.* , px = rank() over(partition by c_id order by s_score desc) from sc t order by t.c_id , px
–s_score重复时合并名次(DENSE_RANK完成)
select t.* , px = DENSE_RANK() over(partition by c_id order by s_score desc) from sc t order by t.c_id , 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.s_id 学生编号 ,
m.s_name 学生姓名 ,
isnull(sum(s_score),0) 总成绩
from Student m left join Score n on m.s_id = n.s_id
group by m.s_id , m.s_name
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.s_id 学生编号 ,
m.s_name 学生姓名 ,
isnull(sum(s_score),0) 总成绩
from Student m left join Score n on m.s_id = n.s_id
group by m.s_id , m.s_name
) t2 where 总成绩 > t1.总成绩) + 1 from
(
select m.s_id 学生编号 ,
m.s_name 学生姓名 ,
isnull(sum(s_score),0) 总成绩
from Student m left join Score n on m.s_id = n.s_id
group by m.s_id , m.s_name
) t1
order by px

select t1.* , px = (select count(distinct 总成绩) from
(
select m.s_id 学生编号 ,
m.s_name 学生姓名 ,
isnull(sum(s_score),0) 总成绩
from Student m left join Score n on m.s_id = n.s_id
group by m.s_id , m.s_name
) t2 where 总成绩 >= t1.总成绩) from
(
select m.s_id 学生编号 ,
m.s_name 学生姓名 ,
isnull(sum(s_score),0) 总成绩
from Student m left join Score n on m.s_id = n.s_id
group by m.s_id , m.s_name
) 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.s_id 学生编号 ,
m.s_name 学生姓名 ,
isnull(sum(s_score),0) 总成绩
from Student m left join Score n on m.s_id = n.s_id
group by m.s_id , m.s_name
) t
order by px

select t.* , px = DENSE_RANK() over(order by 总成绩 desc) from
(
select m.s_id 学生编号 ,
m.s_name 学生姓名 ,
isnull(sum(s_score),0) 总成绩
from Student m left join Score n on m.s_id = n.s_id
group by m.s_id , m.s_name
) 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.t_id , m.t_name , cast(avg(o.s_score) as decimal(18,2)) avg_s_score
from Teacher m , Course n , Score o
where m.t_id = n.t_id and n.c_id = o.c_id
group by m.t_id , m.t_name
order by avg_s_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
–s_score重复时保留名次空缺
select * from (select t.* , px = (select count(1) from Score where c_id = t.c_id and s_score > t.s_score) + 1 from sc t) m where px between 2 and 3 order by m.c_id , m.px
–s_score重复时合并名次
select * from (select t.* , px = (select count(distinct s_score) from Score where c_id = t.c_id and s_score >= t.s_score) from sc t) m where px between 2 and 3 order by m.c_id , m.px
–22.2 sql 2005用rank,DENSE_RANK完成
–s_score重复时保留名次空缺(rank完成)
select * from (select t.* , px = rank() over(partition by c_id order by s_score desc) from sc t) m where px between 2 and 3 order by m.c_id , m.px
–s_score重复时合并名次(DENSE_RANK完成)
select * from (select t.* , px = DENSE_RANK() over(partition by c_id order by s_score desc) from sc t) m where px between 2 and 3 order by m.c_id , 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.c_id 课程编号 , c_name as 课程名称 ,
sum(case when s_score >= 85 then 1 else 0 end) 85-100 ,
sum(case when s_score >= 70 and s_score < 85 then 1 else 0 end) 70-85 ,
sum(case when s_score >= 60 and s_score < 70 then 1 else 0 end) 60-70 ,
sum(case when s_score < 60 then 1 else 0 end) 0-60
from sc , Course
where Score.c_id = Course.c_id
group by Course.c_id , Course.c_name
order by Course.c_id
–纵向显示1(显示存在的分数段)
select m.c_id 课程编号 , m.c_name 课程名称 , 分数段 = (
case when n.s_score >= 85 then'85-100'
when n.s_score >= 70 and n.s_score < 85 then'70-85'
when n.s_score >= 60 and n.s_score < 70 then'60-70'
else'0-60'
end) ,
count(1) 数量
from Course m , sc n
where m.c_id = n.c_id
group by m.c_id , m.c_name , (
case when n.s_score >= 85 then'85-100'
when n.s_score >= 70 and n.s_score < 85 then'70-85'
when n.s_score >= 60 and n.s_score < 70 then'60-70'
else'0-60'
end)
order by m.c_id , m.c_name , 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.c_id 课程编号 , m.c_name 课程名称 , 分数段 = (
case when n.s_score >= 85 then'85-100'
when n.s_score >= 70 and n.s_score < 85 then'70-85'
when n.s_score >= 60 and n.s_score < 70 then'60-70'
else'0-60'
end) ,
count(1) 数量
from Course m , sc n
where m.c_id = n.c_id
group by all m.c_id , m.c_name , (
case when n.s_score >= 85 then'85-100'
when n.s_score >= 70 and n.s_score < 85 then'70-85'
when n.s_score >= 60 and n.s_score < 70 then'60-70'
else'0-60'
end)
order by m.c_id , m.c_name , 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.c_id 课程编号, m.c_name 课程名称,
(select count(1) from Score where c_id = m.c_id and s_score < 60) 0-60 ,
cast((select count(1) from Score where c_id = m.c_id and s_score < 60)100.0 / (select count(1) from Score where c_id = m.c_id) as decimal(18,2)) 百分比 ,
(select count(1) from Score where c_id = m.c_id and s_score >= 60 and s_score < 70) 60-70 ,
cast((select count(1) from Score where c_id = m.c_id and s_score >= 60 and s_score < 70)100.0 / (select count(1) from Score where c_id = m.c_id) as decimal(18,2)) 百分比 ,
(select count(1) from Score where c_id = m.c_id and s_score >= 70 and s_score < 85) 70-85 ,
cast((select count(1) from Score where c_id = m.c_id and s_score >= 70 and s_score < 85)100.0 / (select count(1) from Score where c_id = m.c_id) as decimal(18,2)) 百分比 ,
(select count(1) from Score where c_id = m.c_id and s_score >= 85) 85-100 ,
cast((select count(1) from Score where c_id = m.c_id and s_score >= 85)100.0 / (select count(1) from Score where c_id = m.c_id) as decimal(18,2)) 百分比
from Course m
order by m.c_id
–纵向显示1(显示存在的分数段)
select m.c_id 课程编号 , m.c_name 课程名称 , 分数段 = (
case when n.s_score >= 85 then'85-100'
when n.s_score >= 70 and n.s_score < 85 then'70-85'
when n.s_score >= 60 and n.s_score < 70 then'60-70'
else'0-60'
end) ,
count(1) 数量 ,
cast(count(1) * 100.0 / (select count(1) from sc where c_id = m.c_id) as decimal(18,2)) 百分比
from Course m , sc n
where m.c_id = n.c_id
group by m.c_id , m.c_name , (
case when n.s_score >= 85 then'85-100'
when n.s_score >= 70 and n.s_score < 85 then'70-85'
when n.s_score >= 60 and n.s_score < 70 then'60-70'
else'0-60'
end)
order by m.c_id , m.c_name , 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.c_id 课程编号 , m.c_name 课程名称 , 分数段 = (
case when n.s_score >= 85 then'85-100'
when n.s_score >= 70 and n.s_score < 85 then'70-85'
when n.s_score >= 60 and n.s_score < 70 then'60-70'
else'0-60'
end) ,
count(1) 数量 ,
cast(count(1) * 100.0 / (select count(1) from sc where c_id = m.c_id) as decimal(18,2)) 百分比
from Course m , sc n
where m.c_id = n.c_id
group by all m.c_id , m.c_name , (
case when n.s_score >= 85 then'85-100'
when n.s_score >= 70 and n.s_score < 85 then'70-85'
when n.s_score >= 60 and n.s_score < 70 then'60-70'
else'0-60'
end)
order by m.c_id , m.c_name , 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。
select t1.* , px = (select count(1) from
(
select m.s_id 学生编号 ,
m.s_name 学生姓名 ,
isnull(cast(avg(s_score) as decimal(18,2)),0) 平均成绩
from Student m left join Score n on m.s_id = n.s_id
group by m.s_id , m.s_name
) t2 where 平均成绩 > t1.平均成绩) + 1 from
(
select m.s_id 学生编号 ,
m.s_name 学生姓名 ,
isnull(cast(avg(s_score) as decimal(18,2)),0) 平均成绩
from Student m left join Score n on m.s_id = n.s_id
group by m.s_id , m.s_name
) t1
order by px

select t1.* , px = (select count(distinct 平均成绩) from
(
select m.s_id 学生编号 ,
m.s_name 学生姓名 ,
isnull(cast(avg(s_score) as decimal(18,2)),0) 平均成绩
from Student m left join Score n on m.s_id = n.s_id
group by m.s_id , m.s_name
) t2 where 平均成绩 >= t1.平均成绩) from
(
select m.s_id 学生编号 ,
m.s_name 学生姓名 ,
isnull(cast(avg(s_score) as decimal(18,2)),0) 平均成绩
from Student m left join Score n on m.s_id = n.s_id
group by m.s_id , m.s_name
) 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.s_id 学生编号 ,
m.s_name 学生姓名 ,
isnull(cast(avg(s_score) as decimal(18,2)),0) 平均成绩
from Student m left join Score n on m.s_id = n.s_id
group by m.s_id , m.s_name
) t
order by px

select t.* , px = DENSE_RANK() over(order by 平均成绩 desc) from
(
select m.s_id 学生编号 ,
m.s_name 学生姓名 ,
isnull(cast(avg(s_score) as decimal(18,2)),0) 平均成绩
from Student m left join Score n on m.s_id = n.s_id
group by m.s_id , m.s_name
) 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.c_id , n.s_score from Student m, Score n where m.s_id = n.s_id and n.s_score in
(select top 3 s_score from sc where c_id = n.c_id order by s_score desc) order by n.c_id , n.s_score desc
–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.* , px = (select count(distinct s_score) from Score where c_id = t.c_id and s_score >= t.s_score) from sc t) m where px between 1 and 3 order by m.c_id , m.px
–sql 2005用DENSE_RANK实现
select * from (select t.* , px = DENSE_RANK() over(partition by c_id order by s_score desc) from sc t) m where px between 1 and 3 order by m.c_id , m.px

–26、查询每门课程被选修的学生数
select c_id , count(s_id) 学生数 from sc group by c_id

–27、查询出只有两门课程的全部学生的学号和姓名
select Student.s_id , Student.s_name
from Student , Score
where Student.s_id = Score.s_id
group by Student.s_id , Student.s_name
having count(Score.c_id) = 2
order by Student.s_id

–28、查询男生、女生人数
select count(s_sex) as 男生人数 from Student where s_sex = N'男'
select count(s_sex) as 女生人数 from Student where s_sex = N'女'
select sum(case when s_sex = N'男' then 1 else 0 end) 男生人数 ,sum(case when s_sex = N'女' then 1 else 0 end) 女生人数 from student
select case when s_sex = 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 s_sex = 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 s_name like N'%风%'
select * from student where charindex(N'风' , s_name) > 0

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

–31、查询1990年出生的学生名单(注：Student表中s_birth列的类型是datetime)
select * from Student where year(s_birth) = 1990
select * from Student where datediff(yy,s_birth,'1990-01-01') = 0
select * from Student where datepart(yy,s_birth) = 1990
select * from Student where convert(varchar(4),s_birth,120) ='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.c_id , m.c_name , cast(avg(n.s_score) as decimal(18,2)) avg_s_score
from Course m, Score n
where m.c_id = n.c_id
group by m.c_id , m.c_name
order by avg_s_score desc, m.c_id 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.s_id , a.s_name , cast(avg(b.s_score) as decimal(18,2)) avg_s_score
from Student a , sc b
where a.s_id = b.s_id
group by a.s_id , a.s_name
having cast(avg(b.s_score) as decimal(18,2)) >= 85
order by a.s_id

–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 s_name , s_score
from Student , Score , Course
where Score.s_id = Student.s_id and Score.c_id = Course.c_id and Course.c_name = N'数学' and s_score < 60

–35、查询所有学生的课程及分数情况；
select Student.* , Course.c_name , Score.c_id , Score.s_score
from Student, Score , Course
where Student.s_id = Score.s_id and Score.c_id = Course.c_id
order by Student.s_id , Score.c_id

–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.c_name , Score.c_id , Score.s_score
from Student, Score , Course
where Student.s_id = Score.s_id and Score.c_id = Course.c_id and Score.s_score >= 70
order by Student.s_id , Score.c_id

–37、查询不及格的课程
select Student.* , Course.c_name , Score.c_id , Score.s_score
from Student, Score , Course
where Student.s_id = Score.s_id and Score.c_id = Course.c_id and Score.s_score < 60
order by Student.s_id , Score.c_id

–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.c_name , Score.c_id , Score.s_score
from Student, Score , Course
where Student.s_id = Score.s_id and Score.c_id = Course.c_id and Score.c_id ='01' and Score.s_score >= 80
order by Student.s_id , Score.c_id

–39、求每门课程的学生人数
select Course.c_id , Course.c_name , count(*) 学生人数
from Course , Score
where Course.c_id = Score.c_id
group by Course.c_id , Course.c_name
order by Course.c_id , Course.c_name

–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
select top 1 Student.* , Course.c_name , Score.c_id , Score.s_score
from Student, Score , Course , Teacher
where Student.s_id = Score.s_id and Score.c_id = Course.c_id and Course.t_id = Teacher.t_id and Teacher.t_name = N'张三'
order by Score.s_score desc
–40.2 当最高分出现多个时
select Student.* , Course.c_name , Score.c_id , Score.s_score
from Student, Score , Course , Teacher
where Student.s_id = Score.s_id and Score.c_id = Course.c_id and Course.t_id = Teacher.t_id and Teacher.t_name = N'张三' and
Score.s_score = (select max(Score.s_score) from Score , Course , Teacher where Score.c_id = Course.c_id and Course.t_id = Teacher.t_id and Teacher.t_name = N'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 Score m ,(select c_id , s_score from Score group by c_id , s_score having count(1) > 1) n
where m.c_id= n.c_id and m.s_score = n.s_score order by m.c_id , m.s_score , m.s_id
–方法2
select m.* from Score m where exists (select 1 from (select c_id , s_score from Score group by c_id , s_score having count(1) > 1) n
where m.c_id= n.c_id and m.s_score = n.s_score) order by m.c_id , m.s_score , m.s_id

–42、查询每门功成绩最好的前两名
select t.* from sc t where s_score in (select top 2 s_score from sc where c_id = T.c_id order by s_score desc) order by t.c_id , t.s_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.c_id , Course.c_name , count() 学生人数
from Course , Score
where Course.c_id = Score.c_id
group by Course.c_id , Course.c_name
having count() >= 5
order by 学生人数 desc , Course.c_id

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

–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 s_id in
(select s_id from sc group by s_id 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.s_id not in
(
select distinct m.s_id from
(
select s_id , c_id from student , course
) m where not exists (select 1 from sc n where n.s_id = m.s_id and n.c_id = m.c_id)
)
–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.s_id from
(
select s_id , c_id from student , course
) m where not exists (select 1 from sc n where n.s_id = m.s_id and n.c_id = m.c_id)
) k where k.s_id = t.s_id
)

–46、查询各学生的年龄
–46.1 只按照年份来算
select * , datediff(yy , s_birth , getdate()) 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 * , case when right(convert(varchar(10),getdate(),120),5) < right(convert(varchar(10),s_birth,120),5) then datediff(yy , s_birth , getdate()) - 1 else datediff(yy , s_birth , getdate()) end 年龄 from student

–47、查询本周过生日的学生
select * from student where datediff(week,datename(yy,getdate()) + right(convert(varchar(10),s_birth,120),6),getdate()) = 0

–48、查询下周过生日的学生
select * from student where datediff(week,datename(yy,getdate()) + right(convert(varchar(10),s_birth,120),6),getdate()) = -1

–49、查询本月过生日的学生
select * from student where datediff(mm,datename(yy,getdate()) + right(convert(varchar(10),s_birth,120),6),getdate()) = 0

–50、查询下月过生日的学生
select * from student where datediff(mm,datename(yy,getdate()) + right(convert(varchar(10),s_birth,120),6),getdate()) = -1

疑惑测试

where和having的区别
1. where和having都可以使用的场景
```
select s_idfrom studentwhere s_id > 5;select s_idfrom studenthaving s_id > 5;/*解释：上面的having可以用的前提是我已经筛选出了s_id字段，在这种情况下和where的效果是等效的，但是如果我没有select s_id 就会报错！！因为having是从前筛选的字段再筛选，而where是从数据表中的字段直接进行的筛选的。 */ 
```
1. 只可以用where，不可以用having的情况
```
select s_namefrom studentwhere s_id > 5;select s_namefrom studenthaving s_id > 5; # [42S22][1054] Unknown column 's_id' in 'having clause'/*解释：因为前面并没有筛选出s_id字段 */
```
1. 只可以用having，不可以用where情况
```
# 全班学生的语数英成绩平均分，输出65分以上的科目select avg(s.s_score) sc, c.c_namefrom score s         left join course c on s.c_id = c.c_idgroup by c.c_namehaving sc > 65;select avg(s.s_score) sc, c.c_namefrom score s         left join course c on s.c_id = c.c_idwhere sc > 65group by c.c_name;/*注意:where 后面要跟的是数据表里的字段，如果我把sc换成AVG(s.s_score)也是错误的！因为表里没有该字段。而having只是根据前面查询出来的是什么就可以后面接什么。*/
```
left函数
```
# 查询"李"姓老师的数量select count(t_name) 李姓老师的数量from Teacherwhere left(t_name, 1) = '李';/*sql的left()函数表示的是从字符表达式最左边一个字符开始返回指定数目的字符.若 b 的值大于 a 的长度,则返回字符表达式的全部字符a.如果 b 为负值或 0,则返回空字符串*/select left('12345678aaa',9); #12345678aselect left('李四 - 输出第一位数',1); # 李
```
exists和not exists和select 1的理解

原理解释：

exists（sql返回结果集为真）
```
select *from student awhere exists(select 1 from score s where a.s_id = a.s_name); 
```
not exists（sql不返回结果集为真或返回结果集为假）
```
select *from student awhere not exists(select 1 from score s where a.s_id = a.s_name);
```
select 1 from table;
```
y
```
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原文地址：https://www.cnblogs.com/youngleesin/p/14452088.html

５０道练习题

查询"01"课程比"02"课程成绩高的学生的信息及课程分数

查询"01"课程比"02"课程成绩低的学生的信息及课程分数

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

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

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

查询"李"姓老师的数量

查询学过"张三"老师授课的同学的信息

查询没学过"张三"老师授课的同学的信息

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

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

查询没有学全所有课程的同学的信息

疑惑测试

where和having的区别

left函数

exists和not exists和select 1的理解