重庆航天职业学院,–1.1、查询同时存在”01″课程和”02″课程的情况

发布时间:2020-04-10  栏目:数据  评论:0 Comments

sql server 查询记录平均值及并排序
的语句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# [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.S# = n.S# group by m.S# , m.Sname)
t2 where 平均成绩 t1.平均成绩) + 1 from ( select m.S# [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.S# = n.S# group by
m.S# , m.Sname) t1order by px

青云

 随笔 – 2, 文章 – 0, 评论 – 1, 引用 – 0

/*
FROM CSDN
说明:以下五十个语句都按照测试数据进行过测试,最好每次只单独运行一个语句。
问题及描述:
–1.学生表
Student(S#,Sname,Sage,Ssex) –S# 学生编号,Sname 学生姓名,Sage
出生年月,Ssex 学生性别
–2.课程表
Course(C#,Cname,T#) –C# –课程编号,Cname 课程名称,T# 教师编号
–3.教师表
Teacher(T#,Tname) –T# 教师编号,Tname 教师姓名
–4.成绩表
SC(S#,C#,score) –S# 学生编号,C# 课程编号,score 分数
*/
–创建测试数据
create table Student(S# varchar(10),Sname nvarchar(10),Sage datetime,Ssex nvarchar(10))
insert into Student values(’01’ , N’赵雷’ , ‘1990-01-01′ , N’男’)
insert into Student values(’02’ , N’钱电’ , ‘1990-12-21′ , N’男’)
insert into Student values(’03’ , N’孙风’ , ‘1990-05-20′ , N’男’)
insert into Student values(’04’ , N’李云’ , ‘1990-08-06′ , N’男’)
insert into Student values(’05’ , N’周梅’ , ‘1991-12-01′ , N’女’)
insert into Student values(’06’ , N’吴兰’ , ‘1992-03-01′ , N’女’)
insert into Student values(’07’ , N’郑竹’ , ‘1989-07-01′ , N’女’)
insert into Student values(’08’ , N’王菊’ , ‘1990-01-20′ , N’女’)
create table Course(C# varchar(10),Cname nvarchar(10),T# varchar(10))
insert into Course values(’01’ , N’语文’ , ’02’)
insert into Course values(’02’ , N’数学’ , ’01’)
insert into Course values(’03’ , N’英语’ , ’03’)
create table Teacher(T# varchar(10),Tname nvarchar(10))
insert into Teacher values(’01’ , N’张三’)
insert into Teacher values(’02’ , N’李四’)
insert into Teacher values(’03’ , N’王五’)
create table SC(S# varchar(10),C# 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)
go

 

select t1.* , px = (select count(distinct 平均成绩) from ( select m.S#
[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.S# =
n.S# group by m.S# , m.Sname) t2 where 平均成绩 = t1.平均成绩) from (
select m.S# [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.S# = n.S# group by m.S# , m.Sname) t1order by px

一个项目涉及到的50个Sql语句(整理版)

/*标题:一个项目涉及到的50个Sql语句(整理版)说明:以下五十个语句都按照测试数据进行过测试,最好每次只单独运行一个语句。*/
--1.学生表
Student(S,Sname,Sage,Ssex) --S 学生编号,Sname 学生姓名,Sage 出生年月,Ssex 学生性别--2.课程表 
Course(C,Cname,T) --C --课程编号,Cname 课程名称,T 教师编号
--3.教师表 
Teacher(T,Tname) --T 教师编号,Tname 教师姓名--4.成绩表 
SC(S,C,score) --S 学生编号,C 课程编号,score 分数*/--创建测试数据
create table Student(S varchar(10),Sname nvarchar(10),Sage datetime,Ssex nvarchar(10))insert into Student values('01' , N'赵雷' , '1990-01-01' , N'男')insert into Student values('02' , N'钱电' , '1990-12-21' , N'男')insert into Student values('03' , N'孙风' , '1990-05-20' , N'男')insert into Student values('04' , N'李云' , '1990-08-06' , N'男')insert into Student values('05' , N'周梅' , '1991-12-01' , N'女')insert into Student values('06' , N'吴兰' , '1992-03-01' , N'女')insert into Student values('07' , N'郑竹' , '1989-07-01' , N'女')insert into Student values('08' , N'王菊' , '1990-01-20' , N'女')create table Course(C varchar(10),Cname nvarchar(10),T varchar(10))insert into Course values('01' , N'语文' , '02')insert into Course values('02' , N'数学' , '01')insert into Course values('03' , N'英语' , '03')create table Teacher(T varchar(10),Tname nvarchar(10))insert into Teacher values('01' , N'张三')insert into Teacher values('02' , N'李四')insert into Teacher values('03' , N'王五')create table SC(S varchar(10),C 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)go

–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.S = b.S and a.S = c.S and b.C = ’01’ and c.C = ’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.S = b.S and b.C = ’01’
left join SC c on a.S = c.S and c.C = ’02’
where b.score > isnull(c.score,0)

–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.S = b.S and a.S = c.S and b.C = ’01’ and c.C = ’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.S = b.S and b.C = ’01’
left join SC c on a.S = c.S and c.C = ’02’
where isnull(b.score,0) < c.score

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

–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.S , a.Sname , cast(avg(b.score) as decimal(18,2)) avg_score
from Student a , sc b
where a.S = b.S
group by a.S , a.Sname
having cast(avg(b.score) as decimal(18,2)) < 60 
order by a.S
–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 , a.Sname , isnull(cast(avg(b.score) as decimal(18,2)),0)
avg_score
from Student a left join sc b
on a.S = b.S
group by a.S , a.Sname
having isnull(cast(avg(b.score) as decimal(18,2)),0) < 60 
order by a.S

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

–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
N’李%’
–方法2
select count(Tname) [“李”姓老师的数量] from Teacher where
left(Tname,1) = N’李’
/*
“李”姓老师的数量   
———– 
1
*/

–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.S = SC.S and SC.C = Course.C and Course.T = Teacher.T and
Teacher.Tname = N’张三’
order by Student.S

–8、查询没学过”张三”老师授课的同学的信息 
select m.* from Student m where S not in (select distinct SC.S from SC
, Course , Teacher where SC.C = Course.C and Course.T = Teacher.T and
Teacher.Tname = N’张三’) order by m.S

–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.S = SC.S and SC.C =
’01’ and exists (Select 1 from SC SC_2 where SC_2.S = SC.S and SC_2.C
= ’02’) order by Student.S
–方法2
select Student.* from Student , SC where Student.S = SC.S and SC.C =
’02’ and exists (Select 1 from SC SC_2 where SC_2.S = SC.S and SC_2.C
= ’01’) order by Student.S
–方法3
select m.* from Student m where S in
(
  select S from
  (
    select distinct S from SC where C = ’01’
    union all
    select distinct S from SC where C = ’02’
  ) t group by S having count(1) = 2 
)
order by m.S

–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.S = SC.S and SC.C =
’01’ and not exists (Select 1 from SC SC_2 where SC_2.S = SC.S and
SC_2.C = ’02’) order by Student.S
–方法2
select Student.* from Student , SC where Student.S = SC.S and SC.C =
’01’ and Student.S not in (Select SC_2.S from SC SC_2 where SC_2.S =
SC.S and SC_2.C = ’02’) order by Student.S

–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.S = SC.S 
group by Student.S , Student.Sname , Student.Sage , Student.Ssex having
count(C) < (select count(C) from Course) 
–11.2
select Student.*
from Student left join SC 
on Student.S = SC.S 
group by Student.S , Student.Sname , Student.Sage , Student.Ssex having
count(C) < (select count(C) 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.S = SC.S and
SC.C in (select C from SC where S = ’01’) and Student.S <> ’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 in
(select distinct SC.S from SC where S <> ’01’ and SC.C in (select
distinct C from SC where S = ’01’) 
group by SC.S having count(1) = (select count(1) from SC where S=’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 not in 
(select distinct sc.S from sc , course , teacher where sc.C = course.C
and course.T = teacher.T and teacher.tname = N’张三’)
order by student.S

–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 , student.sname , cast(avg(score) as decimal(18,2))
avg_score from student , sc 
where student.S = SC.S and student.S in (select S from SC where score
< 60 group by S having count(1) >= 2)
group by student.S , 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.C , sc.score from student , sc 
where student.S = SC.S and sc.score < 60 and sc.C = ’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.S 学生编号 , a.Sname 学生姓名 ,
       max(case c.Cname when N’语文’ then b.score else null end)
[语文],
       max(case c.Cname when N’数学’ then b.score else null end)
[数学],
       max(case c.Cname when N’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.S = b.S
left join Course c on b.C = c.C
group by a.S , a.Sname
order by 平均分 desc
–17.2 SQL 2000 动态 
declare @sql nvarchar(4000)
set @sql = ‘select a.S ‘ + N’学生编号’ + ‘ , a.Sname ‘ + N’学生姓名’
select @sql = @sql + ‘,max(case c.Cname when N”’+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)) ‘ + N’平均分’

  • ‘ from Student a left join SC b on a.S = b.S left join Course c on b.C
    = c.C
    group by a.S , a.Sname order by ‘ + N’平均分’ + ‘ desc’
    exec(@sql)
    –17.3 有关sql 2005的动静态写法参见我的文章《普通行列转换(version
    2.0)》或《普通行列转换(version 3.0)》。

 

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#1楼 得分:0回复于:2010-05-17 17:47:07
SQL code
–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 [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 C = m.C and score >= 60)*100.0
/ (select count(1) from SC where C = m.C) as decimal(18,2))
[及格率(%)],
  cast((select count(1) from SC where C = m.C and score >= 70 and
score < 80 )*100.0 / (select count(1) from SC where C = m.C) as
decimal(18,2)) [中等率(%)],
  cast((select count(1) from SC where C = m.C and score >= 80 and
score < 90 )*100.0 / (select count(1) from SC where C = m.C) as
decimal(18,2)) [优良率(%)],
  cast((select count(1) from SC where C = m.C and score >= 90)*100.0
/ (select count(1) from SC where C = m.C) as decimal(18,2))
[优秀率(%)]
from Course m , SC n
where m.C = n.C
group by m.C , m.Cname
order by m.C
–方法2
select m.C [课程编号], m.Cname [课程名称], 
  (select max(score) from SC where C = m.C) [最高分],
  (select min(score) from SC where C = m.C) [最低分],
  (select cast(avg(score) as decimal(18,2)) from SC where C = m.C)
[平均分],
  cast((select count(1) from SC where C = m.C and score >= 60)*100.0
/ (select count(1) from SC where C = m.C) as decimal(18,2))
[及格率(%)],
  cast((select count(1) from SC where C = m.C and score >= 70 and
score < 80 )*100.0 / (select count(1) from SC where C = m.C) as
decimal(18,2)) [中等率(%)],
  cast((select count(1) from SC where C = m.C and score >= 80 and
score < 90 )*100.0 / (select count(1) from SC where C = m.C) as
decimal(18,2)) [优良率(%)],
  cast((select count(1) from SC where C = m.C and score >= 90)*100.0
/ (select count(1) from SC where C = m.C) as decimal(18,2))
[优秀率(%)]
from Course m 
order by m.C

–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.* , px = (select count(1) from SC where C = t.C and score >
t.score) + 1 from sc t order by t.C , px 
–Score重复时合并名次
select t.* , px = (select count(distinct score) from SC where C = t.C
and score >= t.score) from sc t order by t.C , 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 C order by score desc) from
sc t order by t.C , px 
–Score重复时合并名次(DENSE_RANK完成)
select t.* , px = DENSE_RANK() over(partition by C order by score
desc) from sc t order by t.C , 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 [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.S = n.S 
group by m.S , 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.S [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.S = n.S 
  group by m.S , m.Sname
) t2 where 总成绩 > t1.总成绩) + 1 from 
(
  select m.S [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.S = n.S 
  group by m.S , m.Sname
) t1
order by px

select t1.* , px = (select count(distinct 总成绩) from 
(
  select m.S [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.S = n.S 
  group by m.S , m.Sname
) t2 where 总成绩 >= t1.总成绩) from 
(
  select m.S [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.S = n.S 
  group by m.S , 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.S [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.S = n.S 
  group by m.S , m.Sname
) t
order by px

select t.* , px = DENSE_RANK() over(order by [总成绩] desc) from
(
  select m.S [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.S = n.S 
  group by m.S , 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.T , m.Tname , cast(avg(o.score) as decimal(18,2)) avg_score
from Teacher m , Course n , SC o
where m.T = n.T and n.C = o.C
group by m.T , 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
select * from (select t.* , px = (select count(1) from SC where C =
t.C and score > t.score) + 1 from sc t) m where px between 2 and 3
order by m.C , m.px 
–Score重复时合并名次
select * from (select t.* , px = (select count(distinct score) from SC
where C = t.C and score >= t.score) from sc t) m where px between 2
and 3 order by m.C , m.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 C order by
score desc) from sc t) m where px between 2 and 3 order by m.C , m.px 
–Score重复时合并名次(DENSE_RANK完成)
select * from (select t.* , px = DENSE_RANK() over(partition by C
order by score desc) from sc t) m where px between 2 and 3 order by m.C
, 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 [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.C = Course.C 
group by Course.C , Course.Cname
order by Course.C
–纵向显示1(显示存在的分数段)
select m.C [课程编号] , m.Cname [课程名称] , 分数段 = (
  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) , 
  count(1) 数量 
from Course m , sc n
where m.C = n.C 
group by m.C , 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.C , 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.C [课程编号] , m.Cname [课程名称] , 分数段 = (
  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) , 
  count(1) 数量 
from Course m , sc n
where m.C = n.C 
group by all m.C , 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.C , 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.C 课程编号, m.Cname 课程名称,
  (select count(1) from SC where C = m.C and score < 60) [0-60],
  cast((select count(1) from SC where C = m.C and score < 60)*100.0
/ (select count(1) from SC where C = m.C) as decimal(18,2))
[百分比(%)],
  (select count(1) from SC where C = m.C and score >= 60 and score
< 70) [60-70],
  cast((select count(1) from SC where C = m.C and score >= 60 and
score < 70)*100.0 / (select count(1) from SC where C = m.C) as
decimal(18,2)) [百分比(%)],
  (select count(1) from SC where C = m.C and score >= 70 and score
< 85) [70-85],
  cast((select count(1) from SC where C = m.C and score >= 70 and
score < 85)*100.0 / (select count(1) from SC where C = m.C) as
decimal(18,2)) [百分比(%)],
  (select count(1) from SC where C = m.C and score >= 85)
[85-100],
  cast((select count(1) from SC where C = m.C and score >= 85)*100.0
/ (select count(1) from SC where C = m.C) as decimal(18,2))
[百分比(%)]
from Course m 
order by m.C
–纵向显示1(显示存在的分数段)
select m.C [课程编号] , m.Cname [课程名称] , 分数段 = (
  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) , 
  count(1) 数量 ,  
  cast(count(1) * 100.0 / (select count(1) from sc where C = m.C) as
decimal(18,2)) [百分比(%)]
from Course m , sc n
where m.C = n.C 
group by m.C , 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.C , 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.C [课程编号] , m.Cname [课程名称] , 分数段 = (
  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) , 
  count(1) 数量 ,  
  cast(count(1) * 100.0 / (select count(1) from sc where C = m.C) as
decimal(18,2)) [百分比(%)]
from Course m , sc n
where m.C = n.C 
group by all m.C , 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.C , m.Cname , 分数段

 

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#2楼 得分:0回复于:2010-05-17 17:47:22
SQL code
–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 [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.S = n.S 
  group by m.S , m.Sname
) t2 where 平均成绩 > t1.平均成绩) + 1 from 
(
  select m.S [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.S = n.S 
  group by m.S , m.Sname
) t1
order by px

select t1.* , px = (select count(distinct 平均成绩) from 
(
  select m.S [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.S = n.S 
  group by m.S , m.Sname
) t2 where 平均成绩 >= t1.平均成绩) from 
(
  select m.S [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.S = n.S 
  group by m.S , 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.S [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.S = n.S 
  group by m.S , m.Sname
) t
order by px

select t.* , px = DENSE_RANK() over(order by [平均成绩] desc) from
(
  select m.S [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.S = n.S 
  group by m.S , 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.C , n.score from Student m, SC n where m.S = n.S and
n.score in 
(select top 3 score from sc where C = n.C order by score desc) order by
n.C , n.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 score) from SC
where C = t.C and score >= t.score) from sc t) m where px between 1
and 3 order by m.C , m.px 
–sql 2005用DENSE_RANK实现
select * from (select t.* , px = DENSE_RANK() over(partition by C
order by score desc) from sc t) m where px between 1 and 3 order by m.C
, m.px

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

–27、查询出只有两门课程的全部学生的学号和姓名 
select Student.S , Student.Sname
from Student , SC 
where Student.S = SC.S 
group by Student.S , Student.Sname
having count(SC.C) = 2
order by Student.S
 
–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%’
select * from student where charindex(N’风’ , sname) > 0

–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 year(sage) = 1990
select * from Student where datediff(yy,sage,’1990-01-01′) = 0
select * from Student where datepart(yy,sage) = 1990
select * from Student where convert(varchar(4),sage,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 , m.Cname , cast(avg(n.score) as decimal(18,2)) avg_score
from Course m, SC n 
where m.C = n.C    
group by m.C , m.Cname 
order by avg_score desc, m.C 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 , a.Sname , cast(avg(b.score) as decimal(18,2)) avg_score
from Student a , sc b
where a.S = b.S
group by a.S , a.Sname
having cast(avg(b.score) as decimal(18,2)) >= 85 
order by a.S

–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.S = Student.S and SC.C = Course.C 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.C , SC.score  
from Student, SC , Course 
where Student.S = SC.S and SC.C = Course.C 
order by Student.S , SC.C

–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.C , SC.score  
from Student, SC , Course 
where Student.S = SC.S and SC.C = Course.C and SC.score >= 70 
order by Student.S , SC.C

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

–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.C , SC.score  
from Student, SC , Course 
where Student.S = SC.S and SC.C = Course.C and SC.C = ’01’ and SC.score
>= 80 
order by Student.S , SC.C

–39、求每门课程的学生人数 
select Course.C , Course.Cname , count(*) [学生人数]
from Course , SC 
where Course.C = SC.C
group by  Course.C , Course.Cname
order by Course.C , Course.Cname

–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.Cname , SC.C , SC.score  
from Student, SC , Course , Teacher
where Student.S = SC.S and SC.C = Course.C and Course.T = Teacher.T and
Teacher.Tname = N’张三’
order by SC.score desc
–40.2 当最高分出现多个时
select Student.* , Course.Cname , SC.C , SC.score  
from Student, SC , Course , Teacher
where Student.S = SC.S and SC.C = Course.C and Course.T = Teacher.T and
Teacher.Tname = N’张三’ and
SC.score = (select max(SC.score) from SC , Course , Teacher where SC.C =
Course.C and Course.T = Teacher.T and Teacher.Tname = 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 SC m ,(select C , score from SC group by C , score
having count(1) > 1) n 
where m.C= n.C and m.score = n.score order by m.C , m.score , m.S
–方法2
select m.* from SC m where exists (select 1 from (select C , score from
SC group by C , score having count(1) > 1) n 
where m.C= n.C and m.score = n.score) order by m.C , m.score , m.S

–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 C
= T.C order by score desc) order by t.C , 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.C , Course.Cname , count(*) [学生人数]
from Course , SC 
where Course.C = SC.C
group by  Course.C , Course.Cname
having count(*) >= 5
order by [学生人数] desc , Course.C

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

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

–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
–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),sage,120),5) then datediff(yy , sage ,
getdate()) – 1 else datediff(yy , sage , 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),sage,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),sage,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),sage,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),sage,120),6),getdate()) = -1

drop table  Student,Course,Teacher,SC

–1.学生表
Student(S,Sname,Sage,Ssex) –S 学生编号,Sname 学生姓名,Sage
出生年月,Ssex 学生性别
–2.课程表 
Course(C,Cname,T) –C –课程编号,Cname 课程名称,T 教师编号
–3.教师表 
Teacher(T,Tname) –T 教师编号,Tname 教师姓名
–4.成绩表 
SC(S,C,score) –S 学生编号,C 课程编号,score 分数
*/
–创建测试数据
create table Student(S varchar(10),Sname nvarchar(10),Sage datetime,Ssex
nvarchar(10))
insert into Student values(’01’ , N’赵雷’ , ‘1990-01-01′ , N’男’)
insert into Student values(’02’ , N’钱电’ , ‘1990-12-21′ , N’男’)
insert into Student values(’03’ , N’孙风’ , ‘1990-05-20′ , N’男’)
insert into Student values(’04’ , N’李云’ , ‘1990-08-06′ , N’男’)
insert into Student values(’05’ , N’周梅’ , ‘1991-12-01′ , N’女’)
insert into Student values(’06’ , N’吴兰’ , ‘1992-03-01′ , N’女’)
insert into Student values(’07’ , N’郑竹’ , ‘1989-07-01′ , N’女’)
insert into Student values(’08’ , N’王菊’ , ‘1990-01-20′ , N’女’)
create table Course(C varchar(10),Cname nvarchar(10),T varchar(10))
insert into Course values(’01’ , N’语文’ , ’02’)
insert into Course values(’02’ , N’数学’ , ’01’)
insert into Course values(’03’ , N’英语’ , ’03’)
create table Teacher(T varchar(10),Tname nvarchar(10))
insert into Teacher values(’01’ , N’张三’)
insert into Teacher values(’02’ , N’李四’)
insert into Teacher values(’03’ , N’王五’)
create table SC(S varchar(10),C 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)
go

–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.S = b.S and a.S = c.S and b.C = ’01’ and c.C = ’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.S = b.S and b.C = ’01’
left join SC c on a.S = c.S and c.C = ’02’
where b.score > isnull(c.score,0)

–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.S = b.S and a.S = c.S and b.C = ’01’ and c.C = ’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.S = b.S and b.C = ’01’
left join SC c on a.S = c.S and c.C = ’02’
where isnull(b.score,0) < c.score

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

–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.S , a.Sname , cast(avg(b.score) as decimal(18,2)) avg_score
from Student a , sc b
where a.S = b.S
group by a.S , a.Sname
having cast(avg(b.score) as decimal(18,2)) < 60 
order by a.S
–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 , a.Sname , isnull(cast(avg(b.score) as decimal(18,2)),0)
avg_score
from Student a left join sc b
on a.S = b.S
group by a.S , a.Sname
having isnull(cast(avg(b.score) as decimal(18,2)),0) < 60 
order by a.S

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

–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
N’李%’
–方法2
select count(Tname) [“李”姓老师的数量] from Teacher where
left(Tname,1) = N’李’
/*
“李”姓老师的数量   
———– 
1
*/

–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.S = SC.S and SC.C = Course.C and Course.T = Teacher.T and
Teacher.Tname = N’张三’
order by Student.S

–8、查询没学过”张三”老师授课的同学的信息 
select m.* from Student m where S not in (select distinct SC.S from SC
, Course , Teacher where SC.C = Course.C and Course.T = Teacher.T and
Teacher.Tname = N’张三’) order by m.S

–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.S = SC.S and SC.C =
’01’ and exists (Select 1 from SC SC_2 where SC_2.S = SC.S and SC_2.C
= ’02’) order by Student.S
–方法2
select Student.* from Student , SC where Student.S = SC.S and SC.C =
’02’ and exists (Select 1 from SC SC_2 where SC_2.S = SC.S and SC_2.C
= ’01’) order by Student.S
–方法3
select m.* from Student m where S in
(
  select S from
  (
    select distinct S from SC where C = ’01’
    union all
    select distinct S from SC where C = ’02’
  ) t group by S having count(1) = 2 
)
order by m.S

–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.S = SC.S and SC.C =
’01’ and not exists (Select 1 from SC SC_2 where SC_2.S = SC.S and
SC_2.C = ’02’) order by Student.S
–方法2
select Student.* from Student , SC where Student.S = SC.S and SC.C =
’01’ and Student.S not in (Select SC_2.S from SC SC_2 where SC_2.S =
SC.S and SC_2.C = ’02’) order by Student.S

–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.S = SC.S 
group by Student.S , Student.Sname , Student.Sage , Student.Ssex having
count(C) < (select count(C) from Course) 
–11.2
select Student.*
from Student left join SC 
on Student.S = SC.S 
group by Student.S , Student.Sname , Student.Sage , Student.Ssex having
count(C) < (select count(C) 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.S = SC.S and
SC.C in (select C from SC where S = ’01’) and Student.S <> ’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 in
(select distinct SC.S from SC where S <> ’01’ and SC.C in (select
distinct C from SC where S = ’01’) 
group by SC.S having count(1) = (select count(1) from SC where S=’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 not in 
(select distinct sc.S from sc , course , teacher where sc.C = course.C
and course.T = teacher.T and teacher.tname = N’张三’)
order by student.S

–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 , student.sname , cast(avg(score) as decimal(18,2))
avg_score from student , sc 
where student.S = SC.S and student.S in (select S from SC where score
< 60 group by S having count(1) >= 2)
group by student.S , 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.C , sc.score from student , sc 
where student.S = SC.S and sc.score < 60 and sc.C = ’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.S 学生编号 , a.Sname 学生姓名 ,
       max(case c.Cname when N’语文’ then b.score else null end)
[语文],
       max(case c.Cname when N’数学’ then b.score else null end)
[数学],
       max(case c.Cname when N’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.S = b.S
left join Course c on b.C = c.C
group by a.S , a.Sname
order by 平均分 desc
–17.2 SQL 2000 动态 
declare @sql nvarchar(4000)
set @sql = ‘select a.S ‘ + N’学生编号’ + ‘ , a.Sname ‘ + N’学生姓名’
select @sql = @sql + ‘,max(case c.Cname when N”’+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)) ‘ + N’平均分’

  • ‘ from Student a left join SC b on a.S = b.S left join Course c on b.C
    = c.C
    group by a.S , a.Sname order by ‘ + N’平均分’ + ‘ desc’
    exec(@sql)
    –17.3 有关sql 2005的动静态写法参见我的文章《普通行列转换(version
    2.0)》或《普通行列转换(version 3.0)》。

 

对我有用[9]丢个板砖[0]引用举报管理TOP 回复次数:1043

dawugui
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#1楼 得分:0回复于:2010-05-17 17:47:07
SQL code
–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 [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 C = m.C and score >= 60)*100.0
/ (select count(1) from SC where C = m.C) as decimal(18,2))
[及格率(%)],
  cast((select count(1) from SC where C = m.C and score >= 70 and
score < 80 )*100.0 / (select count(1) from SC where C = m.C) as
decimal(18,2)) [中等率(%)],
  cast((select count(1) from SC where C = m.C and score >= 80 and
score < 90 )*100.0 / (select count(1) from SC where C = m.C) as
decimal(18,2)) [优良率(%)],
  cast((select count(1) from SC where C = m.C and score >= 90)*100.0
/ (select count(1) from SC where C = m.C) as decimal(18,2))
[优秀率(%)]
from Course m , SC n
where m.C = n.C
group by m.C , m.Cname
order by m.C
–方法2
select m.C [课程编号], m.Cname [课程名称], 
  (select max(score) from SC where C = m.C) [最高分],
  (select min(score) from SC where C = m.C) [最低分],
  (select cast(avg(score) as decimal(18,2)) from SC where C = m.C)
[平均分],
  cast((select count(1) from SC where C = m.C and score >= 60)*100.0
/ (select count(1) from SC where C = m.C) as decimal(18,2))
[及格率(%)],
  cast((select count(1) from SC where C = m.C and score >= 70 and
score < 80 )*100.0 / (select count(1) from SC where C = m.C) as
decimal(18,2)) [中等率(%)],
  cast((select count(1) from SC where C = m.C and score >= 80 and
score < 90 )*100.0 / (select count(1) from SC where C = m.C) as
decimal(18,2)) [优良率(%)],
  cast((select count(1) from SC where C = m.C and score >= 90)*100.0
/ (select count(1) from SC where C = m.C) as decimal(18,2))
[优秀率(%)]
from Course m 
order by m.C

–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.* , px = (select count(1) from SC where C = t.C and score >
t.score) + 1 from sc t order by t.C , px 
–Score重复时合并名次
select t.* , px = (select count(distinct score) from SC where C = t.C
and score >= t.score) from sc t order by t.C , 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 C order by score desc) from
sc t order by t.C , px 
–Score重复时合并名次(DENSE_RANK完成)
select t.* , px = DENSE_RANK() over(partition by C order by score
desc) from sc t order by t.C , 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 [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.S = n.S 
group by m.S , 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.S [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.S = n.S 
  group by m.S , m.Sname
) t2 where 总成绩 > t1.总成绩) + 1 from 
(
  select m.S [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.S = n.S 
  group by m.S , m.Sname
) t1
order by px

select t1.* , px = (select count(distinct 总成绩) from 
(
  select m.S [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.S = n.S 
  group by m.S , m.Sname
) t2 where 总成绩 >= t1.总成绩) from 
(
  select m.S [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.S = n.S 
  group by m.S , 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.S [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.S = n.S 
  group by m.S , m.Sname
) t
order by px

select t.* , px = DENSE_RANK() over(order by [总成绩] desc) from
(
  select m.S [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.S = n.S 
  group by m.S , 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.T , m.Tname , cast(avg(o.score) as decimal(18,2)) avg_score
from Teacher m , Course n , SC o
where m.T = n.T and n.C = o.C
group by m.T , 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
select * from (select t.* , px = (select count(1) from SC where C =
t.C and score > t.score) + 1 from sc t) m where px between 2 and 3
order by m.C , m.px 
–Score重复时合并名次
select * from (select t.* , px = (select count(distinct score) from SC
where C = t.C and score >= t.score) from sc t) m where px between 2
and 3 order by m.C , m.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 C order by
score desc) from sc t) m where px between 2 and 3 order by m.C , m.px 
–Score重复时合并名次(DENSE_RANK完成)
select * from (select t.* , px = DENSE_RANK() over(partition by C
order by score desc) from sc t) m where px between 2 and 3 order by m.C
, 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 [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.C = Course.C 
group by Course.C , Course.Cname
order by Course.C
–纵向显示1(显示存在的分数段)
select m.C [课程编号] , m.Cname [课程名称] , 分数段 = (
  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) , 
  count(1) 数量 
from Course m , sc n
where m.C = n.C 
group by m.C , 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.C , 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.C [课程编号] , m.Cname [课程名称] , 分数段 = (
  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) , 
  count(1) 数量 
from Course m , sc n
where m.C = n.C 
group by all m.C , 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.C , 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.C 课程编号, m.Cname 课程名称,
  (select count(1) from SC where C = m.C and score < 60) [0-60],
  cast((select count(1) from SC where C = m.C and score < 60)*100.0
/ (select count(1) from SC where C = m.C) as decimal(18,2))
[百分比(%)],
  (select count(1) from SC where C = m.C and score >= 60 and score
< 70) [60-70],
  cast((select count(1) from SC where C = m.C and score >= 60 and
score < 70)*100.0 / (select count(1) from SC where C = m.C) as
decimal(18,2)) [百分比(%)],
  (select count(1) from SC where C = m.C and score >= 70 and score
< 85) [70-85],
  cast((select count(1) from SC where C = m.C and score >= 70 and
score < 85)*100.0 / (select count(1) from SC where C = m.C) as
decimal(18,2)) [百分比(%)],
  (select count(1) from SC where C = m.C and score >= 85)
[85-100],
  cast((select count(1) from SC where C = m.C and score >= 85)*100.0
/ (select count(1) from SC where C = m.C) as decimal(18,2))
[百分比(%)]
from Course m 
order by m.C
–纵向显示1(显示存在的分数段)
select m.C [课程编号] , m.Cname [课程名称] , 分数段 = (
  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) , 
  count(1) 数量 ,  
  cast(count(1) * 100.0 / (select count(1) from sc where C = m.C) as
decimal(18,2)) [百分比(%)]
from Course m , sc n
where m.C = n.C 
group by m.C , 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.C , 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.C [课程编号] , m.Cname [课程名称] , 分数段 = (
  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) , 
  count(1) 数量 ,  
  cast(count(1) * 100.0 / (select count(1) from sc where C = m.C) as
decimal(18,2)) [百分比(%)]
from Course m , sc n
where m.C = n.C 
group by all m.C , 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.C , m.Cname , 分数段

 

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#2楼 得分:0回复于:2010-05-17 17:47:22
SQL code
–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 [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.S = n.S 
  group by m.S , m.Sname
) t2 where 平均成绩 > t1.平均成绩) + 1 from 
(
  select m.S [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.S = n.S 
  group by m.S , m.Sname
) t1
order by px

select t1.* , px = (select count(distinct 平均成绩) from 
(
  select m.S [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.S = n.S 
  group by m.S , m.Sname
) t2 where 平均成绩 >= t1.平均成绩) from 
(
  select m.S [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.S = n.S 
  group by m.S , 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.S [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.S = n.S 
  group by m.S , m.Sname
) t
order by px

select t.* , px = DENSE_RANK() over(order by [平均成绩] desc) from
(
  select m.S [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.S = n.S 
  group by m.S , 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.C , n.score from Student m, SC n where m.S = n.S and
n.score in 
(select top 3 score from sc where C = n.C order by score desc) order by
n.C , n.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 score) from SC
where C = t.C and score >= t.score) from sc t) m where px between 1
and 3 order by m.C , m.px 
–sql 2005用DENSE_RANK实现
select * from (select t.* , px = DENSE_RANK() over(partition by C
order by score desc) from sc t) m where px between 1 and 3 order by m.C
, m.px

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

–27、查询出只有两门课程的全部学生的学号和姓名 
select Student.S , Student.Sname
from Student , SC 
where Student.S = SC.S 
group by Student.S , Student.Sname
having count(SC.C) = 2
order by Student.S
 
–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%’
select * from student where charindex(N’风’ , sname) > 0

–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 year(sage) = 1990
select * from Student where datediff(yy,sage,’1990-01-01′) = 0
select * from Student where datepart(yy,sage) = 1990
select * from Student where convert(varchar(4),sage,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 , m.Cname , cast(avg(n.score) as decimal(18,2)) avg_score
from Course m, SC n 
where m.C = n.C    
group by m.C , m.Cname 
order by avg_score desc, m.C 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 , a.Sname , cast(avg(b.score) as decimal(18,2)) avg_score
from Student a , sc b
where a.S = b.S
group by a.S , a.Sname
having cast(avg(b.score) as decimal(18,2)) >= 85 
order by a.S

–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.S = Student.S and SC.C = Course.C 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.C , SC.score  
from Student, SC , Course 
where Student.S = SC.S and SC.C = Course.C 
order by Student.S , SC.C

–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.C , SC.score  
from Student, SC , Course 
where Student.S = SC.S and SC.C = Course.C and SC.score >= 70 
order by Student.S , SC.C

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

–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.C , SC.score  
from Student, SC , Course 
where Student.S = SC.S and SC.C = Course.C and SC.C = ’01’ and SC.score
>= 80 
order by Student.S , SC.C

–39、求每门课程的学生人数 
select Course.C , Course.Cname , count(*) [学生人数]
from Course , SC 
where Course.C = SC.C
group by  Course.C , Course.Cname
order by Course.C , Course.Cname

–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.Cname , SC.C , SC.score  
from Student, SC , Course , Teacher
where Student.S = SC.S and SC.C = Course.C and Course.T = Teacher.T and
Teacher.Tname = N’张三’
order by SC.score desc
–40.2 当最高分出现多个时
select Student.* , Course.Cname , SC.C , SC.score  
from Student, SC , Course , Teacher
where Student.S = SC.S and SC.C = Course.C and Course.T = Teacher.T and
Teacher.Tname = N’张三’ and
SC.score = (select max(SC.score) from SC , Course , Teacher where SC.C =
Course.C and Course.T = Teacher.T and Teacher.Tname = 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 SC m ,(select C , score from SC group by C , score
having count(1) > 1) n 
where m.C= n.C and m.score = n.score order by m.C , m.score , m.S
–方法2
select m.* from SC m where exists (select 1 from (select C , score from
SC group by C , score having count(1) > 1) n 
where m.C= n.C and m.score = n.score) order by m.C , m.score , m.S

–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 C
= T.C order by score desc) order by t.C , 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.C , Course.Cname , count(*) [学生人数]
from Course , SC 
where Course.C = SC.C
group by  Course.C , Course.Cname
having count(*) >= 5
order by [学生人数] desc , Course.C

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

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

–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
–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),sage,120),5) then datediff(yy , sage ,
getdate()) – 1 else datediff(yy , sage , 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),sage,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),sage,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),sage,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),sage,120),6),getdate()) = -1

drop table  Student,Course,Teacher,SC

 

–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.S# = b.S# and a.S# = c.S# and b.C# = ’01’ and c.C# = ’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.S# = b.S# and b.C# = ’01’
left join SC c on a.S# = c.S# and c.C# = ’02’
where b.score > isnull(c.score,0)

/*
标题:一个项目涉及到的50个Sql语句(整理版)
作者:爱新觉罗.毓华(十八年风雨,守得冰山雪莲花开)
时间:2010-05-10
地点:重庆航天职业学院
说明:以下五十个语句都按照测试数据进行过测试,最好每次只单独运行一个语句。
问题及描述:
–1.学生表
Student(S#,Sname,Sage,Ssex) –S# 学生编号,Sname 学生姓名,Sage
出生年月,Ssex 学生性别
–2.课程表
Course(C#,Cname,T#) –C# –课程编号,Cname 课程名称,T# 教师编号
–3.教师表
Teacher(T#,Tname) –T# 教师编号,Tname 教师姓名
–4.成绩表
SC(S#,C#,score) –S# 学生编号,C# 课程编号,score 分数
*/
–创建测试数据
create table Student(S# varchar(10),Sname nvarchar(10),Sage
datetime,Ssex nvarchar(10))
insert into Student values(’01’ , N’赵雷’ , ‘1990-01-01′ , N’男’)
insert into Student values(’02’ , N’钱电’ , ‘1990-12-21′ , N’男’)
insert into Student values(’03’ , N’孙风’ , ‘1990-05-20′ , N’男’)
insert into Student values(’04’ , N’李云’ , ‘1990-08-06′ , N’男’)
insert into Student values(’05’ , N’周梅’ , ‘1991-12-01′ , N’女’)
insert into Student values(’06’ , N’吴兰’ , ‘1992-03-01′ , N’女’)
insert into Student values(’07’ , N’郑竹’ , ‘1989-07-01′ , N’女’)
insert into Student values(’08’ , N’王菊’ , ‘1990-01-20′ , N’女’)
create table Course(C# varchar(10),Cname nvarchar(10),T#
varchar(10))
insert into Course values(’01’ , N’语文’ , ’02’)
insert into Course values(’02’ , N’数学’ , ’01’)
insert into Course values(’03’ , N’英语’ , ’03’)
create table Teacher(T# varchar(10),Tname nvarchar(10))
insert into Teacher values(’01’ , N’张三’)
insert into Teacher values(’02’ , N’李四’)
insert into Teacher values(’03’ , N’王五’)
create table SC(S# varchar(10),C# 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)
go

/*数据表结构

–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.S# = b.S# and a.S# = c.S# and b.C# = ’01’ and c.C# = ’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.S# = b.S# and b.C# = ’01’
left join SC c on a.S# = c.S# and c.C# = ’02’
where isnull(b.score,0) < c.score

 

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