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50个SQL语句

时间：2022-03-13 22:51

一个项目用到的50个SQL语句

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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(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

createtable Student(Sid varchar(10),Sname nvarchar(10),Sage datetime,Ssex nvarchar(10))

insertinto Student values(‘01‘ , N‘赵雷‘ , ‘1990-01-01‘ , N‘男‘)

insertinto Student values(‘02‘ , N‘钱电‘ , ‘1990-12-21‘ , N‘男‘)

insertinto Student values(‘03‘ , N‘孙风‘ , ‘1990-05-20‘ , N‘男‘)

insertinto Student values(‘04‘ , N‘李云‘ , ‘1990-08-06‘ , N‘男‘)

insertinto Student values(‘05‘ , N‘周梅‘ , ‘1991-12-01‘ , N‘女‘)

insertinto Student values(‘06‘ , N‘吴兰‘ , ‘1992-03-01‘ , N‘女‘)

insertinto Student values(‘07‘ , N‘郑竹‘ , ‘1989-07-01‘ , N‘女‘)

insertinto Student values(‘08‘ , N‘王菊‘ , ‘1990-01-20‘ , N‘女‘)

createtable Course(Cid varchar(10),Cname nvarchar(10),Tid varchar(10))

insertinto Course values(‘01‘ , N‘语文‘ , ‘02‘)

insertinto Course values(‘02‘ , N‘数学‘ , ‘01‘)

insertinto Course values(‘03‘ , N‘英语‘ , ‘03‘)

createtable Teacher(Tid varchar(10),Tname nvarchar(10))

insertinto Teacher values(‘01‘ , N‘张三‘)

insertinto Teacher values(‘02‘ , N‘李四‘)

insertinto Teacher values(‘03‘ , N‘王五‘)

createtable SC(Sid varchar(10),Cid varchar(10),score decimal(18,1))

insertinto SC values(‘01‘ , ‘01‘ , 80)

insertinto SC values(‘01‘ , ‘02‘ , 90)

insertinto SC values(‘01‘ , ‘03‘ , 99)

insertinto SC values(‘02‘ , ‘01‘ , 70)

insertinto SC values(‘02‘ , ‘02‘ , 60)

insertinto SC values(‘02‘ , ‘03‘ , 80)

insertinto SC values(‘03‘ , ‘01‘ , 80)

insertinto SC values(‘03‘ , ‘02‘ , 80)

insertinto SC values(‘03‘ , ‘03‘ , 80)

insertinto SC values(‘04‘ , ‘01‘ , 50)

insertinto SC values(‘04‘ , ‘02‘ , 30)

insertinto SC values(‘04‘ , ‘03‘ , 20)

insertinto SC values(‘05‘ , ‘01‘ , 76)

insertinto SC values(‘05‘ , ‘02‘ , 87)

insertinto SC values(‘06‘ , ‘01‘ , 31)

insertinto SC values(‘06‘ , ‘03‘ , 34)

insertinto SC values(‘07‘ , ‘02‘ , 89)

insertinto 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

思路：课程01（一个记录集合），课程02（一个记录集合），STUDENT表（一个记录集合），包含在这三个记录集合里，并且01分数>02分数的记录。

select * fromstudent s inner join(select * from sc where cid=‘01‘) a

on s.sid=a.sid innerjoin (select * from sc where cid=‘02‘) b

on s.sid=b.sid wherea.score>b.score

select a.*,b.*,c.* fromstudent a inner joinsc b

on a.sid=b.sid andb.cid=‘01‘ inner join sc c

on a.sid=c.sid andc.cid=‘02‘ where b.score>c.score

--1.1、查询同时存在"01"课程和"02"课程的情况

思路：课程01（一个记录集合），课程02（一个记录集合），STUDENT表（一个记录集合），包含在这三个记录集合里的记录。

select * fromstudent s inner join(select * from sc where cid=‘01‘) a

on s.sid=a.sid innerjoin (select * from sc where cid=‘02‘) b

on s.sid=b.sid wherea.sid=b.sid

select s.*,a.*,b.* fromstudent s inner joinsc a

on s.sid=a.sid anda.cid=‘01‘ inner joinsc b

on s.sid=b.sid andb.cid=‘02‘

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)

思路：课程01（一个记录集合），课程02可能有，可能不存在（cid=’02’ or cid is null）,STUDENT表（一个记录集合）

select * fromstudent s inner joinsc a

on s.sid=a.sid anda.cid=‘01‘ left join sc b

on s.sid=b.sid and(b.cid=‘02‘ or b.cid is null) where a.score>isnull(b.score,0)

select a.* , b.score [课程"01"的分数],c.score [课程"02"的分数]from Student a leftjoin SC b

on a.Sid = b.Sid and b.Cid =‘01‘leftjoin SC c

on a.Sid = c.Sid and c.Cid =‘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

select * fromstudent s inner joinsc a

on s.sid=a.sid anda.cid=‘01‘ inner join sc b

on s.sid=b.sid andb.cid=‘02‘ where a.score<b.score

--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 * fromstudent s left joinsc a

on s.sid=a.sid and(a.cid=‘01‘ or a.cid is null) inner join sc b

on s.sid=b.sid andb.cid=‘02‘

select * fromstudent s inner join

(select * from sc where cid=‘02‘) a on s.sid=a.sid left join

(select * from sc where (cid=‘01‘ or cid is null)) b on s.sid=b.sid

select a.* , b.score [课程"01"的分数],c.score [课程"02"的分数]from Student a

leftjoin SC b on a.Sid = b.Sid and b.Cid =‘01‘

leftjoin SC c on a.Sid = c.Sid and c.Cid =‘02‘

whereisnull(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

思路：平均成绩大于等于60分（一个记录集合），STUDENT表（一个记录集合）

select s.sid,s.sname,b.[平均成绩] fromstudent s inner join

(select sid,convert(decimal(18,2),avg(score)) as ‘平均成绩‘ from sc group by sid having avg(score)>=60) b

on s.sid=b.sid

select * fromstudent s inner join

(select sid,avg(score) as avgscore from sc group by sid having avg(score)>=60) a

on s.sid=a.sid

select a.Sid , a.Sname , cast(avg(b.score) asdecimal(18,2)) avg_score

from Student a , sc b

where a.Sid = b.Sid

groupby a.Sid , a.Sname

havingcast(avg(b.score) asdecimal(18,2)) >=60

orderby a.Sid

--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

思路：平均成绩小于60分（一个记录集合），STUDENT（一个记录集合）

select s.sid,s.sname,b.[平均成绩] fromstudent s inner join

(select sid,convert(decimal(18,2),avg(score)) as ‘平均成绩‘ from sc group by sid having avg(score)>60) b

on s.sid=b.sid

--4.1、查询在sc表存在成绩的学生信息的SQL语句。

思路：STUDENT表（一个记录集合）是否有记录包含在SC表（一个记录集合）

select * fromstudent where sid in(select sid from sc)

select * fromstudent s where exists(select 1 from sc a where s.sid=a.sid)

select a.Sid , a.Sname , cast(avg(b.score) asdecimal(18,2)) avg_score

from Student a , sc b

where a.Sid = b.Sid

groupby a.Sid , a.Sname

havingcast(avg(b.score) asdecimal(18,2)) <60

orderby a.Sid

--4.2、查询在sc表中不存在成绩的学生信息的SQL语句。

select * from student where sid not in (select distinct sid from sc)

select * fromstudent s where notexists (select 1 from sc a where s.sid=a.sid)

select a.Sid , a.Sname , isnull(cast(avg(b.score) asdecimal(18,2)),0) avg_score

from Student a leftjoin sc b

on a.Sid = b.Sid

groupby a.Sid , a.Sname

havingisnull(cast(avg(b.score) asdecimal(18,2)),0) <60

orderby 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

思路：SC表的选课总数、总成绩（一个记录集合），STUDENT表（一个记录集合）

select s.sid,s.sname,a.[选课总数],a.[总成绩] fromstudent s inner join

(select sid,count(*) as ‘选课总数‘,sum(score) as ‘总成绩‘ from sc group by sid) a

on s.sid=a.sid

select * fromstudent s inner join

(select sid,count(cid) as ‘课程总数‘,sum(score) as ‘课程总成绩‘ from sc group by sid) a

on s.sid=a.sid

select s.sid,s.sname,count(a.cid) as ‘课程总数‘,sum(a.score) as ‘课程总成绩‘ from student s inner join sc a

on s.sid=a.sid groupby s.sid,s.sname

--5.1、查询所有有成绩的SQL。

select s.sid,s.sname,a.[选课总数],a.[总成绩] fromstudent s inner join

(select sid,count(*) as ‘选课总数‘,sum(score) as ‘总成绩‘ from sc group by sid) a

on s.sid=a.sid

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

groupby a.Sid,a.Sname

orderby a.Sid

--5.2、查询所有(包括有成绩和无成绩)的SQL。

select s.sid,s.sname,a.[选课总数],a.[总成绩] fromstudent s left join

(select sid,count(*) as ‘选课总数‘,sum(score) as ‘总成绩‘ from sc group by sid) a

on s.sid=a.sid

select * fromstudent s left join

(select sid,count(cid) as ‘课程总数‘,sum(score) as ‘课程总成绩‘ from sc group by sid) a

on s.sid=a.sid orderby s.sid

select s.sid,s.sname,count(a.cid) as ‘课程总数‘,sum(a.score) as ‘课程总成绩‘ from student s left join sc a

on s.sid=a.sid groupby s.sid,s.sname order by s.sid

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 leftjoin SC b

on a.Sid = b.Sid

groupby a.Sid,a.Sname

orderby a.Sid

--6、查询"李"姓老师的数量

select count(*) as ‘数量‘ fromteacher where left(tname,1)=‘李‘

--方法1

selectcount(Tname) ["李"姓老师的数量]from Teacher where Tname like N‘李%‘

--方法2

selectcount(Tname) ["李"姓老师的数量]from Teacher whereleft(Tname,1) = N‘李‘

--7、查询学过"张三"老师授课的同学的信息

思路： STUDENT（一个记录集合），张三老师（一个记录集合），张三老师上的课（一个记录集合），张三老师上的课的成绩（一个记录集合）

select * fromstudent s inner joinsc a

on s.sid=a.sid innerjoin course c

on a.cid=c.cid innerjoin teacher t

on c.tid=t.tid wheret.tname=‘张三‘

思路：从全部学生中（一个记录集合）提取上过张三老师课的学生（一个记录集合）

select * fromstudent where sid in(

select sid from sc a inner join course b

on a.cid=b.cid innerjoin teacher c

on b.tid=c.tid andc.tname=‘张三‘)

selectdistinct Student.*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‘

orderby Student.Sid

--8☆、查询没学过"张三"老师授课的同学的信息

思路：从全部学生中（一个记录集合）删除上过张三老师课的学生（一个记录集合）。

select * fromstudent where sid notin ( select distinct sid from sc a inner join course c

on a.cid=c.cid innerjoin teacher t

on c.tid=t.tid wheret.tname=‘张三‘)

select m.*from Student m where Sid notin (selectdistinct SC.Sid from SC , Course , Teacher where SC.Cid = Course.Cid and Course.Tid = Teacher.Tid and Teacher.Tname = N‘张三‘) orderby 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

思路：上过课程01（一个记录集合），上过课程02（一个记录集合），STUDENT表（一个记录集合）

select * fromstudent s inner joinsc a

on s.sid=a.sid anda.cid=‘01‘ inner join sc b

on s.sid=b.sid andb.cid=‘02‘

思路：上过课程01的学生（一个记录集合）并且存在上过课程02的学生（一个记录集合）

select * fromstudent s inner joinsc a

on s.sid=a.sid anda.cid=‘01‘ and exists (select 1 from sc b where s.sid=b.sid and b.cid=‘02‘)

--方法1

select Student.*from Student , SC where Student.Sid = SC.Sid and SC.Cid =‘01‘andexists (Select1from SC SC_2 where SC_2.Sid = SC.Sid and SC_2.Cid =‘02‘) orderby Student.Sid

--方法2

select Student.*from Student , SC where Student.Sid = SC.Sid and SC.Cid =‘02‘andexists (Select1from SC SC_2 where SC_2.Sid = SC.Sid and SC_2.Cid =‘01‘) orderby Student.Sid

--方法3

select m.*from Student m where Sid in

(

select Sid from

(

selectdistinctSid from SC where Cid =‘01‘

unionall

selectdistinctSid from SC where Cid =‘02‘

) t groupby Sid havingcount(1) =2

)

orderby 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

思路：上过课程01的学生（一个记录集合）并且不存在上过课程02的学生（一个记录集合）

select * fromstudent s inner joinsc a

on s.sid=a.sid anda.cid=‘01‘ and not exists (select 1 from sc b where s.sid=b.sid and b.cid=‘02‘)

思路：从全部学生中（一个记录集合）先提取上过课程01的学生记录（一个记录集合）再排除没上过课程02的学生记录（一个记录集合）

select * fromstudent where sid in

(select sid from sc where cid=‘01‘) and sid not in (

select sid from sc where cid=‘02‘)

select * fromstudent s inner joinsc a

on s.sid=a.sid anda.cid=‘01‘ where s.sid not in (select sid from sc where cid=‘02‘)

--方法1

select Student.*from Student , SC where Student.Sid = SC.Sid and SC.Cid =‘01‘andnotexists (Select1from SC SC_2 where SC_2.Sid = SC.Sid and SC_2.Cid =‘02‘) orderby Student.Sid

--方法2

select Student.*from Student , SC where Student.Sid = SC.Sid and SC.Cid =‘01‘and Student.Sid notin (Select SC_2.Sid from SC SC_2 where SC_2.Sid = SC.Sid and SC_2.Cid =‘02‘) orderby Student.Sid

--11、查询没有学全所有课程的同学的信息

思路：从全部学生中（一个记录集合）提取在SC表中课程总数不是全部的学生（一个记录集合）

select * fromstudent where sid in

(select sid from

(select sid,count(*) as abc from sc group by sid having count(*)<(select count(*) from course)) t)

该方法只列出有课程分数的学生，一个课程分数也没有的学生不存在第二个记录集合中。

思路：从全部学生中（一个记录集合）排除在SC表中有全部课程分数的学生（一个记录集合）

select * fromstudent where sid notin

(select sid from

(select sid,count(*) as abc from sc group by sid having count(*)=(select count(*) from course)) t)

该方法还会列出一个课程分数都没有的学生。

--11.1、

select Student.*

from Student , SC

where Student.Sid = SC.Sid

groupby Student.Sid , Student.Sname ,Student.Sage , Student.Ssex havingcount(Cid)< (selectcount(Cid) from Course)

--11.2

select Student.*

from Student leftjoin SC

on Student.Sid = SC.Sid

groupby Student.Sid , Student.Sname ,Student.Sage , Student.Ssex havingcount(Cid)< (selectcount(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

思路：从全部学生中（一个记录集合）提取所学课程中至少有一门和学生01所学课程相同（一个记录集合）（也就是课程ID至少有一个存在于学生01的课程ID中）并排除学生01

select * fromstudent where sid in

(select distinct sid from sc where cid in

(select cid from sc where sid=‘01‘) and sid<>‘01‘)

selectdistinct 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

思路：从全部学生中（一个记录集合）提取所学全部课程ID存在于学生01的课程ID中并且课程总数等于学生01的课程总数（一个记录集合）

select * fromstudent where sid in

(select distinct sid from sc where cid in

(select cid from sc where sid=‘01‘) and sid<>‘01‘ group by sid

having count(*)=(select count(*) from sc where sid=‘01‘))

select Student.*from Student where Sid in

(selectdistinct SC.Sid from SC where Sid <>‘01‘and SC.Cid in (selectdistinct Cid from SC where Sid =‘01‘)

groupby SC.Sid havingcount(1) = (selectcount(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

思路：从全部学生中（一个记录集合）排除学过老师张三上过的课的学生（一个记录集合）（就是在SC表中有张三老师上过的课的分数）

select * fromstudent where sid notin

(select distinct a.sid from sc a inner join course b

on a.cid=b.cid innerjoin teacher c

on b.tid=c.tid wherec.tname=‘张三‘)

select student.*from student where student.Sid notin

(selectdistinct sc.Sid from sc , course , teacher where sc.Cid = course.Cid and course.Tid = teacher.Tid and teacher.tname = N‘张三‘)

orderby 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

思路：全部学生（一个记录集合），两门及以上不及格课程（一个记录集合）

select * fromstudent s inner join

(select sid,count(*) as ‘不及格课程总数‘,convert(decimal(18,2),avg(score)) as ‘平均分数‘ from sc where score<60 group by sid having count(*)>=2) b

on s.sid=b.sid

select s.sid,s.sname,convert(decimal(5,2),avg(a.score)) as average fromstudent s inner joinsc a

on s.sid=a.sid groupby s.sid,s.sname having s.sid in

(select sid from

(select sid,count(*) as times from sc where score<60 group by sid having count(*)>=2) t)

select student.Sid ,student.sname , cast(avg(score) asdecimal(18,2)) avg_score from student , sc

where student.Sid = SC.Sid and student.Sid in (select Sid from SC where score <60groupby Sid havingcount(1) >=2)

groupby 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

思路：全部学生（一个记录集合），课程01分数小于60（一个记录集合）

select * fromstudent s inner joinsc a

on s.sid=a.sid wherecid=‘01‘ and score<60 order by score desc

select * fromstudent s inner join(select * from sc where cid=‘01‘ and score<60) a

on s.sid=a.sid orderby a.score

select student.* , sc.Cid , sc.score from student , sc

where student.Sid = SC.Sid and sc.score <60and sc.Cid =‘01‘

orderby 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

思路：全部学生（一个记录集合），全部课程分数和平均分（一个记录集合），两个记录集合进行合并行转列（新的一个记录集合）

select s.sid,s.sname,max(case b.cname when N‘语文‘ then a.score else null end) as ‘语文‘,

max(case b.cname when N‘数学‘ then a.score else null end) as ‘数学‘,

max(case b.cname when N‘英语‘ then a.score else null end) as ‘英语‘,

convert(decimal(18,2),avg(a.score)) as ‘平均成绩‘

from student s left join sc a

on s.sid=a.sid leftjoin course b

on a.cid=b.cid groupby s.sid,s.sname

order by [平均成绩]desc

--17.1 SQL 2000 静态

select a.Sid 学生编号 , a.Sname 学生姓名 ,

max(case c.Cname when N‘语文‘then b.score elsenullend) [语文],

max(case c.Cname when N‘数学‘then b.score elsenullend) [数学],

max(case c.Cname when N‘英语‘then b.score elsenullend) [英语],

cast(avg(b.score) asdecimal(18,2)) 平均分

from Student a

leftjoin SC b on a.Sid = b.Sid

leftjoin Course c on b.Cid = c.Cid

groupby a.Sid , a.Sname

orderby平均分desc

--17.2 SQL 2000 动态

declare@sqlnvarchar(4000)

set@sql=‘select a.Sid ‘+ 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 (selectdistinct 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.Sid= b.Sid left join Course c on b.Cid = c.Cid

groupby a.Sid , a.Sname order by ‘+ N‘平均分‘+‘ desc‘

exec(@sql)

--17.3 有关sql2005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)》。

--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

思路：SC表和COURSE表联合查询，每一个字段要求都可以看作是一个子查询，一个一个子查询单独做出来后，再拼接在一起。

select b.cid,b.cname,max(score) as ‘最高分‘,min(score) as ‘最低分‘,convert(decimal(5,2),avg(score)) as ‘平均分‘,

convert(varchar,convert(decimal(5,2),convert(decimal(5,2),count(case when a.score>=60 then 1 else null end))/count(1)*100))+‘%‘ as ‘及格率‘,

convert(varchar,convert(decimal(5,2),convert(decimal(5,2),count(case when a.score>=70 and a.score<80 then 1 else null end))/count(1)*100))+‘%‘ as ‘中等率‘,

convert(varchar,convert(decimal(5,2),convert(decimal(5,2),count(case when a.score>=80 and a.score<90 then 1 else null end))/count(1)*100))+‘%‘ as ‘优良率‘,

convert(varchar,convert(decimal(5,2),convert(decimal(5,2),count(case when a.score>=90 then 1 else null end))/count(1)*100))+‘%‘ as ‘优秀率‘

from sc a inner join course b on a.cid=b.cid group by b.cid,b.cname

--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)asdecimal(18,2)) [平均分],

cast((selectcount(1) from SC where Cid = m.Cid and score >=60)*100.0/ (selectcount(1) from SC where Cid = m.Cid) asdecimal(18,2)) [及格率(%)],

cast((selectcount(1) from SC where Cid = m.Cid and score >=70and score <80 )*100.0/ (selectcount(1) from SC where Cid = m.Cid) asdecimal(18,2)) [中等率(%)],

cast((selectcount(1) from SC where Cid = m.Cid and score >=80and score <90 )*100.0/ (selectcount(1) from SC where Cid = m.Cid) asdecimal(18,2)) [优良率(%)],

cast((selectcount(1) from SC where Cid = m.Cid and score >=90)*100.0/ (selectcount(1) from SC where Cid = m.Cid) asdecimal(18,2)) [优秀率(%)]

from Course m , SC n

where m.Cid = n.Cid

groupby m.Cid , m.Cname

orderby m.Cid

--方法2

select m.Cid [课程编号], m.Cname [课程名称],

(selectmax(score) from SC where Cid = m.Cid) [最高分],

(selectmin(score)from SC where Cid = m.Cid) [最低分],

(selectcast(avg(score) asdecimal(18,2)) from SC where Cid = m.Cid) [平均分],

cast((selectcount(1) from SC where Cid = m.Cid and score >=60)*100.0/ (selectcount(1) from SC where Cid = m.Cid) asdecimal(18,2)) [及格率(%)],

cast((selectcount(1) from SC where Cid = m.Cid and score >=70and score <80 )*100.0/ (selectcount(1) from SC where Cid = m.Cid) asdecimal(18,2)) [中等率(%)],

cast((selectcount(1) from SC where Cid = m.Cid and score >=80and score <90 )*100.0/ (selectcount(1) from SC where Cid = m.Cid) asdecimal(18,2)) [优良率(%)],

cast((selectcount(1) from SC where Cid = m.Cid and score >=90)*100.0/ (selectcount(1) from SC where Cid = m.Cid) asdecimal(18,2)) [优秀率(%)]

from Course m

orderby m.Cid

--19、按各科成绩进行排序，并显示排名

思路：利用over(partition by 字段名order by 字段名)函数。

正常排序：1，2，3

select row_number() over(partition by cid order by cid,score desc) as sort,* from sc

合并重复不保留空缺：1，1，2，3

select dense_rank() over(partition by cid order by cid,score desc) as sort,* from sc

合并重复保留空缺：1，1，3

select rank() over(partitionby cid order by cid,score desc) as sort,* from sc

--19.1 sql 2000用子查询完成

--Score重复时保留名次空缺

select t.* , px = (selectcount(1) from SC where Cid = t.Cid and score > t.score) +1from sc t orderby t.cid , px

--Score重复时合并名次

select t.* , px = (selectcount(distinct score) from SC where Cid = t.Cid and score >= t.score) from sc t orderby 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 orderby score desc) from sc t orderby t.Cid , px

--Score重复时合并名次(DENSE_RANK完成)

select t.* , px = DENSE_RANK() over(partition by cid orderby score desc) from sc t orderby t.Cid , px

--20、查询学生的总成绩并进行排名

思路：所有学生的总成绩（一个记录集合），再使用函数进行排序。

select rank() over(order by sum(a.score) desc) as ranking,s.sid,s.sname,sum(a.score) as ‘总成绩‘ from student s inner join sc a

on s.sid=a.sid groupby s.sid,s.sname

这个查询只能查询到有成绩的7名学生。

select dense_rank() over(order by isnull(sum(a.score),0) desc) as ranking,s.sid,s.sname,

isnull(sum(a.score),0) as ‘总成绩‘

from student s left join sc a on s.sid=a.sid group by s.sid,s.sname

用了leftjoin就可以查询到所有的8名学生了，包括没有成绩的1名学生。

--20.1 查询学生的总成绩

select m.Sid [学生编号] ,

m.Sname [学生姓名] ,

isnull(sum(score),0) [总成绩]

from Student m leftjoin SC n on m.Sid = n.Sid

groupby m.Sid , m.Sname

orderby[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 = (selectcount(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 leftjoin SC non m.Sid = n.Sid

groupby m.Sid, m.Sname

)t2 where总成绩> t1.总成绩) +1from

(

select m.Sid [学生编号] ,

m.Sname [学生姓名] ,

isnull(sum(score),0) [总成绩]

from Student m leftjoin SC non m.Sid = n.Sid

groupby m.Sid, m.Sname

)t1

orderby px

select t1.* , px = (selectcount(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 leftjoin SC non m.Sid = n.Sid

groupby 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 leftjoin SC non m.Sid = n.Sid

groupby m.Sid, m.Sname

)t1

orderby 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(orderby[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 leftjoin SC non m.Sid = n.Sid

groupby m.Sid, m.Sname

orderby px

select t.* , px = DENSE_RANK() over(orderby[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 leftjoin SC non m.Sid = n.Sid

groupby m.Sid, m.Sname

orderby px

--21、查询不同老师所教不同课程平均分从高到低显示

思路：不同老师所教不同课程的平均分（一个记录集合），再使用函数over(order by 字段名)

select rank() over(order by convert(decimal(5,2),avg(score)) desc) as ranking,c.tid,c.tname,b.cid,b.cname,

convert(decimal(5,2),avg(score)) as ‘平均分‘ from sc a

inner join course b on a.cid=b.cid inner join teacher c on b.tid=c.tid group by c.tid,c.tname,b.cid,b.cname

select m.Tid , m.Tname , cast(avg(o.score) asdecimal(18,2)) avg_score

from Teacher m , Course n , SCo

where m.Tid = n.Tid and n.Cid = o.Cid

groupby m.Tid , m.Tname

orderby 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

思路：所有课程成绩的学生及课程信息（一个记录集合），再利用函数排序（一个记录集合），选择第2名和第3名的记录。

;with abc as

(select row_number() over(partition by a.cidorder by a.score desc)as ranking,s.sid,s.sname,a.cid,b.cname,

a.score from student s inner join sc a on s.sid=a.sid inner join course b on a.cid=b.cid)

select * from abc where ranking in (2,3)

select * from

(select row_number() over(partition by a.cidorder by a.score desc)as ranking,s.sid,s.sname,a.cid,b.cname,

a.score from student s inner join sc a on s.sid=a.sid inner join course b on a.cid=b.cid) t

where t.ranking in(2,3)

--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 = (selectcount(1) from SC where Cid = t.Cid and score > t.score) +1from sc t) m where px between2and3orderby m.cid , m.px

--Score重复时合并名次

select*from (select t.* , px = (selectcount(distinct score) from SC where Cid = t.Cid and score >= t.score) from sc t) m where px between2and3orderby m.cid , 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 cid orderby score desc) from sc t) m where px between2and3orderby 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 orderby score desc) from sc t) m where px between2and3orderby 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

思路：SC表和COURSE表联合查询（一个记录集合），然后每个字段都看做是一个子查询，最后将这些子查询拼接起来。

select b.cid as ‘课程编号‘,b.cname as ‘课程名称‘,

count(1) as‘总人数‘,

count(case when a.score<60 then 1 else null end) as ‘不及格人数‘,

convert(decimal(5,2),count(case when a.score>=0 and a.score<60 then 1 else null end)*100/count(1)) as ‘不及格率%‘,

count(case when a.score>=60 and a.score<70 then 1 else null end) as ‘及格人数‘,

convert(decimal(5,2),count(case when a.score>=60 and a.score<70 then 1 else null end)*100/count(1)) as ‘及格率%‘,

count(case when a.score>=70 and a.score<85 then 1 else null end) as ‘优良人数‘,

convert(decimal(5,2),count(case when a.score>=70 and a.score<85 then 1 else null end)*100/count(1)) as ‘优良率%‘,

count(case when a.score>=85 then 1 else null end) as ‘优秀人数‘,

convert(decimal(5,2),count(case when a.score>=85 then 1 else null end)*100/count(1)) as ‘优秀率%‘

from sc a inner join course b on a.cid=b.cid group by b.cid,b.cname

以上方法为横向显示。

select b.cid as ‘课程编号‘,b.cname as ‘课程名称‘,(case when score<60 then ‘0-59‘

when score>=60 and score<70 then ‘60-69‘

when score>=70 and score<85 then ‘70-85‘

else ‘85-100‘ end) as ‘分数段‘,

count(1) as‘人数‘,

convert(decimal(18,2),count(1)*100/(select count(1) from sc where cid=b.cid)) as ‘百分比‘

from sc a inner join course b on a.cid=b.cid group by all b.cid,b.cname,(case when score<60 then ‘0-59‘

when score>=60 and score<70 then ‘60-69‘

when score>=70 and score<85 then ‘70-85‘

else ‘85-100‘ end)

order by b.cid,b.cname,‘分数段‘

以上方法为纵向显示，但为0的就不显示了。

--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(casewhen score >=85then1else0end) [85-100],

sum(casewhen score >=70and score <85then1else0end) [70-85],

sum(casewhen score >=60and score <70then1else0end) [60-70],

sum(casewhen score <60then1else0end) [0-60]

from sc , Course

where SC.Cid = Course.Cid

groupby Course.Cid , Course.Cname

orderby 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] , 分数段= (

casewhenn.score >=85then‘85-100‘

when n.score >=70and n.score <85then‘70-85‘

when n.score >=60and n.score <70then‘60-70‘

else‘0-60‘

end) ,

count(1) 数量

from Course m , sc n

where m.Cid = n.Cid

groupby m.Cid , m.Cname , (

casewhenn.score >=85then‘85-100‘

when n.score >=70and n.score <85then‘70-85‘

when n.score >=60and n.score <70then‘60-70‘

else‘0-60‘

end)

orderby 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] , 分数段= (

casewhenn.score >=85then‘85-100‘

when n.score >=70and n.score <85then‘70-85‘

when n.score >=60and n.score <70then‘60-70‘

else‘0-60‘

end) ,

count(1) 数量

from Course m , sc n

where m.Cid = n.Cid

groupbyall m.Cid , m.Cname , (

casewhenn.score >=85then‘85-100‘

when n.score >=70and n.score <85then‘70-85‘

when n.score >=60and n.score <70then‘60-70‘

else‘0-60‘

end)

orderby 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,

(selectcount(1) from SC where Cid = m.Cid and score <60) [0-60],

cast((selectcount(1) from SC where Cid = m.Cid and score <60)*100.0/ (selectcount(1) from SC where Cid = m.Cid) asdecimal(18,2)) [百分比(%)],

(selectcount(1) from SC where Cid = m.Cid and score >=60and score <70) [60-70],

cast((selectcount(1) from SC where Cid = m.Cid and score >=60and score <70)*100.0/ (selectcount(1) from SC where Cid = m.Cid) asdecimal(18,2)) [百分比(%)],

(selectcount(1) from SC where Cid = m.Cid and score >=70and score <85) [70-85],

cast((selectcount(1) from SC where Cid = m.Cid and score >=70and score <85)*100.0/ (selectcount(1) from SC where Cid = m.Cid) asdecimal(18,2)) [百分比(%)],

(selectcount(1) from SC where Cid = m.Cid and score >=85) [85-100],

cast((selectcount(1) from SC where Cid = m.Cid and score >=85)*100.0/ (selectcount(1) from SC where Cid = m.Cid) asdecimal(18,2)) [百分比(%)]

from Course m

orderby 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] , 分数段= (

casewhenn.score >=85then‘85-100‘

when n.score >=70and n.score <85then‘70-85‘

when n.score >=60and n.score <70then‘60-70‘

else‘0-60‘

end) ,

count(1) 数量 ,

cast(count(1) *100.0/ (selectcount(1) from sc where Cid = m.Cid) asdecimal(18,2)) [百分比(%)]

from Course m , sc n

where m.Cid = n.Cid

groupby m.Cid , m.Cname , (

casewhenn.score >=85then‘85-100‘

when n.score >=70and n.score <85then‘70-85‘

when n.score >=60and n.score <70then‘60-70‘

else‘0-60‘

end)

orderby 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] , 分数段= (

casewhenn.score >=85then‘85-100‘

when n.score >=70and n.score <85then‘70-85‘

when n.score >=60and n.score <70then‘60-70‘

else‘0-60‘

end) ,

count(1) 数量 ,

cast(count(1) *100.0/ (selectcount(1) from sc where Cid = m.Cid) asdecimal(18,2)) [百分比(%)]

from Course m , sc n

where m.Cid = n.Cid

groupbyall m.Cid , m.Cname , (

casewhenn.score >=85then‘85-100‘

when n.score >=70and n.score <85then‘70-85‘

when n.score >=60and n.score <70then‘60-70‘

else‘0-60‘

end)

orderby m.Cid , m.Cname , 分数段

--24、查询学生平均成绩及其名次

思路：所有学生的平均成绩（一个记录集合），再使用函数进行排序。

select s.sid,s.sname,row_number() over(order by avg(score) desc) as ranking,convert(decimal(18,2),

avg(score)) as ‘平均成绩‘ from student s inner join sc a on s.sid=a.sid group by s.sid,s.sname

只显示有成绩的学生。

select s.sid,s.sname,row_number() over(order by avg(score) desc) as ranking,convert(decimal(18,2),

avg(score)) as ‘平均成绩‘ from student s left join sc a on s.sid=a.sid group by s.sid,s.sname

显示所有学生。

--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 = (selectcount(1) from

(

select m.Sid [学生编号] ,

m.Sname [学生姓名] ,

isnull(cast(avg(score) asdecimal(18,2)),0) [平均成绩]

from Student m leftjoin SC non m.Sid = n.Sid

groupby m.Sid, m.Sname

)t2 where平均成绩> t1.平均成绩) +1from

(

select m.Sid [学生编号] ,

m.Sname [学生姓名] ,

isnull(cast(avg(score) asdecimal(18,2)),0) [平均成绩]

from Student m leftjoin SC non m.Sid = n.Sid

groupby m.Sid, m.Sname

)t1

orderby px

select t1.* , px = (selectcount(distinct平均成绩) from

(

select m.Sid [学生编号] ,

m.Sname [学生姓名] ,

isnull(cast(avg(score) asdecimal(18,2)),0) [平均成绩]

from Student m leftjoin SC non m.Sid = n.Sid

groupby m.Sid, m.Sname

)t2 where平均成绩>= t1.平均成绩) from

(

select m.Sid [学生编号] ,

m.Sname [学生姓名] ,

isnull(cast(avg(score) asdecimal(18,2)),0) [平均成绩]

from Student m leftjoin SC non m.Sid = n.Sid

groupby m.Sid, m.Sname

)t1

orderby 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(orderby[平均成绩]desc) from

(

select m.Sid [学生编号] ,

m.Sname [学生姓名] ,

isnull(cast(avg(score) asdecimal(18,2)),0) [平均成绩]

from Student m leftjoin SC non m.Sid = n.Sid

groupby m.Sid, m.Sname

orderby px

select t.* , px = DENSE_RANK() over(orderby[平均成绩]desc) from

(

select m.Sid [学生编号] ,

m.Sname [学生姓名] ,

isnull(cast(avg(score) asdecimal(18,2)),0) [平均成绩]

from Student m leftjoin SC non m.Sid = n.Sid

groupby m.Sid, m.Sname

orderby px

--25、查询各科成绩前三名的记录

思路：各学科成绩排序（一个记录集合），再取前3。

select * from

(select row_number() over(partition by a.cidorder by a.score desc)as ranking,

s.sid,s.sname,a.score from student s innerjoin sc a on s.sid=a.sid) t where ranking in (1,2,3)

--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

(selecttop3 score from sc where Cid = n.Cid orderby score desc) orderby n.Cid , 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 = (selectcount(distinct score) from SC where Cid = t.Cid and score >= t.score) from sc t) m where px between1and3orderby m.cid , m.px

--sql 2005用DENSE_RANK实现

select*from (select t.* , px = DENSE_RANK() over(partition by cid orderby score desc) from sc t) m where px between1and3orderby m.Cid , m.px

--26、查询每门课程被选修的学生数

思路：每门课被选修的学生数（一个记录集合）。

select * fromcourse a inner join

(select cid,count(*) as ‘人数‘ from sc group by cid) b

on a.cid=b.cid

select a.cid,a.cname,count(1) as ‘人数‘ from course a inner join sc b

on a.cid=b.cid groupby a.cid,a.cname

select cid , count(Sid)[学生数]from sc groupby 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

groupby Student.Sid , Student.Sname

havingcount(SC.Cid) =2

orderby Student.Sid

--28、查询男生、女生人数

思路：

select ssex,count(1)as ‘人数‘ from student group by ssex

selectcount(Ssex) as男生人数from Student where Ssex = N‘男‘

selectcount(Ssex) as女生人数from Student where Ssex = N‘女‘

selectsum(casewhen Ssex = N‘男‘then1else0end) [男生人数],sum(casewhen Ssex = N‘女‘then1else0end) [女生人数]from student

selectcasewhen 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 groupbycasewhen 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 * fromstudent where sname like‘%风%‘

select*from student where sname like N‘%风%‘

select*from student wherecharindex(N‘风‘ , sname) >0

--30、查询同名同性学生名单，并统计同名人数

思路：按照姓名字段进行GROUP BY，同时计算人数，只要大于1，就是同姓同名。

select sname,count(1) as ‘人数‘ from student group by sname having count(1)>1

select Sname [学生姓名], count(*) [人数]from Student groupby Sname havingcount(*) >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 * fromstudent where datepart(year,sage)=‘1990‘

select*from Student whereyear(sage) =1990

select*from Student wheredatediff(yy,sage,‘1990-01-01‘) =0

select*from Student wheredatepart(yy,sage) =1990

select*from Student whereconvert(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

思路：每门课程的平均成绩（一个记录集合），再使用函数排序，排序时根据平均成绩、课程编号。

select row_number() over(order by convert(decimal(18,2),avg(a.score)) desc,b.cid) as ‘排名‘,b.cid,b.cname,convert(decimal(18,2),avg(a.score)) as ‘平均成绩‘ from sc a inner join course b

on a.cid=b.cid groupby b.cid,b.cname

select m.Cid , m.Cname , cast(avg(n.score) asdecimal(18,2)) avg_score

from Course m, SC n

where m.Cid = n.Cid

groupby m.Cid , m.Cname

orderby 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 s.sid,s.sname,convert(decimal(18,2),avg(a.score)) as ‘平均成绩‘ from student s inner join sc a

on s.sid=a.sid groupby s.sid,s.sname having avg(a.score)>=85

select a.Sid , a.Sname , cast(avg(b.score) asdecimal(18,2)) avg_score

from Student a , sc b

where a.Sid = b.Sid