學(xué)生成績(jī)表(stuscore)：
姓名：name     課程：subject     分?jǐn)?shù)：score     學(xué)號(hào)：stuid
張三     數(shù)學(xué)     89     1
張三     語(yǔ)文     80     1
張三     英語(yǔ)     70     1
李四     數(shù)學(xué)     90     2
李四     語(yǔ)文     70     2
李四     英語(yǔ)     80     2

1.計(jì)算每個(gè)人的總成績(jī)并排名(要求顯示字段：姓名,，總成績(jī))

答案：select name,sum(score) as allscore from stuscore group by name order by allscore

2.計(jì)算每個(gè)人的總成績(jī)并排名(要求顯示字段: 學(xué)號(hào),，姓名,，總成績(jī))

答案：select distinct t1.name,t1.stuid,t2.allscore from  stuscore t1,(    select stuid,sum(score) as allscore from stuscore group by stuid)t2where t1.stuid=t2.stuidorder by t2.allscore desc

3.計(jì)算每個(gè)人單科的最高成績(jī)(要求顯示字段: 學(xué)號(hào)，姓名，課程,，最高成績(jī))

答案：select t1.stuid,t1.name,t1.subject,t1.score from stuscore t1,(select stuid,max(score) as maxscore from stuscore group by stuid) t2where t1.stuid=t2.stuid and t1.score=t2.maxscore

4.計(jì)算每個(gè)人的平均成績(jī)（要求顯示字段: 學(xué)號(hào),，姓名，平均成績(jī)）

答案：select distinct t1.stuid,t1.name,t2.avgscore from stuscore t1,(select stuid,avg(score) as avgscore from stuscore group by stuid) t2where t1.stuid=t2.stuid

5.列出各門(mén)課程成績(jī)最好的學(xué)生(要求顯示字段: 學(xué)號(hào),，姓名,科目,，成績(jī))

答案：select  t1.stuid,t1.name,t1.subject,t2.maxscore from stuscore t1,(select subject,max(score) as maxscore from stuscore group by subject) t2where t1.subject=t2.subject and t1.score=t2.maxscore

6.列出各門(mén)課程成績(jī)最好的兩位學(xué)生(要求顯示字段: 學(xué)號(hào)，姓名,科目,，成績(jī))

答案：select distinct t1.* from stuscore t1 where t1.id in (select top 2 stuscore.id from stuscore where subject = t1.subject order by score desc) order by t1.subject

7.統(tǒng)計(jì)如下：學(xué)號(hào)     姓名     語(yǔ)文     數(shù)學(xué)     英語(yǔ)     總分     平均分

答案：select stuid as 學(xué)號(hào),name as 姓名,sum(case when subject=’語(yǔ)文’ then score else 0 end) as 語(yǔ)文,sum(case when subject=’數(shù)學(xué)’ then score else 0 end) as 數(shù)學(xué),sum(case when subject=’英語(yǔ)’ then score else 0 end) as 英語(yǔ),sum(score) as 總分,(sum(score)/count(*)) as 平均分from stuscoregroup by stuid,name order by 總分desc

8．列出各門(mén)課程的平均成績(jī)（要求顯示字段：課程,，平均成績(jī)）

答案：select subject,avg(score) as avgscore from stuscoregroup by subject

9．列出數(shù)學(xué)成績(jī)的排名（要求顯示字段：學(xué)號(hào)，姓名,，成績(jī),，排名）

答案：

declare @tmp table(pm int,name varchar(50),score int,stuid int)
insert into @tmp select null,name,score,stuid from stuscore where subject=’數(shù)學(xué)’ order by score desc
declare @id int
set @id=0;
update @tmp set @id=@id+1,pm=@id
select * from @tmp

oracle:
select  DENSE_RANK () OVER(order by score desc) as row,name,subject,score,stuid from stuscore where subject=’數(shù)學(xué)’order by score desc
ms sql(最佳選擇)
select (select count(*) from stuscore t1 where subject =’數(shù)學(xué)’ and t1.score>t2.score)+1 as row ,stuid,name,score from stuscore t2 where subject =’數(shù)學(xué)’ order by score desc

10．列出數(shù)學(xué)成績(jī)?cè)?-3名的學(xué)生（要求顯示字段：學(xué)號(hào)，姓名,科目,，成績(jī)）

答案：select t3.*  from(select top 2 t2.*  from (select top 3 name,subject,score,stuid from stuscore where subject=’數(shù)學(xué)’order by score desc) t2 order by t2.score) t3 order by t3.score desc

11．求出李四的數(shù)學(xué)成績(jī)的排名

答案：

declare @tmp table(pm int,name varchar(50),score int,stuid int)insert into @tmp select null,name,score,stuid from stuscore where subject=’數(shù)學(xué)’ order by score descdeclare @id intset @id=0;update @tmp set @id=@id+1,pm=@idselect * from @tmp where name=’李四’

12．統(tǒng)計(jì)如下：課程     不及格（0-59）個(gè)     良（60-80）個(gè)     優(yōu)（81-100）個(gè)

答案：select subject, (select count(*) from stuscore where score<60 and subject=t1.subject) as 不及格,(select count(*) from stuscore where score between 60 and 80 and subject=t1.subject) as 良,(select count(*) from stuscore where score >80 and subject=t1.subject) as 優(yōu)from stuscore t1 group by subject

13．統(tǒng)計(jì)如下：數(shù)學(xué):張三(50分),李四(90分),王五(90分),趙六(76分)

答案：

declare @s varchar(1000)set @s=”select @s =@s+’,’+name+'(‘+convert(varchar(10),score)+’分)’ from stuscore where subject=’數(shù)學(xué)’ set @s=stuff(@s,1,1,”)print ‘?dāng)?shù)學(xué):’+@s

14.計(jì)算科科及格的人的平均成績(jī)

答案： select distinct t1.stuid,t2.avgscore  from stuscore t1,(select stuid,avg(score) as avgscore from stuscore   group by stuid  ) t2,(select stuid from stuscore where score<60 group by stuid) t3 where t1.stuid=t2.stuid and t1.stuid!=t3.stuid;
select  name,avg(score) as avgscore   from stuscore s  where (select sum(case when i.score>=60 then 1 else 0 end) from stuscore i where  i.name= s.name)=3   group by name