學生各門課程成績統計SQL語句大全(面試題)

建立表

SET ANSI_NULLS ONGOSET QUOTED_IDENTIFIER ONGOSET ANSI_PADDING ONGOCREATE TABLE [dbo].[stuscore]
( [name] [varchar](50)   COLLATE Chinese_PRC_CI_AS   NULL,    
[subject] [varchar](50)   COLLATE Chinese_PRC_CI_AS   NULL,    
[score] [int]   NULL,   
[stuid] [int]    NULL) 
ON [PRIMARY] 
GO
SET ANSI_PADDING OFF

插入數據

insert into dbo.stuscore values ('張三','數學',89,1)；
insert into dbo.stuscore values ('張三','語文',80,1)；
insert into dbo.stuscore values ('張三','英語',70,1)；
insert into dbo.stuscore values ('李四','數學',90,2)；
insert into dbo.stuscore values ('李四','語文',70,2)；
insert into dbo.stuscore values ('李四','英語',80,2)；

查詢結果顯示，以下截圖：

問題：

 

1.計算每一個人的總成績並排名(要求顯示字段：姓名，總成績) 

select name,SUM(score) as allscore from dbo.stuscore
group by name 
order by allscore;

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2.計算每一個人的總成績並排名(要求顯示字段: 學號，姓名，總成績) 

select stuid,name,SUM(score) as allscore from dbo.stuscore 
group by name,stuid 
order by allscore;

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3.計算每一個人單科的最高成績(要求顯示字段: 學號，姓名，課程，最高成績)

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

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4.計算每一個人的平均成績（要求顯示字段: 學號，姓名，平均成績）

select stuid,name,AVG(score) avgscore from dbo.stuscore 
group by stuid,name;

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5.列出各門課程成績最好的學生(要求顯示字段: 學號，姓名,科目，成績) 

select t1.stuid,t1.name,t1.subject,t1.score from stuscore t1,(
select subject,MAX(score) as maxscore from stuscore group by subject)t2
where t1.subject = t2.subject and t1.score = t2.maxscore;

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6.列出各門課程成績最好的兩位學生(要求顯示字段: 學號，姓名,科目，成績) 

select  t1.* from stuscore t1 where t1.stuid in (
select top 2 stuid from stuscore where subject = t1.subject order by score desc)
order by t1.subject;

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7.統計以下： 

學號	姓名	語文	數學	英語	總分	平均分

select stuid 學號,name 姓名,sum(case when subject='語文' then score else 0 end )as 語文,
sum(case when subject='數學' then score else 0 end )as 數學,
sum(case when subject='英語' then score else 0 end )as 英語,
SUM(score)總分,avg(score)平均分 from stuscore
group by stuid,name order by 總分;

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8．列出各門課程的平均成績（要求顯示字段：課程，平均成績）

select subject,AVG(score)平均成績 from stuscore 
group by subject;

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9．列出數學成績的排名（要求顯示字段：學號，姓名，成績，排名）

select stuid,name,score,
(select count(*) from stuscore t1 where subject ='數學' and t1.score > t2.score)+1 as 名次 from stuscore t2  
where subject='數學' order by score desc;
 --註釋：排序，比較大小，比較的次數+1 = 排名。

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10．列出數學成績在2-3名的學生（要求顯示字段：學號，姓名,科目，成績） 

select t3.* from (
select top 2  t2.* from (
select top 3 stuid,name,subject,score from stuscore where 
subject = '數學' order by score desc) t2 order by t2.score) t3
order by t3.score desc;

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 select t3.*  from (
 select top 100 percent stuid,name,subject,score,
(select count(*) from stuscore t1 where subject ='數學' and t1.score > t2.score)+1 as 名次 from
 stuscore t2  where subject='數學' order by t2.score desc) t3 
 where t3.名次 between 2 and 3 order by t3.score desc;

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 select t3.*  from (
 select stuid,name,subject,score,
(select count(*) from stuscore t1 where subject ='數學' and t1.score > t2.score)+1 as 名次 from
 stuscore t2  where subject='數學') t3 
 where t3.名次 between 2 and 3 order by t3.score desc;

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後面兩個方法的不一樣之處能夠參見：http://blog.csdn.net/wrm_nancy/article/details/17170115

11．求出李四的數學成績的排名 

declare @tmp table(pm int,name varchar(50),score int,stuid int)insert into @tmp 
select null,name,score,stuid from stuscore where subject='數學' 
order by score desc declare @id int set @id=0;
update @tmp set @id=@id+1,pm=@id select * from @tmp where name='李四'

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select stuid,name,subject,score,(select count(*) from stuscore t1 where subject ='數學' and t1.score > t2.score)+1 as 名次
 from stuscore t2  where subject='數學' and name = '李四' order by score desc;

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12．統計以下： 

課程	不及格（0-59）個	良（60-80）個	優（81-100）個

 select subject 科目,sum(case when score between 0 and 59 then 1 else 0 end) as 不及格,
 sum(case when score between 60 and 80 then 1 else 0 end) as 良,
 sum(case when score between 81 and 100 then 1 else 0 end) as 優秀 from stuscore
 group by subject;

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13．統計以下：

 

數學: 張三(50分),李四(90分),王五(90分),趙六(76分) 

 

 declare @s nvarchar(1000)
 set @s=''
 select @s =@s+','+name+'('+convert(nvarchar(10),score)+'分)' from 
 stuscore where subject='數學'
 set @s=stuff(@s,1,1,' ')print '數學:'+@s

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