学生各门课程成绩统计SQL语句大全

学生成绩表(stuscore)：

姓名：name课程：subject分数：score学号：stuid张三数学891张三语文801张三英语701李四数学902李四语文702李四英语802

创建表

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] GOSET ANSI_PADDING OFF

问题：

1.计算每个人的总成绩并排名(要求显示字段：姓名，总成绩)
2.计算每个人的总成绩并排名(要求显示字段: 学号，姓名，总成绩)
3.计算每个人单科的最高成绩(要求显示字段: 学号，姓名，课程，最高成绩)
4.计算每个人的平均成绩（要求显示字段: 学号，姓名，平均成绩）
5.列出各门课程成绩最好的学生(要求显示字段: 学号，姓名,科目，成绩)
6.列出各门课程成绩最好的两位学生(要求显示字段: 学号，姓名,科目，成绩)
7.统计如下：

学号姓名语文数学英语总分平均分

8．列出各门课程的平均成绩（要求显示字段：课程，平均成绩）
9．列出数学成绩的排名（要求显示字段：学号，姓名，成绩，排名）
10．列出数学成绩在2-3名的学生（要求显示字段：学号，姓名,科目，成绩）
11．求出李四的数学成绩的排名
12．统计如下：

课程不及格（0-59）个良（60-80）个优（81-100）个

13．统计如下：数学:张三(50分),李四(90分),王五(90分),赵六(76分)

答案：




1.计算每个人的总成绩并排名

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


2.计算每个人的总成绩并排名

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. 计算每个人单科的最高成绩

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.计算每个人的平均成绩

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.列出各门课程成绩最好的学生

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.列出各门课程成绩最好的两位学生

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.学号     姓名     语文      数学      英语      总分   平均分

select stuid as 学号,name as 姓名,sum(case when subject='语文' then score else 0 end) as 语文,sum(case when subject='数学' then score else 0 end) as 数学,sum(case when subject='英语' then score else 0 end) as 英语,sum(score) as 总分,(sum(score)/count(*)) as 平均分from stuscoregroup by stuid,name order by 总分desc


8.列出各门课程的平均成绩

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


9.列出数学成绩的排名

declare @tmp table(pm int,name varchar(50),score int,stuid int)
insert into @tmp select null,name,score,stuid from stuscore where subject='数学' 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='数学'order by score desc

ms sql(最佳选择)

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


10. 列出数学成绩在2-3名的学生

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


11. 求出李四的数学成绩的排名

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


12. 课程     不及格（-59）    良（-80）    优（-100）

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 优from stuscore t1 group by subject


13. 数学:张三(50分),李四(90分),王五(90分),赵六(76分)

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


14.计算科科及格的人的平均成绩

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