标签:
学生成绩表(stuscore):
姓名:name 课程:subject 分数:score 学号:stuid
张三 数学 89 1
张三 语文 80 1
张三 英语 70 1
李四 数学 90 2
李四 语文 70 2
李四 英语 80 2
答案:select name,sum(score) as allscore from stuscore group by name order by allscore
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
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
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
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
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
答案: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
答案:select subject,avg(score) as avgscore from stuscoregroup by subject
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=\’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
答案: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
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=\’数学\’ order by score descdeclare @id intset @id=0;update @tmp set @id=@id+1,pm=@idselect * from @tmp where name=\’李四\’
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 优from stuscore t1 group by subject
答案:
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
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
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原文地址:http://www.cnblogs.com/taz372436/p/5590637.html