SQL 面试标题及答案ITeye - 牛牛娱乐

SQL 面试标题及答案ITeye

2019-01-10 13:44:58 | 作者: 念波 | 标签: 答案,成果,数学 | 浏览: 571

学生成果表(stuscore):
名字: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

1.核算每个人的总成果并排名(要求显现字段:名字,总成果)

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

2.核算每个人的总成果并排名(要求显现字段: 学号,名字,总成果)

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.核算每个人单科的最高成果(要求显现字段: 学号,名字,课程,最高成果)

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.核算每个人的均匀成果(要求显现字段: 学号,名字,均匀成果)

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.列出各门课程成果最好的学生(要求显现字段: 学号,名字,科目,成果)

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.列出各门课程成果最好的两位学生(要求显现字段: 学号,名字,科目,成果)

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.核算如下:学号     名字     语文     数学     英语     总分     均匀分

答案:select stuid as 学号,name as 名字,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 均匀分from stuscoregroup by stuid,name order by 总分desc

8.列出各门课程的均匀成果(要求显现字段:课程,均匀成果)

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

9.列出数学成果的排名(要求显现字段:学号,名字,成果,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=\’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(最佳挑选)
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.列出数学成果在2-3名的学生(要求显现字段:学号,名字,科目,成果)

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

11.求出李四的数学成果的排名

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=\’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.核算如下:课程     不及格(0-59)个     良(60-80)个     优(81-100)个

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

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

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

14.核算科科及格的人的均匀成果

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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