sql笔试
2016-06-16 12:49
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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
t1,( select stuid,sum(score) as allscore from stuscore group by
stuid)t2where t1.stuid=t2.stuidorder by t2.allscore desc
t1,(select stuid,max(score) as maxscore from stuscore group by stuid)
t2where t1.stuid=t2.stuid and t1.score=t2.maxscore
t1,(select stuid,avg(score) as avgscore from stuscore group by stuid)
t2where t1.stuid=t2.stuid
t1,(select subject,max(score) as maxscore from stuscore group by
subject) t2where t1.subject=t2.subject and t1.score=t2.maxscore
stuscore.id from stuscore where subject = t1.subject order by score
desc) order by t1.subject
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
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
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
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\’
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
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
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
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
1.计算每个人的总成绩并排名(要求显示字段:姓名,总成绩)
答案:select name,sum(score) as allscore from stuscore group by name order by allscore2.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)
答案:select distinct t1.name,t1.stuid,t2.allscore from stuscoret1,( 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(要求显示字段: 学号,姓名,课程,最高成绩)
答案:select t1.stuid,t1.name,t1.subject,t1.score from stuscoret1,(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(要求显示字段: 学号,姓名,平均成绩)
答案:select distinct t1.stuid,t1.name,t2.avgscore from stuscoret1,(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(要求显示字段: 学号,姓名,科目,成绩)
答案:select t1.stuid,t1.name,t1.subject,t2.maxscore from stuscoret1,(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(要求显示字段: 学号,姓名,科目,成绩)
答案:select distinct t1.* from stuscore t1 where t1.id in (select top 2stuscore.id from stuscore where subject = t1.subject order by score
desc) order by t1.subject
7.统计如下:学号 姓名 语文 数学 英语 总分 平均分
答案:select stuid as 学号,name as 姓名,sum(case when subject=\’语文\’ thenscore 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(要求显示字段:课程,平均成绩)
答案:select subject,avg(score) as avgscore from stuscoregroup by subject9.列出数学成绩的排名(要求显示字段:学号,姓名,成绩,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(要求显示字段:学号,姓名,科目,成绩)
答案:select t3.* from(select top 2 t2.* from (select top 3name,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
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.统计如下: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<60and 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.统计如下: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.计算科科及格的人的平均成绩
答案: select distinct t1.stuid,t2.avgscore from stuscore t1,(selectstuid,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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