1.
学生表(学生id，姓名，性别，分数)student（s_id，name,sex,score）
班级表（班级id，班级名称）class（c_id，c_name）
学生班级表（班级id，学生id）student_class(s_id,c_id)
1.查询一班得分在80分以上的学生
2.查询所有班级的名称，和所有版中女生人数和女生的平均分
题解:
2.一道SQL语句面试题，关于group by表内容：
info 表
date result
2005-05-09 win
2005-05-09 lose
2005-05-09 lose
2005-05-09 lose
2005-05-10 win
2005-05-10 lose
2005-05-10 lose
如果要生成下列结果, 该如何写sql语句?
win lose
2005-05-09 2 2
2005-05-10 1 2
答案：
( 1 ) select date , sum ( case when result = “win” then 1 else 0 end ) as “win”,
sum ( case when result = “lose” then 1 else 0 end ) as “lose” from info group by date ;
( 2 ) select a.date, a.result as win, b.result as lose from ( select date , count ( result ) as result from info where result = “win” group by date ) as a
join ( select date , count ( result ) as result from info where result = “lose” group by date ) as b
on a.date = b.date; 2.学生成绩表(stuscore)：
3.表中有A B C三列,用SQL语句实现：当A列大于B列时选择A列否则选择B列，当B列大于C列时选择B列否则选择C列
select ( case when a > b then a else b end ), ( case when b > c then b else c end ) from table ;
4.
有一张表，里面有3个字段：语文，数学，英语。其中有3条记录分别表示语文70分，数学80分，英语58分，请用一条sql语句查询出这三条记录并按以下条件显示出来（并写出您的思路）：?
大于或等于80表示优秀，大于或等于60表示及格，小于60分表示不及格。?
显示格式：?
及格 优秀 不及格?
——————————————
select
(case when 语文>=80 then ‘优秀’
when 语文>=60 then ‘及格’
else ‘不及格’) as 语文,
(case when 数学>=80 then ‘优秀’
when 数学>=60 then ‘及格’
else ‘不及格’) as 数学,
(case when 英语>=80 then ‘优秀’
when 英语>=60 then ‘及格’
else ‘不及格’) as 英语,
from table
5.姓名：name 课程：subject 分数：score 学 ：stuid
张三 数学 89 1
张三 英语 70 1
李四 数学 90 2
李四 英语 80 2
题解：
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 allscore
2.计算每个人的总成绩并排名(要求显示字段: 学 ，姓名，总成绩)
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.计算每个人单科的最高成绩(要求显示字段: 学 ，姓名，课程，最高成绩)
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.计算每个人的平均成绩（要求显示字段: 学 ，姓名，平均成绩）
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.列出各门课程成绩最好的学生(要求显示字段: 学 ，姓名,科目，成绩)
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.列出各门课程成绩最好的两位学生(要求显示字段: 学 ，姓名,科目，成绩)
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 ,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（要求显示字段：课程，平均成绩）
答案：select subject,avg(score) as avgscore from stuscoregroup by subject
9．列出数学成绩的排名（要求显示字段：学 ，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 ，姓名,科目，成绩）
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 score80 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,(select stuid,avg(score) as avgscore from stuscore group by stuid ) t2,(select stuid from stuscore where score
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
1. 用一条SQL 语句 查询出每门课都大于80 分的学生姓名
name kecheng fenshu
张三 数学 75
李四 数学 90
王五 数学 100
王五 英语 90
A: select distinct name from table where name not in (select distinct name from table where fenshu
select name from table group by name having min(fenshu)>80
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