sql server 查詢練習

需要建的三個表：

　　學生表

　　create table Student

　　(
　　Sno varchar(20) not null primary key,
　　Sname varchar(20) not null,
　　Ssex varchar(20) not null,
　　Sbirthday datetime,
　　Class varchar(20)
　　)

　   課程表

　　create table Course
　　(
　　Cno varchar(20) not null primary key,
　　Cname varchar(20) not null,
　　Tno varchar(20) not null references Teacher(Tno)
　　)

　　成績表：　

　　create table Score
　　(
　　Sno varchar(20) not null references Student(Sno),
　　Cno varchar(20) not null references Course(Cno),
　　Degree Decimal(4,1)
　　)

插入數據：

　　學生表:　　

　　insert into student values (108,'曾華','男',1977-09-01,95033);
　　insert into student values (105,'匡明','男',1975-10-02,95031);
　　insert into student values (107,'王麗','女',1976-01-23,95033);
　　insert into student values (101,'李軍','男',1976-01-23,95033);
　　insert into student values (109,'王芳','女',1975-02-10,95031);
　　insert into student values (103,'陸君','女',1974-036-03,95031);

　　 課程表:　

　　insert into Course values (3-105,'計算機導論',825);
　　insert into Course values (3-245,'操作系統',804);
　　insert into Course values (6-166,'數據電路',856);
　　insert into Course values (9-888,'高等數學',831);

　　 成績表:

　　insert into Score values (103,3-245,86);
　　insert into Score values (105,3-245,75);
　　insert into Score values (109,3-245,68);
　　insert into Score values (103,3-105,92);
　　insert into Score values (105,3-105,88);
　　insert into Score values (109,3-105,76);
　　insert into Score values (101,3-105,64);
　　insert into Score values (107,3-105,91);
　　insert into Score values (108,3-105,78);
　　insert into Score values (101,6-166,85);
　　insert into Score values (107,6-166,79);
　　insert into Score values (108,6-166,81);

 

查詢題目：　　

--1) 查詢java 課程比C#分數高的學生

--2)查詢平均分成績大於 70 分的同學的姓名和平均成績

--3)查詢所有同學的學號、姓名、選課數、總成績

--5)查詢沒有學過 java 課的學生的學號、姓名
　　--學過java

　　--沒學過java

--6)查詢學過“C#”課程並且也學過“sql”課程的學生的學號、姓名

--7)查詢所有課程的平均分、及格率

---8)查詢所有課程成績小於 60 分的同學的學號、姓名、性別

--9)查詢沒有學全所有課的同學的學號、姓名、性別

--10)查詢至少有一門課與學號為“002”的同學所學相同的同學的學號和姓名

--13)查詢和“002”號的同學學習的課程完全相同的其他同學學號和姓名

--15)按平均成績從高到低顯示所有學生的“sql”、“java”、“c#”三門的課程 成績，按如下形式顯示：學生 ID,sql,java,c#,有效課程數,有效平均分

--16)查詢各科成績最高和最低的分：以如下形式顯示：課程 ID，最高分，最低分

--17)查詢不同班級所教不同課程平均分從高到低顯示

--18)查詢各科成績前三名的記錄:(不考慮成績並列情況)

/*
row_number() over( order by sc.mark desc)
*/

--19)查詢每門課程被選修的學生數

--20)查詢出只選修了一門課程的全部學生的學號和姓名

--21)查詢男生、女生人數

--22)查詢姓“張”的學生名單

--23)查詢同名同性學生名單，並統計同名人數

--24)查詢每門課程的平均成績，結果按平均成績升序排列，平均成績相同時， 按課程號降序排列

--25)查詢平均成績大於70的所有學生的學號、姓名和平均成績

--26)檢索至少選修兩門課程的學生學號

--27)查詢兩門以上不及格課程的同學的學號及其平均成績

--28)檢索“java”課程分數小於 60，按分數降序排列的同學姓名

 

查詢題目答案：

1：select * from Student stu 
left join Score sco on stu.Sno=sco.Sno and sco.Cno=-102
left join Score sco2 on stu.Sno=sco2.Sno and sco2.Cno=-242
where sco.Degree>sco2.Degree;

2：select Sname,AVG(sco.Degree) from Student stu 
left join Score sco on stu.Sno=sco.Sno
group by stu.Sname,sco.Sno  having AVG(sco.Degree)>70;

3：select stu.Sno,stu.Sname,count(sco.Sno) as '選課數' ,sum(sco.Degree) '總成績'
from Student stu inner join  Score sco on stu.Sno=sco.Sno
group by stu.Sname,stu.Sno,sco.Sno;

5.1:select * from Student where Sno  in(select Sno from score  where Sno  not in (select sno from Score where Cno=-242));

5.2:select * from Student where Sno in (select Sno from score where  Sno in (select sno from Score where Cno=-242));

6:select  * from Student stu  
left join Score sco on stu.Sno=sco.Sno and sco.Cno=-102
left join Score sco2 on stu.Sno=sco.Sno and sco2.Cno=-242
where sco.Sno=sco2.Sno;

7:select a.Cno,avg(a.Degree) as 'avg', 
CONVERT(float,SUM(case when Degree>0  then 1 else 0 end))  as '總人數',
cONVERT(float,SUM(case when Degree>60  then 1 else 0 end))  as '每門的及格人數',
CONVERT(varchar(20),CONVERT(decimal(18,2),cONVERT(float,SUM(case when Degree>60  then 1.0 else 0.0 end))/
SUM(case when Degree>0  then 1 else 0 end))*100 )+'%'
 as '及格率'
from Score a
group  by a.Cno 

8:select * from Student where Sno in(select Sno from Score where  Degree<60);

9:select sco.Sno,stu.Sname,stu.Ssex from Student stu
left join Score sco on stu.Sno=sco.Sno
group by sco.Sno,stu.Sname,stu.Ssex having COUNT(sco.Sno)!=(select COUNT(*) from Course)

10:select * from Student where sno in (select Sno from Score where Cno in (select Cno from Score where Score.Sno=108));



13:select sco.Sno from Score  sco where sco.Sno not  in (select Sno from Score where Cno not in (select Cno from Score where Sno=103)) 
 group by sco.Sno having COUNT(*)=(select COUNT(*) from Score where Sno=103)
 and sco.Sno<>103 

15:select stu.Sno,stu.Sname,
sum(case when sco.cno=-102 then sco.Degree else 0 end) '計算機導論',
sum(case when sco.cno=-160 then sco.Degree else 0 end) '數據電路',
sum(case when sco.cno=-242 then sco.Degree else 0 end) '操作系統',
sum(case when sco.cno=-879 then sco.Degree else 0 end) '高等數學',
COUNT(*) as '有效課程數' ,AVG(sco.Degree) as '有效平均分'
from Student stu 
left join Score sco on stu.Sno=sco.Sno
group by stu.Sno,stu.Sname order by AVG(sco.Degree)desc;

16:select Sno,MAX(Degree) as '最高分' ,MIN(Degree) as '最低分' from Score group by Sno;

17:select cou.Cno,stu.Class,avg(sco.Degree)from Score sco
left join Course cou
on sco.Cno=cou.Cno
left join Student stu
on stu.Sno=sco.Sno
group by cou.Cno,stu.Class 
order by AVG(sco.Degree) desc;

18:select * from (select *, ROW_NUMBER() over (partition by cno order by Degree desc  ) ev   from Score sco) t
where t.ev<4 order by t.Cno,t.Degree desc

19:select Sno,COUNT(cno) '選修的課程數' from Score group by Sno;

20:select * from Student stu
left join  Score sco  on stu.Sno=sco.Sno
where(select COUNT(*) from Score sco2 where sco.Sno=sco2.Sno)=3; 

21:select SUM(case when ssex='男' then 1 else 0 end )as '男',
SUM(case when ssex='女' then 1 else 0 end )as '女'
from Student;

22:select * from Student where sname like '張%'; 

23:select sname,COUNT(*) from Student group by Sname having COUNT(*)>1;

24:select Cno,AVG(Degree) from Score group by Cno order by AVG(Degree) desc ,Cno  ;

25:select Sno,AVG(Degree) as '平均分' from Score group by Sno having AVG(Degree)>70;

26:select sno ,count(Cno) as '選修課程數' from  Score group by Sno having COUNT(Cno)>2 or COUNT(Cno)=2 ;

27:select Sno,SUM(Case when Degree<60 then 1 else 0 end ) as '不及格人數' , 
avg(Degree) from Score group by Sno having SUM(Case when Degree<60 then 1 else 0 end )=1;

28:select * from Student stu left join Score sco on stu.Sno=sco.Sno
where Degree>60 and Cno=-242 order by Degree desc ;