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sql基礎題目測試及正確答案

學習 creat 統計 完全 當前 前三 ear style 合並

在網上做了一套基本的sql題目,以下是我的寫的答案,適合基礎人員練練

--創建測試數據

use test

create table Student(S# varchar(10),Sname nvarchar(10),Sage datetime,Ssex nvarchar(10))

insert into Student values(‘01‘ , N‘趙雷‘ , ‘1990-01-01‘ , N‘‘)

insert into Student values(‘02‘ , N‘錢電‘ , ‘1990-12-21‘ , N‘‘)

insert into Student values(‘03‘ , N‘孫風‘ , ‘1990-05-20‘ , N‘

‘)

insert into Student values(‘04‘ , N‘李雲‘ , ‘1990-08-06‘ , N‘‘)

insert into Student values(‘05‘ , N‘周梅‘ , ‘1991-12-01‘ , N‘‘)

insert into Student values(‘06‘ , N‘吳蘭‘ , ‘1992-03-01‘ , N‘‘)

insert into Student values(‘07‘ , N‘鄭竹‘ , ‘1989-07-01‘ , N‘‘)

insert into Student values(‘08‘ , N‘王菊‘ , ‘1990-01-20‘ , N‘

‘)

create table Course(C# varchar(10),Cname nvarchar(10),T# varchar(10))

insert into Course values(‘01‘ , N‘語文‘ , ‘02‘)

insert into Course values(‘02‘ , N‘數學‘ , ‘01‘)

insert into Course values(‘03‘ , N‘英語‘ , ‘03‘)

create table Teacher(T# varchar(10),Tname nvarchar(10))

insert into Teacher values(‘01‘ , N‘張三‘)

insert into Teacher values(‘02‘ , N‘李四

‘)

insert into Teacher values(‘03‘ , N‘王五‘)

create table SC(S# varchar(10),C# varchar(10),score decimal(18,1))

insert into SC values(‘01‘ , ‘01‘ , 80)

insert into SC values(‘01‘ , ‘02‘ , 90)

insert into SC values(‘01‘ , ‘03‘ , 99)

insert into SC values(‘02‘ , ‘01‘ , 70)

insert into SC values(‘02‘ , ‘02‘ , 60)

insert into SC values(‘02‘ , ‘03‘ , 80)

insert into SC values(‘03‘ , ‘01‘ , 80)

insert into SC values(‘03‘ , ‘02‘ , 80)

insert into SC values(‘03‘ , ‘03‘ , 80)

insert into SC values(‘04‘ , ‘01‘ , 50)

insert into SC values(‘04‘ , ‘02‘ , 30)

insert into SC values(‘04‘ , ‘03‘ , 20)

insert into SC values(‘05‘ , ‘01‘ , 76)

insert into SC values(‘05‘ , ‘02‘ , 87)

insert into SC values(‘06‘ , ‘01‘ , 31)

insert into SC values(‘06‘ , ‘03‘ , 34)

insert into SC values(‘07‘ , ‘02‘ , 89)

insert into SC values(‘07‘ , ‘03‘ , 98)

go

註意:有些題目為了方便測試,可以自行修改表中的數據

--1、查詢"01"課程比"02"課程成績高的學生的信息及課程分數

select s.* ,a.score as ‘01‘ ,b.score as ‘02‘ from Student s

left join sc a on a.S#=s.S# and a.C#=‘01‘

left join sc b on b.S#=s.S# and b.C#=‘02‘

where a.score>b.score

select a.* ,b.C# , b.score from Student a,SC b where a.S#=b.S#

and a.S# in ( select S# from SC c1 where C#=‘01‘ and score>

( select score from SC c2 where C#=‘02‘ and c2.S#=c1.S# )

);

--1.1、查詢同時存在"01"課程和"02"課程的情況

select a.S#,a.Sname,b.score,c.score from Student a

left join sc b on b.s#=a.s# and b.c#=‘01‘

left join sc c on c.s#=a.s# and c.c#=‘02‘

where b.score>0 and c.score>0;

--1.2、查詢同時存在"01"課程和"02"課程的情況和存在"01"課程但可能不存在"02"課程的情況(不存在時顯示為null)(以下存在相同內容時不再解釋)

select a.S#,a.Sname,b.score as ‘01‘,c.score as ‘02‘ from Student a

left join sc b on b.s#=a.s# and b.c#=‘01‘

left join sc c on c.s#=a.s# and c.c#=‘02‘

Where b.score>isnull(c.score,0);

--1.2.1存在01,但是可能不存在02

--這個表示左連接後c#編號為null的也包含進去

select * from student s left join sc a on s.S#=a.S# and a.C#=‘02‘

--這個表示左連接後c#編號必須為‘02‘,不包含null

select * from student s left join sc a on s.S#=a.S# where a.C#=‘02‘

--這兩個結起來可以是01必須有,02可以不要

select s.*,a.score as ‘01‘,b.score as ‘02‘ from student s

left join sc a on a.S#=s.S#

left join sc b on b.s#=s.s# and b.C#=‘02‘

where a.C#=‘01‘

--2、查詢"01"課程比"02"課程成績低的學生的信息及課程分數

select a.S#,a.Sname,b.score ,c.score from Student a

left join sc b on b.s#=a.s# and b.c#=’01’

left join sc c on c.s#=a.s# and c.c#=’02’

Where b.score<c.score;

--2.1、查詢同時存在"01"課程和"02"課程的情況

select a.S#,a.Sname,b.score as ‘01‘,c.score as ‘02‘ from Student a

inner join sc b on b.s#=a.s# and b.c#=‘01‘

inner join sc c on c.s#=a.s# and c.c#=‘02‘;

--2.2、查詢同時存在"01"課程和"02"課程的情況和不存在"01"課程但存在"02"課程的情況

select a.S#,a.Sname,b.score as ‘01‘,c.score as ‘02‘ from Student a

left join sc b on b.s#=a.s# and b.c#=‘01‘

left join sc c on c.s#=a.s# and c.c#=‘02‘

Where c.score>isnull(b.score,0);

2.2.1存在01,但是不存在02課程的人

select distinct s.* from student s

left join sc a on a.s#=s.s#

left join sc b on b.s#=s.s#

--"<>"是指不等於的意思

where a.c#=‘02‘ and b.C#<>‘01‘

--3、查詢平均成績大於等於60分的同學的學生編號和學生姓名和平均成績

select s.S# ,s.Sname ,avgs from Student s ,

(select s# ,cast(avg(score) as decimal(18,2)) as avgs from sc group by s# having AVG(score)>=60) c

where c.S#=s.S#

--4、查詢平均成績小於60分的同學的學生編號和學生姓名和平均成績

select s.S# ,s.Sname ,avgs from Student s ,

(select s# ,cast(avg(score) as decimal(18,2)) as avgs from sc group by s# having AVG(score)<60) c

where c.S#=s.S#

--4.1、查詢在sc表存在成績的學生信息的SQL語句。

select * from Student where s# in (select s# from sc);

--4.2、查詢在sc表中不存在成績的學生信息的SQL語句。

select * from Student where s# not in (select s# from sc);

--5、查詢所有同學的學生編號、學生姓名、選課總數、所有課程的總成績

select s.s# ,s.Sname ,b.c as 課程數 ,b.d as 總分數 from Student s

left join (select s# ,count(c#) as c,sum(score) as d from sc group by s# ) b

on s.s#=b.S#

--5.1、查詢所有有成績的SQL

select s.* ,sc.C# ,sc.score from Student s ,sc

where s.S#=sc.S#

--5.2、查詢所有(包括有成績和無成績)SQL

select s.* ,sc.C# ,sc.score from Student s left join

sc on s.S#=sc.S#

--6、查詢""姓老師的數量

select count(1) as 數量 from Teacher t where t.Tname=‘李%‘

--7、查詢學過"張三"老師授課的同學的信息

Select s.s#, s.sname from student s,sc where s.s#=sc.s# and

Sc.c# in(select c.c# from course c,teacher t where c.t#=t.t# and t.tname=‘張三‘);

--8、查詢沒學過"張三"老師授課的同學的信息

Select distinct(s.s#) , s.sname from student s,sc where s.s#=sc.s# and

Sc.c# not in(select c.c# from course c,teacher t where c.t#=t.t# and t.tname=‘張三‘);

--9、查詢學過編號為"01"並且也學過編號為"02"的課程的同學的信息

select a.S#,a.Sname,b.score ,c.score from Student a

inner join sc b on b.s#=a.s# and b.c#=‘01‘

inner join sc c on c.s#=a.s# and c.c#=‘02‘

--10、查詢學過編號為"01"但是沒有學過編號為"02"的課程的同學的信息

select a.S#,a.Sname,b.score,c.score from Student a

left join sc b on b.s#=a.s# and b.c#=‘01‘

left join sc c on c.s#=a.s# and c.c#=‘02‘

where c.score is null and b.score>0;

--11、查詢沒有學全所有課程的同學的信息

select a.S#,a.Sname,b.score,c.score ,d.score from Student a

left join sc b on b.s#=a.s# and b.c#=‘01‘

left join sc c on c.s#=a.s# and c.c#=‘02‘

left join sc d on d.S#=a.S# and d.C#=‘03‘

where c.score is null or b.score is null or d.score is null;

select s.* from Student s

where s# not in (select s# from sc group by s# having count(1)=3)

--12、查詢至少有一門課與學號為"01"的同學所學相同的同學的信息

select distinct(a.S#),a.Sname from Student a,sc

where a.S#=sc.S# and sc.C#=any(select C# from Student a,sc

where a.S#=sc.S# and a.S#=‘01‘) and sc.S#!=‘01‘;

select s.* from Student s

where s# in

( select s# from sc where c# in (select C# from sc where S#=‘01‘and s#!=‘01‘))

select distinct s.* from student s

left join sc on s.s# = sc.s#

--先將學號為01 的去處掉,然後再將查找有這幾門課程的學生

where sc.c# in ( select c# from sc where sc.s#=‘01‘) and sc.s# <> ‘01‘

--13、查詢和"01"號的同學學習的課程完全相同的其他同學的信息

select S# from SC where S# not in

(select distinct(S#) from SC where C# not in (select C# from SC where S#=‘01‘))

group by S# having count(*)=(select count(*) from SC where S#=‘01‘);

select s.* from Student s

where s# in

( select distinct(s#) from sc

where c# in (select C# from sc where S#=‘01‘) and s#!=‘01‘

group by s# having count(1)=3

)

--更全面些

select * from student s

where s# in (

select s# from sc

where c# in (select c# from sc where s#=‘01‘)

group by s# having count(1)=(select count(1) from sc where s#=‘01‘))

--14、查詢沒學過"張三"老師講授的任一門課程的學生姓名

Select distinct(s.s#) , s.sname from student s,sc where s.s#=sc.s# and

Sc.c# not in(select c.c# from course c,teacher t where c.t#=t.t# and t.tname=‘張三‘);

--15、查詢兩門及其以上不及格課程的同學的學號,姓名及其平均成績

select a.S#,a.Sname,avg(sc.score) from Student a,sc

where a.S#=sc.S# and sc.S# in(

select s# from sc where score>=60

group by s# having count(*)>=2 )

group by a.S#,a.Sname;

--16、檢索"01"課程分數小於60,按分數降序排列的學生信息

select a.S#,a.Sname,sc.score from Student a,sc

where a.S#=sc.S# and sc.score<60 and sc.c#=‘01‘

order by sc.score desc;

--17、按平均成績從高到低顯示所有學生的所有課程的成績以及平均成績

select *from

(select s.S#, avg(f.score) as avg from Student s,sc f

where s.S#=f.s#

group by s.S#) a left join

(select a.S#,a.Sname,b.score as bs ,c.score as cs ,d.score as ds from Student a

left join sc b on b.s#=a.s# and b.c#=‘01‘

left join sc c on c.s#=a.s# and c.c#=‘02‘

left join sc d on d.S#=a.S# and d.C#=‘03‘)b on a.S#=b.S#;

--用左連接將一個一個數連接起來,然後再顯示出來

select s.*,b.score as ‘01‘,c.score as ‘02‘ , d.score as ‘03‘,a.avg_score from student s

--求出平均分,不計算null值

left join (select s# ,avg (score) as avg_score from sc group by s# ) a on s.S# =a.S#

left join sc b on b.S#=s.s# and b.c#=‘01‘

left join sc c on c.s#=s.s# and c.c#=‘02‘

left join sc d on d.s#= s.s# and d.c#=‘03‘

order by a.avg_score desc

--18、查詢各科成績最高分、最低分和平均分:以如下形式顯示:課程ID,課程name,最高分,最低分,平均分,及格率,中等率,優良率,優秀率

--及格為>=60,中等為:70-80,優良為:80-90,優秀為:>=90

select m.C# [課程編號], m.Cname [課程名稱],

max(n.score) [最高分],

min(n.score) [最低分],

cast(avg(n.score) as decimal(18,2)) [平均分]--,

,cast((select count(1) from SC where C# = m.C# and score >= 60)*100.00 / (select count(1) from SC where C# = m.C#) as decimal(18,2)) as 合格率

,cast((select count(1) from SC where C# = m.C# and score >= 70 and score < 80 )*100.0 / (select count(1) from SC where C# = m.C#) as decimal(18,2)) as 中等率

,cast((select count(1) from SC where C# = m.C# and score >= 80 and score < 90 )*100.0 / (select count(1) from SC where C# = m.C#) as decimal(18,2)) as 優良率

,cast((select count(1) from SC where C# = m.C# and score >= 90)*100.00/ (select count(1) from SC where C# = m.C#) as decimal(18,2)) as 優秀率

from Course m , SC n

where m.C# = n.C#

group by m.C# , m.Cname

--19、按各科成績進行排序,並顯示排名

select c#,a.Sname ,score ,rank() over(partition by c# order by score) as ‘排名‘ from sc,student a

where a.S#=sc.S#;

--20、查詢學生的總成績並進行排名

select sc.s#,a.Sname,sum(sc.score) as ‘總成績‘ from sc,Student a

where sc.S#=a.S#

group by sc.s#,a.Sname order by sum(sc.score)desc;

--20.1 查詢學生的總成績

select sc.s#,a.Sname,sum(sc.score) as ‘總成績‘ from sc,Student a

where sc.S#=a.S#

group by sc.s#,a.Sname;

--20.2 查詢學生的總成績並進行排名,sql 2000用子查詢完成,分總分重復時保留名次空缺和不保留名次空缺兩種。

--20.3 查詢學生的總成績並進行排名,sql 2005rank,DENSE_RANK完成,分總分重復時保留名次空缺和不保留名次空缺兩種。

select sc.s#,a.Sname,sum(sc.score) as ‘sscore‘, rank() over(order by sum(sc.score))

from sc,Student a

where sc.S#=a.S#

group by sc.s#,a.Sname

select sc.s#,a.Sname,sum(sc.score) as ‘sscore‘, dense_rank() over(order by sum(sc.score))

from sc,Student a

where sc.S#=a.S#

group by sc.s#,a.Sname

--21、查詢不同老師所教不同課程平均分從高到低顯示

select a.C#,a.Cname,b.Tname ,avg(sc.score) as ‘avg‘ from sc,Course a,Teacher b

where a.T#=b.T# and sc.C#=a.C#

group by a.C#,a.Cname,b.Tname

order by avg(score) desc

--22、查詢所有課程的成績第2名到第3名的學生信息及該課程成績

select *from (

select c#,a.Sname ,score ,DENSE_RANK() over(partition by c# order by score) as ‘srank‘ from sc,student a

where a.S#=sc.S#) a

where a.srank between 2 and 3;

--23、統計各科成績各分數段人數:課程編號,課程名稱,[100-85],[85-70],[70-60],[0-60]及所占百分比

select t1.*,round(t1.num1*100/t2.num,2) as ‘百分比‘ from

( select s.C#,c.Cname,

(case when s.score>=85 then ‘100-85‘

when s.score>=70 and s.score<85 then ‘85-70‘

when s.score>=60 and s.score<70 then ‘70-60‘

else ‘0-60‘

end) as px,count(1) as num1

from sc s,Course c

where c.C#=s.C#

group by s.C#,c.Cname,(case when s.score>=85 then ‘100-85‘

when s.score>=70 and s.score<85 then ‘85-70‘

when s.score>=60 and s.score<70 then ‘70-60‘

else ‘0-60‘

end)) t1,

(select s.C#,c.Cname,count(1)as num from sc s,Course c

where s.C#=c.C#

group by s.C#,c.Cname

) t2

where t1.C#=t2.C#;

select

s.c#,c.cname,(select count(1) from sc where C#=s.c# and score between 85 and 100) as [85-100],

(select count(1) from sc where c# = s.c# and score between 70 and 85) as [70-85],

(select count(1) from sc where c# = s.c# and score between 60 and 70) as [60-70],

(select count(1) from sc where c# = s.c# and score<60) as [0-60],

cast(cast((select count(1) from sc where c# = s.c# and score between 85 and 100) as decimal(18,2))*100/

cast((select count(1) from sc where c# = s.c#) as decimal(18,2)) as decimal(18,2)) as[85-100百分比],

cast(cast((select count(1) from sc where c# = s.c# and score between 70 and 85) as decimal(18,2))*100/

cast((select count(1) from sc where c# = s.c#) as decimal(18,2)) as decimal(18,2)) as[70-85百分比],

cast(cast((select count(1) from sc where c# = s.c# and score between 60 and 70) as decimal(18,2))*100/

cast((select count(1) from sc where c# = s.c#) as decimal(18,2)) as decimal(18,2)) as[60-70百分比],

cast(cast((select count(1) from sc where c# = s.c# and score<60) as decimal(18,2))*100/

cast((select count(1) from sc where c# = s.c#) as decimal(18,2)) as decimal(18,2)) as[0-60百分比]

from sc s inner join course c on c.c#=s.c#

group by s.c#,c.cname

--23.1 統計各科成績各分數段人數:課程編號,課程名稱,[100-85],[85-70],[70-60],[0-60]

select s.C#,c.Cname,

( case when s.score>=85 then ‘100-85‘

when s.score>=70 and s.score<85 then ‘85-70‘

when s.score>=60 and s.score<70 then ‘70-60‘

else ‘0-60‘

end

) as px,count(1) as num from sc s,Course c

where s.C#=c.C#

group by s.C#,c.Cname, ( case when s.score>=85 then ‘100-85‘

when s.score>=70 and s.score<85 then ‘85-70‘

when s.score>=60 and s.score<70 then ‘70-60‘

else ‘0-60‘

end

)

order by s.C#;

--23.2 統計各科成績各分數段人數:課程編號,課程名稱,[100-85],[85-70],[70-60],[<60]及所占百分比

select t1.*,round(t1.num1*100/t2.num,2) as ‘百分比‘ from

( select s.C#,c.Cname,

(case when s.score>=85 then ‘100-85‘

when s.score>=70 and s.score<85 then ‘85-70‘

when s.score>=60 and s.score<70 then ‘70-60‘

else ‘0-60‘

end) as px,count(1) as num1

from sc s,Course c

where c.C#=s.C#

group by s.C#,c.Cname,(case when s.score>=85 then ‘100-85‘

when s.score>=70 and s.score<85 then ‘85-70‘

when s.score>=60 and s.score<70 then ‘70-60‘

else ‘0-60‘

end)) t1,

(select s.C#,c.Cname,count(1)as num from sc s,Course c

where s.C#=c.C#

group by s.C#,c.Cname

) t2

where t1.C#=t2.C#;

--簡化版

select s.c#,c.cname,

(case when s.score between 85 and 100 then ‘85-100‘

when s.score between 70 and 85 then ‘70-85‘

when s.score between 60 and 70 then ‘60-70‘

else ‘0-60‘ end) as fshu,count(1) as num,

cast(cast(count(1)*100 as decimal(18,2))

/cast((select count(1) from sc where c#=s.c#) as decimal(18,2))as decimal(18,2)) as ‘百分比‘

from sc s

inner join course c on s.C# = c.C#

group by s.c#,c.cname,

(case when s.score between 85 and 100 then ‘85-100‘

when s.score between 70 and 85 then ‘70-85‘

when s.score between 60 and 70 then ‘60-70‘

else ‘0-60‘ end)

--24、查詢學生平均成績及其名次

select sc.S#,s.Sname,cast(avg(sc.score) as decimal(18,2)) as 平均分,rank() over(order by avg(sc.score)) as 排名 from sc,Student s

where sc.S#=s.S#

group by sc.S#,s.Sname;

select s.s#,s.sname,convert(decimal(18,2),avg(sc.score)) as [分數],

rank() over(order by avg(sc.score)desc ) as 排名 from student s

left join sc on s.s#=sc.s#

group by s.s#,s.sname

--24.1 查詢學生的平均成績並進行排名,sql 2000用子查詢完成,分平均成績重復時保留名次空缺和不保留名次空缺兩種。

--24.2 查詢學生的平均成績並進行排名,sql 2005rank,DENSE_RANK完成,分平均成績重復時保留名次空缺和不保留名次空缺兩種。

select sc.S#,s.Sname,cast(avg(sc.score) as decimal(18,2)) as 平均分, isnull (rank() over(order by avg(sc.score)),null) as 排名 from sc,Student s

where sc.S#=s.S#

group by sc.S#,s.Sname;

select sc.S#,s.Sname,cast(avg(sc.score) as decimal(18,2)) as 平均分,rank() over(order by avg(sc.score)) as 排名 from sc,Student s

where sc.S#=s.S#

group by sc.S#,s.Sname;

--不保留名次空缺

select s.s#,s.sname,convert(decimal(18,2),avg(sc.score)) as [平均成績],

dense_rank() over(order by avg(sc.score) desc) as [排名] from student s

left join sc on s.s#=sc.s#

group by s.s#,s.sname

--25、查詢各科成績前三名的記錄

--25.1 分數重復時保留名次空缺 ???????????????????????????????

select sc.c# ,sc.s# ,s.Sname , sc.score ,rank() over(partition by sc.c# order by sc.score desc)

from sc ,Student s

where sc.S#=s.S#

--25.2 分數重復時不保留名次空缺,合並名次

select * from(

select c.c#,c.cname,sc.score,dense_rank() over(partition by c.c# order by sc.score desc) as [排名]

from sc left join course c on c.c#=sc.c#

left join student s on sc.s#=s.s# ) t

where t.排名<4

--26、查詢每門課程被選修的學生數

select c.C#,c.Cname,count(1) as 數量

from Course c,sc

where c.C#=sc.C#

group by c.C#,c.Cname;

--27、查詢出只有兩門課程的全部學生的學號和姓名

select s.S#,s.Sname,count(1) as 課程數 from Student s,sc

where s.S#=sc.S#

group by s.S#,s.Sname

having count(1)=2

--28、查詢男生、女生人數

select Ssex, count(1) from Student

group by Ssex;

--29、查詢名字中含有""字的學生信息

select *from student

where sname like ‘%風%‘;

--30、查詢同名同性學生名單,並統計同名人數

select Sname, count(1) as 數量 from Student

group by Sname

having COUNT(1)>1;

--31、查詢1990年出生的學生名單(註:Student表中Sage列的類型是datetime)

select * from Student where year(sage)=1990;

--32、查詢每門課程的平均成績,結果按平均成績降序排列,平均成績相同時,按課程編號升序排列

select c#, cast(avg(score ) as decimal(18,2)) as avgscore from sc

group by c#

order by avgscore desc,C#

--33、查詢平均成績大於等於85的所有學生的學號、姓名和平均成績

select s.S# ,s.Sname ,cast(AVG(sc.score) as decimal(18,2)) as avgscore from Student s, sc

where s.S#=sc.S#

group by s.S#,s.Sname

having avg(score)>=85

--34、查詢課程名稱為"數學",且分數低於60的學生姓名和分數

select s.Sname ,sc.score from Student s ,Course c ,sc

where s.S#=sc.S# and c.C#=sc.C# and sc.score<60 and c.Cname=‘數學‘;

--35、查詢所有學生的課程及分數情況;

select s.Sname ,c.Cname ,sc.score from Student s ,Course c ,sc

where s.S#=sc.S# and sc.C#=c.C#

order by score

--36、查詢任何一門課程成績在70分以上的姓名、課程名稱和分數;

select s.S# ,s.Sname ,c.Cname ,sc.score from sc ,Course c ,Student s

where sc.S#=s.S# and c.C#=sc.C# and s.S# in (

select s# from sc a where score>70 group by s#

having count(1)=(select count(1) from sc where s#=a.S# group by s#))

--37、查詢不及格的課程

select s.S# ,s.Sname ,c.Cname ,sc.score from sc ,Course c ,Student s

where sc.S#=s.S# and c.C#=sc.C# and sc.score<60

--38、查詢課程編號為01且課程成績在80分以上的學生的學號和姓名;

select s.S# ,s.Sname ,sc.score from Student s ,sc

where s.S#=sc.S# and sc.C#=‘01‘ and sc.score>=80

--39、求每門課程的學生人數

select sc.C# ,count(1) as 人數 from sc

group by sc.C#

--40、查詢選修"張三"老師所授課程的學生中,成績最高的學生信息及其成績

select sc.C# ,s.* ,sc.score from Student s ,sc ,(

select sc.C# ,max(sc.score) as maxscore from Teacher t ,Course c ,sc

where t.T#=c.T# and t.Tname=‘張三‘ and sc.C#=c.C#

group by sc.C# ) t

where s.S#=sc.S# and t.C#=sc.C# and sc.score=t.maxscore

order by sc.C#

--將所有的表連接起來,排列出該老所教課程的所有學生的分數,根據分數從高到低進行排序

--取第一條數據,便是成績最高的學生了

select top 1 s.s#, s.sname, sc.score from student s inner join sc on s.s#=sc.s#

inner join course c on c.c#=sc.c#

inner join teacher t on c.t#=t.t#

where t.tname=‘張三‘

order by sc.score desc

--40.1 當最高分只有一個時

select s.* ,sc.score from student s ,sc ,

(select sc.C#,t.maxscore ,count(1) as aount from sc , (

select sc.C# ,max(sc.score) as maxscore from Teacher t ,Course c ,sc

where t.T#=c.T# and t.Tname=‘張三‘ and sc.C#=c.C#

group by sc.C# ) t

where t.C#=sc.C# and sc.score=t.maxscore

group by sc.C# ,t.maxscore

having count(1)<2 ) a

where s.S#=sc.S# and a.C#=sc.C# and a.maxscore=sc.score

--將所有的表連接起來,排列出該老所教課程的所有學生的分數,根據分數從高到低進行排序

--取第一條數據,便是成績最高的學生了

select top 1 s.s#, s.sname, sc.score from student s inner join sc on s.s#=sc.s#

inner join course c on c.c#=sc.c#

inner join teacher t on c.t#=t.t#

where t.tname=‘張三‘

order by sc.score desc

--40.2 當最高分出現多個時

select sc.C# ,s.* ,sc.score from student s ,sc ,

(select sc.C#,t.maxscore ,count(1) as aount from sc , (

select sc.C# ,max(sc.score) as maxscore from Teacher t ,Course c ,sc

where t.T#=c.T# and t.Tname=‘張三‘ and sc.C#=c.C#

group by sc.C# ) t --算出每一門科目的最高分及科目號

where t.C#=sc.C# and sc.score=t.maxscore

group by sc.C# ,t.maxscore

having count(1)>1 ) a --算出多人得分的科目及最高分

where s.S#=sc.S# and a.C#=sc.C# and a.maxscore=sc.score --根據科目號和最高分匹配學生

--先將最高分和課程取出來,然後再將將表進行連接,group by

select t.s#,t.sname,sc.score from student t inner join sc on t.S#=sc.S#

inner join

(

select c.c#, max(sc.score) as maxscore from sc inner join course c on c.C#=sc.C#

inner join teacher t on t.t#=c.t#

where t.tname=‘張三‘

group by c.C#

)a on a.C#=sc.C# and a.maxscore = sc.score

--先將最高分和課程取出來,然後再將將表進行連接,order by

select t.s#,t.sname,sc.score from student t inner join sc on t.S#=sc.S#

inner join

(

select top 1 sc.c#, sc.score as maxscore from sc inner join course c on c.C#=sc.C#

inner join teacher t on t.t#=c.t#

where t.tname=‘張三‘

order by sc.score desc

)a on a.C#=sc.C# and a.maxscore = sc.score

--41、查詢不同課程成績相同的學生的學生編號、課程編號、學生成績

--查詢某個學生的課程不同,但是成績同的情況

select s.* ,a.score as ‘01‘ ,b.score as ‘02‘,c.score as ‘03‘ from Student s

left join sc a on s.S#=a.S# and a.C#=‘01‘

left join sc b on s.S#=b.S# and b.C#=‘02‘

left join sc c on s.S#=c.S# and c.C#=‘03‘

where a.score=b.score or a.score=c.score or b.score=c.score

--42、查詢每門功成績最好的前兩名

select * from (

select sc.C# ,s.Sname , sc.score ,

rank() over( partition by sc.c# order by sc.score desc) as scorank

from sc,Student s

where s.S#=sc.S# ) t

where t.scorank<3;

--43、統計每門課程的學生選修人數(超過5人的課程才統計)。要求輸出課程號和選修人數,查詢結果按人數降序排列,若人數相同,按課程號升序排列

select c# ,count(1) as acount from sc

group by c#

having count(1)>5

order by acount desc ,C#

--44、檢索至少選修兩門課程的學生學號

select s# as 學生學號,count(1) as 課程數 from sc

group by s#

having count(1)>1

--45、查詢選修了全部課程的學生信息

select sc.s# ,count(1) as 課程數 from sc

group by sc.S#

having count(1)=(select count(1) from Course )

select distinct(s.S#), s.* from sc,Student s

where sc.S#=s.S# and s.S#=(

select sc.s# as 課程數 from sc

where sc.S#=s.S#

group by sc.S#

having count(1)=(select count(1) from Course ) )

--46、查詢各學生的年齡

select s# ,sname ,( year(GETDATE())-year(Sage)) as 年齡 ,Ssex from Student

--46.1 只按照年份來算

select s# ,sname ,( year(GETDATE())-year(Sage)) as 年齡 ,Ssex from Student

select s# ,sname ,DATEDIFF(yy ,sage ,getdate()) as 年齡 from student

--46.2 按照出生日期來算,當前月日 < 出生年月的月日則,年齡減一

--dateadd(month,2,getdate())--日期相加 :比如getdate2017-12-27 ,顯示是:2018-02-27

--datediff(month,sage, getdate()) --日期相減

  1. 日期指定的部分:月month(這年的第幾個月) ,年year, 天打day(這個月的第幾天)
  2. 開始的日期
  3. 結束的日期

--datename(week,getdate()) --返回日期的指定部分,放回該日期的第幾個星期,nvarchar

註釋:這個返回的是這個年的第幾個星期

--datepart(week,getdate()) --同上,結果返回int

--convert(varchar(10),getdate(), 120) --varchar的形式顯示前10個字符,顯示格式是120

即:2017-12-27

--cast(getdate() as varchar(10)) --與上相同,但是沒有顯示格式,所以顯示出來不好看

select s#, sname,

case when month(getdate())<month(sage) then datediff(year,sage,getdate())-1

when month(getdate())>month(sage) then datediff(year,sage,getdate())

else (case when day(getdate())<=day(sage) then datediff(year,sage,getdate())-1

else datediff(year,sage,getdate())

end)

end

from student

--47、查詢本周過生日的學生

--datename(week,getdate()) --返回日期的指定部分,放回該日期的第幾個星期,nvarchar

註釋:這個返回的是這個年的第幾個星期

--datepart(week,getdate()) --同上,結果返回int

本周的星期一

  1. SELECT DATEADD(wk, DATEDIFF(wk,0,getdate()), 0)

--***wk == week ,從系統指定時間一共過了多少個禮拜

--***datediff只會顯示數量,不會顯示日期

--顯示系統判定時間到現在一共有多少個禮拜了

select datediff(wk,0,getdate())--wk代表星期,getdate()代表現在日期

--***dateadd( [指定格式(wk,day,year)] , [要加入的數量],[從哪天開始加 (getdate())] )

--***dateadd只會顯示日期,不會顯示數量

--這周星期一的日期,從日期是0開始加入禮拜數量

select dateadd(wk,datediff(wk,0,getdate()),0)

select day(dateadd(wk,datediff(wk,0,getdate()),0))--這天是幾號

--這周星期日的日期

select dateadd(wk,datediff(wk,0,getdate())+1,0)-1

--計算是這周過生日的

--流程:1、計算這周的第一天和最後一天,取這兩天的幾月幾號,下面都是這種情況

--2、判斷是否存在跨年的,即判斷第一天和最後一天(12-25,1-1)的大小,可能存在2017-12-25到2018-01-01這種情況

--如果第一天比最後一天小,則生日就是大於第一天,小於最後一天,否則生日大於最後一天,小於第一天。

select s# , sname,convert(varchar(10),sage,120) as [生日] from student

where convert(varchar(5),sage,110) >=

case when convert(varchar(5),dateadd(wk,datediff(wk,0,getdate()),0),110) < convert(varchar(5),dateadd(wk,datediff(wk,0,getdate())+1,0)-1,110)

then convert(varchar(5),dateadd(wk,datediff(wk,0,getdate()),0),110)

else convert(varchar(5),dateadd(wk,datediff(wk,0,getdate())+1,0)-1,110)

end

and convert(varchar(5),sage,110) <=

case when convert(varchar(5),dateadd(wk,datediff(wk,0,getdate()),0),110) > convert(varchar(5),dateadd(wk,datediff(wk,0,getdate())+1,0)-1,110)

then convert(varchar(5),dateadd(wk,datediff(wk,0,getdate()),0),110)

else convert(varchar(5),dateadd(wk,datediff(wk,0,getdate())+1,0)-1,110)

end

--48、查詢下周過生日的學生

select s.* from Student s

where datename(week,day(sage))=DateName(week,day(Getdate()))+1

--這周是月底怎麽搞,還有這周是指第1-7天 為第一周,所以有問題

--49、查詢本月過生日的學生

select s.* from Student s

where month(sage)=month(getdate())

--50、查詢下月過生日的學生

select s.* from Student s

where month(sage)=month(getdate())+1

但是這個月是12月怎麽去算

所以,修改下

--**dateadd加上一個月的日期,就能跳到下個月了,然後下個月的月數

select s#,sname, convert(varchar(10), sage, 120) as [生日] from student

where month(sage) = month(dateadd(month , 1, getdate()))

--用case...when 函數,當這個月為12月時就將他改為1

select s# , sname, convert(varchar(10) ,sage, 120) from student s

where datepart(month,sage) =

case when datepart(month,getdate()) = 12 then 1

else datepart(month,getdate())+1 end

1、以下純取值,不關聯表了

  1. 月初

--datediff從初始到現在一共有多少個月,然後將這些月數相加換成日期

select dateadd(month,datediff(month,0,getdate()),0);

  1. 月末

--datediff從初始日期到現在一共有多少個月,然後+1就多一個月,最後—1就是下個月的月初變為月末

select dateadd(month,datediff(month, 0, getdate())+1,0)-1

  1. 下個月的月初

select dateadd(month,datediff(month,0,getdate())+1,0)

  1. 下個月的月末

select dateadd(month,datediff(month, 0, getdate())+2,0)-1

sql基礎題目測試及正確答案