sql程序設計題

1. 急！關於用SQL語句表達資料庫查詢

1、select Cno,Cname from Course where Teacher='劉老師';

2、select Sno,Sname from Student where Sage < 22;

3、select Sname from Student where Sno not in(
select Sno from sc inner join Course on sc.Cno=Course.Cno where Course.Teacher='劉老師');

4、select Sname from Student where Sno in(
select Sno from sc inner join Course on sc.Cno=Course.Cno where sc.Grade>90 and Course.Cname='程序設計');

5、select Sname from Student where Sno not in(
select Sno from sc inner join Course on sc.Cno=Course.Cno where Course.Cname='C3課');

6、select Sno from Student where Sno not in(
select Sno from sc inner join Course on sc.Cno=Course.Cno where Course.Cname='C1課' or Course.Cname='C2課')
group by Sno having count(*) >= 2;

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以上，希望對你有所幫助。

2. SQL查詢面試題與答案

SQL查詢面試題與答案

SQL語言是一種資料庫查詢和程序設計語言，用於存取數據以及查詢、更新和管理關系資料庫系統;同時也是資料庫腳本文件的擴展名。下面是我搜集的SQL查詢面試題與答案，歡迎大家閱讀。

SQL查詢面試題與答案一

1.一道SQL語句面試題，關於group by表內容：

2005-05-09 勝

2005-05-09 勝

2005-05-09 負

2005-05-09 負

2005-05-10 勝

2005-05-10 負

2005-05-10 負

如果要生成下列結果, 該如何寫sql語句?

勝 負

2005-05-09 2 2

2005-05-10 1 2

------------------------------------------

create table #tmp(rq varchar(10),shengfu nchar(1))

insert into #tmp values('2005-05-09','勝')

insert into #tmp values('2005-05-09','勝')

insert into #tmp values('2005-05-09','負')

insert into #tmp values('2005-05-09','負')

insert into #tmp values('2005-05-10','勝')

insert into #tmp values('2005-05-10','負')

insert into #tmp values('2005-05-10','負')

1)select rq, sum(case when shengfu='勝' then 1 else 0 end)'勝',sum(case when shengfu='負' then 1 else 0 end)'負' from #tmp group by rq

2) select N.rq,N.勝,M.負 from (

select rq,勝=count(*) from #tmp where shengfu='勝'group by rq)N inner join

(select rq,負=count(*) from #tmp where shengfu='負'group by rq)M on N.rq=M.rq

3)select a.col001,a.a1 勝,b.b1 負 from

(select col001,count(col001) a1 from temp1 where col002='勝' group by col001) a,

(select col001,count(col001) b1 from temp1 where col002='負' group by col001) b

where a.col001=b.col001

2.請教一個面試中遇到的SQL語句的查詢問題

表中有A B C三列,用SQL語句實現：當A列大於B列時選擇A列否則選擇B列，當B列大於C列時選擇B列否則選擇C列。

------------------------------------------

select (case when a>b then a else b end ),

(case when b>c then b esle c end)

from table_name

3.面試題：一個日期判斷的sql語句?

請取出tb_send表中日期(SendTime欄位)為當天的所有記錄?(SendTime欄位為datetime型，包含日期與時間)

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select * from tb where datediff(dd,SendTime,getdate())=0

4.有一張表，裡面有3個欄位：語文，數學，英語。其中有3條記錄分別表示語文70分，數學80分，英語58分，請用一條sql語句查詢出這三條記錄並按以下條件顯示出來(並寫出您的思路)：

大於或等於80表示優秀，大於或等於60表示及格，小於60分表示不及格。

顯示格式：

語文 數學 英語

及格 優秀 不及格

------------------------------------------

select

(case when 語文>=80 then '優秀'

when 語文>=60 then '及格'

else '不及格') as 語文,

(case when 數學>=80 then '優秀'

when 數學>=60 then '及格'

else '不及格') as 數學,

(case when 英語>=80 then '優秀'

when 英語>=60 then '及格'

else '不及格') as 英語,

from table

5.在sqlserver2000中請用sql創建一張用戶臨時表和系統臨時表，裡麵包含兩個欄位ID和IDValues,類型都是int型，並解釋下兩者的區別?

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用戶臨時表:create table #xx(ID int, IDValues int)

系統臨時表:create table ##xx(ID int, IDValues int)

區別:

用戶臨時表只對創建這個表的用戶的Session可見,對其他進程是不可見的.

當創建它的進程消失時這個臨時表就自動刪除.

全局臨時表對整個SQL Server實例都可見,但是所有訪問它的Session都消失的時候,它也自動刪除.

6.sqlserver2000是一種大型資料庫，他的`存儲容量只受存儲介質的限制，請問它是通過什麼方式實現這種無限容量機制的。

------------------------------------------

它的所有數據都存儲在數據文件中(*.dbf),所以只要文件夠大,SQL Server的存儲容量是可以擴大的.

SQL Server 2000 資料庫有三種類型的文件：

主要數據文件

主要數據文件是資料庫的起點，指向資料庫中文件的其它部分。每個資料庫都有一個主要數據文件。主要數據文件的推薦文件擴展名是 .mdf。

次要數據文件

次要數據文件包含除主要數據文件外的所有數據文件。有些資料庫可能沒有次要數據文件，而有些資料庫則有多個次要數據文件。次要數據文件的推薦文件擴展名是 .ndf。

日誌文件

日誌文件包含恢復資料庫所需的所有日誌信息。每個資料庫必須至少有一個日誌文件，但可以不止一個。日誌文件的推薦文件擴展名是 .ldf。

7.請用一個sql語句得出結果

從table1,table2中取出如table3所列格式數據，注意提供的數據及結果不準確，只是作為一個格式向大家請教。

如使用存儲過程也可以。

table1

月份mon 部門dep 業績yj

-------------------------------

一月份 01 10

一月份 02 10

一月份 03 5

二月份 02 8

二月份 04 9

三月份 03 8

table2

部門dep 部門名稱dname

--------------------------------

01 國內業務一部

02 國內業務二部

03 國內業務三部

04 國際業務部

table3 (result)

部門dep 一月份 二月份 三月份

--------------------------------------

01 10 null null

02 10 8 null

03 null 5 8

04 null null 9

------------------------------------------

1)

select a.部門名稱dname,b.業績yj as '一月份',c.業績yj as '二月份',d.業績yj as '三月份'

from table1 a,table2 b,table2 c,table2 d

where a.部門dep = b.部門dep and b.月份mon = '一月份' and

a.部門dep = c.部門dep and c.月份mon = '二月份' and

a.部門dep = d.部門dep and d.月份mon = '三月份' and

2)

select a.dep,

sum(case when b.mon=1 then b.yj else 0 end) as '一月份',

sum(case when b.mon=2 then b.yj else 0 end) as '二月份',

sum(case when b.mon=3 then b.yj else 0 end) as '三月份',

sum(case when b.mon=4 then b.yj else 0 end) as '四月份',

sum(case when b.mon=5 then b.yj else 0 end) as '五月份',

sum(case when b.mon=6 then b.yj else 0 end) as '六月份',

sum(case when b.mon=7 then b.yj else 0 end) as '七月份',

sum(case when b.mon=8 then b.yj else 0 end) as '八月份',

sum(case when b.mon=9 then b.yj else 0 end) as '九月份',

sum(case when b.mon=10 then b.yj else 0 end) as '十月份',

sum(case when b.mon=11 then b.yj else 0 end) as '十一月份',

sum(case when b.mon=12 then b.yj else 0 end) as '十二月份',

from table2 a left join table1 b on a.dep=b.dep

8.華為一道面試題

一個表中的Id有多個記錄，把所有這個id的記錄查出來，並顯示共有多少條記錄數。

------------------------------------------

select id, Count(*) from tb group by id having count(*)>1

select * from(select count(ID) as count from table group by ID)T where T.count>1

SQL查詢面試題與答案二

1、查詢不同老師所教不同課程平均分從高到低顯示

SELECT max(Z.T#) AS 教師ID,MAX(Z.Tname) AS 教師姓名,C.C# AS 課程ID,MAX(C.Cname) AS 課程名稱,AVG(Score) AS 平均成績

FROM SC AS T,Course AS C ,Teacher AS Z

where T.C#=C.C# and C.T#=Z.T#

GROUP BY C.C#

ORDER BY AVG(Score) DESC

2、查詢如下課程成績第 3 名到第 6 名的學生成績單：企業管理(001)，馬克思(002)，UML (003)，資料庫(004)

[學生ID],[學生姓名],企業管理,馬克思,UML,資料庫,平均成績

SELECT DISTINCT top 3

SC.S# As 學生學號,

Student.Sname AS 學生姓名 ,

T1.score AS 企業管理,

T2.score AS 馬克思,

T3.score AS UML,

T4.score AS 資料庫,

ISNULL(T1.score,0) + ISNULL(T2.score,0) + ISNULL(T3.score,0) + ISNULL(T4.score,0) as 總分

FROM Student,SC LEFT JOIN SC AS T1

ON SC.S# = T1.S# AND T1.C# = '001'

LEFT JOIN SC AS T2

ON SC.S# = T2.S# AND T2.C# = '002'

LEFT JOIN SC AS T3

ON SC.S# = T3.S# AND T3.C# = '003'

LEFT JOIN SC AS T4

ON SC.S# = T4.S# AND T4.C# = '004'

WHERE student.S#=SC.S# and

ISNULL(T1.score,0) + ISNULL(T2.score,0) + ISNULL(T3.score,0) + ISNULL(T4.score,0)

NOT IN

(SELECT

DISTINCT

TOP 15 WITH TIES

ISNULL(T1.score,0) + ISNULL(T2.score,0) + ISNULL(T3.score,0) + ISNULL(T4.score,0)

FROM sc

LEFT JOIN sc AS T1

ON sc.S# = T1.S# AND T1.C# = 'k1'

LEFT JOIN sc AS T2

ON sc.S# = T2.S# AND T2.C# = 'k2'

LEFT JOIN sc AS T3

ON sc.S# = T3.S# AND T3.C# = 'k3'

LEFT JOIN sc AS T4

ON sc.S# = T4.S# AND T4.C# = 'k4'

ORDER BY ISNULL(T1.score,0) + ISNULL(T2.score,0) + ISNULL(T3.score,0) + ISNULL(T4.score,0) DESC);

3、統計列印各科成績,各分數段人數:課程ID,課程名稱,[100-85],[85-70],[70-60],[ <60]

SELECT SC.C# as 課程ID, Cname as 課程名稱

,SUM(CASE WHEN score BETWEEN 85 AND 100 THEN 1 ELSE 0 END) AS [100 - 85]

,SUM(CASE WHEN score BETWEEN 70 AND 85 THEN 1 ELSE 0 END) AS [85 - 70]

,SUM(CASE WHEN score BETWEEN 60 AND 70 THEN 1 ELSE 0 END) AS [70 - 60]

,SUM(CASE WHEN score < 60 THEN 1 ELSE 0 END) AS [60 -]

FROM SC,Course

where SC.C#=Course.C#

GROUP BY SC.C#,Cname;

4、查詢學生平均成績及其名次

SELECT 1+(SELECT COUNT( distinct 平均成績)

FROM (SELECT S#,AVG(score) AS 平均成績

FROM SC

GROUP BY S#

) AS T1

WHERE 平均成績 > T2.平均成績) as 名次,

S# as 學生學號,平均成績

FROM (SELECT S#,AVG(score) 平均成績

FROM SC

GROUP BY S#

) AS T2

ORDER BY 平均成績 desc;

5、查詢各科成績前三名的記錄:(不考慮成績並列情況)

SELECT t1.S# as 學生ID,t1.C# as 課程ID,Score as 分數

FROM SC t1

WHERE score IN (SELECT TOP 3 score

FROM SC

WHERE t1.C#= C#

ORDER BY score DESC

)

ORDER BY t1.C#;

6、查詢每門課程被選修的學生數

select c#,count(S#) from sc group by C#;

7、查詢出只選修了一門課程的全部學生的學號和姓名

select SC.S#,Student.Sname,count(C#) AS 選課數

from SC ,Student

where SC.S#=Student.S# group by SC.S# ,Student.Sname having count(C#)=1;

8、查詢課程編號“002”的成績比課程編號“001”課程低的所有同學的學號、姓名;

Select S#,Sname from (select Student.S#,Student.Sname,score ,(select score from SC SC_2 where SC_2.S#=Student.S# and SC_2.C#='002') score2

from Student,SC where Student.S#=SC.S# and C#='001') S_2 where score2

9、查詢所有課程成績小於60分的同學的學號、姓名;

select S#,Sname

from Student

where S# not in (select Student.S# from Student,SC where S.S#=SC.S# and score>60);

10、查詢沒有學全所有課的同學的學號、姓名;

select Student.S#,Student.Sname

from Student,SC

where Student.S#=SC.S# group by Student.S#,Student.Sname having count(C#) <(select count(C#) from Course);

11、查詢至少有一門課與學號為“1001”的同學所學相同的同學的學號和姓名;

select S#,Sname from Student,SC where Student.S#=SC.S# and C# in select C# from SC where S#='1001';

12、查詢至少學過學號為“001”同學所有一門課的其他同學學號和姓名;

select distinct SC.S#,Sname

from Student,SC

where Student.S#=SC.S# and C# in (select C# from SC where S#='001');

13、把“SC”表中“葉平”老師教的課的成績都更改為此課程的平均成績;

update SC set score=(select avg(SC_2.score)

from SC SC_2

where SC_2.C#=SC.C# ) from Course,Teacher where Course.C#=SC.C# and Course.T#=Teacher.T# and Teacher.Tname='葉平');

14、查詢和“1002”號的同學學習的課程完全相同的其他同學學號和姓名;

select S# from SC where C# in (select C# from SC where S#='1002')

group by S# having count(*)=(select count(*) from SC where S#='1002');

15、刪除學習“葉平”老師課的SC表記錄;

Delect SC

from course ,Teacher

where Course.C#=SC.C# and Course.T#= Teacher.T# and Tname='葉平';

16、向SC表中插入一些記錄，這些記錄要求符合以下條件：沒有上過編號“003”課程的同學學號、2、

號課的平均成績;

Insert SC select S#,'002',(Select avg(score)

from SC where C#='002') from Student where S# not in (Select S# from SC where C#='002');

17、按平均成績從高到低顯示所有學生的“資料庫”、“企業管理”、“英語”三門的課程成績，按如下形式顯示： 學生ID,,資料庫,企業管理,英語,有效課程數,有效平均分

SELECT S# as 學生ID

,(SELECT score FROM SC WHERE SC.S#=t.S# AND C#='004') AS 資料庫

,(SELECT score FROM SC WHERE SC.S#=t.S# AND C#='001') AS 企業管理

,(SELECT score FROM SC WHERE SC.S#=t.S# AND C#='006') AS 英語

,COUNT(*) AS 有效課程數, AVG(t.score) AS 平均成績

FROM SC AS t

GROUP BY S#

ORDER BY avg(t.score)

18、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

SELECT L.C# As 課程ID,L.score AS 最高分,R.score AS 最低分

FROM SC L ,SC AS R

WHERE L.C# = R.C# and

L.score = (SELECT MAX(IL.score)

FROM SC AS IL,Student AS IM

WHERE L.C# = IL.C# and IM.S#=IL.S#

GROUP BY IL.C#)

AND

R.Score = (SELECT MIN(IR.score)

FROM SC AS IR

WHERE R.C# = IR.C#

GROUP BY IR.C#

);

19、按各科平均成績從低到高和及格率的百分數從高到低順序

SELECT t.C# AS 課程號,max(course.Cname)AS 課程名,isnull(AVG(score),0) AS 平均成績

,100 * SUM(CASE WHEN isnull(score,0)>=60 THEN 1 ELSE 0 END)/COUNT(*) AS 及格百分數

FROM SC T,Course

where t.C#=course.C#

GROUP BY t.C#

ORDER BY 100 * SUM(CASE WHEN isnull(score,0)>=60 THEN 1 ELSE 0 END)/COUNT(*) DESC

20、查詢如下課程平均成績和及格率的百分數(用"1行"顯示): 企業管理(001)，馬克思(002)，OO&UML (003)，資料庫(004)

SELECT SUM(CASE WHEN C# ='001' THEN score ELSE 0 END)/SUM(CASE C# WHEN '001' THEN 1 ELSE 0 END) AS 企業管理平均分

,100 * SUM(CASE WHEN C# = '001' AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN C# = '001' THEN 1 ELSE 0 END) AS 企業管理及格百分數

,SUM(CASE WHEN C# = '002' THEN score ELSE 0 END)/SUM(CASE C# WHEN '002' THEN 1 ELSE 0 END) AS 馬克思平均分

,100 * SUM(CASE WHEN C# = '002' AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN C# = '002' THEN 1 ELSE 0 END) AS 馬克思及格百分數

,SUM(CASE WHEN C# = '003' THEN score ELSE 0 END)/SUM(CASE C# WHEN '003' THEN 1 ELSE 0 END) AS UML平均分

,100 * SUM(CASE WHEN C# = '003' AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN C# = '003' THEN 1 ELSE 0 END) AS UML及格百分數

,SUM(CASE WHEN C# = '004' THEN score ELSE 0 END)/SUM(CASE C# WHEN '004' THEN 1 ELSE 0 END) AS 資料庫平均分

,100 * SUM(CASE WHEN C# = '004' AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN C# = '004' THEN 1 ELSE 0 END) AS 資料庫及格百分數

FROM SC

;