SQL查詢面試題與答案
SQL語言是一種數(shù)據(jù)庫查詢和程序設(shè)計語言，用于存取數(shù)據(jù)以及查詢、更新和管理關(guān)系數(shù)據(jù)庫系統(tǒng);同時也是數(shù)據(jù)庫腳本文件的擴展名。下面是小編搜集的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(guān)于group by表內(nèi)容：
2005-05-09 勝
2005-05-09 勝
2005-05-09 負(fù)
2005-05-09 負(fù)
2005-05-10 勝
2005-05-10 負(fù)
2005-05-10 負(fù)
如果要生成下列結(jié)果, 該如何寫sql語句?
勝 負(fù)
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(fù)')
insert into #tmp values('2005-05-09','負(fù)')
insert into #tmp values('2005-05-10','勝')
insert into #tmp values('2005-05-10','負(fù)')
insert into #tmp values('2005-05-10','負(fù)')
1)select rq, sum(case when shengfu='勝' then 1 else 0 end)'勝',sum(case when shengfu='負(fù)' then 1 else 0 end)'負(fù)' from #tmp group by rq
2) select N.rq,N.勝,M.負(fù) from (
select rq,勝=count(*) from #tmp where shengfu='勝'group by rq)N inner join
(select rq,負(fù)=count(*) from #tmp where shengfu='負(fù)'group by rq)M on N.rq=M.rq
3)select a.col001,a.a1 勝,b.b1 負(fù) 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(fù)' 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(xiàn)：當(dāng)A列大于B列時選擇A列否則選擇B列，當(dāng)B列大于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字段)為當(dāng)天的所有記錄?(SendTime字段為datetime型，包含日期與時間)
------------------------------------------
select * from tb where datediff(dd,SendTime,getdate())=0
4.有一張表，里面有3個字段：語文，數(shù)學(xué)，英語。其中有3條記錄分別表示語文70分，數(shù)學(xué)80分，英語58分，請用一條sql語句查詢出這三條記錄并按以下條件顯示出來(并寫出您的思路)：
大于或等于80表示優(yōu)秀，大于或等于60表示及格，小于60分表示不及格。
顯示格式：
語文 數(shù)學(xué) 英語
及格 優(yōu)秀 不及格
------------------------------------------
select
(case when 語文>=80 then '優(yōu)秀'
when 語文>=60 then '及格'
else '不及格') as 語文,
(case when 數(shù)學(xué)>=80 then '優(yōu)秀'
when 數(shù)學(xué)>=60 then '及格'
else '不及格') as 數(shù)學(xué),
(case when 英語>=80 then '優(yōu)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(chuàng)建一張用戶臨時表和系統(tǒng)臨時表，里面包含兩個字段ID和IDValues,類型都是int型，并解釋下兩者的區(qū)別?
------------------------------------------
用戶臨時表:create table #xx(ID int, IDValues int)
系統(tǒng)臨時表:create table ##xx(ID int, IDValues int)
區(qū)別:
用戶臨時表只對創(chuàng)建這個表的用戶的Session可見,對其他進程是不可見的.
當(dāng)創(chuàng)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是一種大型數(shù)據(jù)庫，他的`存儲容量只受存儲介質(zhì)的限制，請問它是通過什么方式實現(xiàn)這種無限容量機制的。
------------------------------------------
它的所有數(shù)據(jù)都存儲在數(shù)據(jù)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 數(shù)據(jù)庫有三種類型的文件：
主要數(shù)據(jù)文件
主要數(shù)據(jù)文件是數(shù)據(jù)庫的起點，指向數(shù)據(jù)庫中文件的其它部分。每個數(shù)據(jù)庫都有一個主要數(shù)據(jù)文件。主要數(shù)據(jù)文件的推薦文件擴展名是 .mdf。
次要數(shù)據(jù)文件
次要數(shù)據(jù)文件包含除主要數(shù)據(jù)文件外的所有數(shù)據(jù)文件。有些數(shù)據(jù)庫可能沒有次要數(shù)據(jù)文件，而有些數(shù)據(jù)庫則有多個次要數(shù)據(jù)文件。次要數(shù)據(jù)文件的推薦文件擴展名是 .ndf。
日志文件
日志文件包含恢復(fù)數(shù)據(jù)庫所需的所有日志信息。每個數(shù)據(jù)庫必須至少有一個日志文件，但可以不止一個。日志文件的推薦文件擴展名是 .ldf。
7.請用一個sql語句得出結(jié)果
從table1,table2中取出如table3所列格式數(shù)據(jù)，注意提供的數(shù)據(jù)及結(jié)果不準(zhǔn)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(yè)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(nèi)業(yè)務(wù)一部
02 國內(nèi)業(yè)務(wù)二部
03 國內(nèi)業(yè)務(wù)三部
04 國際業(yè)務(wù)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(yè)績yj as '一月份',c.業(yè)績yj as '二月份',d.業(yè)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，并顯示共有多少條記錄數(shù)。
------------------------------------------
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?nèi)缦抡n程成績第 3 名到第 6 名的學(xué)生成績單：企業(yè)管理(001)，馬克思(002)，UML (003)，數(shù)據(jù)庫(004)
[學(xué)生ID],[學(xué)生姓名],企業(yè)管理,馬克思,UML,數(shù)據(jù)庫,平均成績
SELECT DISTINCT top 3
SC.S# As 學(xué)生學(xué)號,
Student.Sname AS 學(xué)生姓名 ,
T1.score AS 企業(yè)管理,
T2.score AS 馬克思,
T3.score AS UML,
T4.score AS 數(shù)據(jù)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(tǒng)計列印各科成績,各分?jǐn)?shù)段人數(shù):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(xué)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(xué)生學(xué)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:(不考慮成績并列情況)
SELECT t1.S# as 學(xué)生ID,t1.C# as 課程ID,Score as 分?jǐn)?shù)
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(xué)生數(shù)
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(xué)生的學(xué)號和姓名
select SC.S#,Student.Sname,count(C#) AS 選課數(shù)
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(xué)的學(xué)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于60分的同學(xué)的學(xué)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(xué)全所有課的同學(xué)的學(xué)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(xué)號為“1001”的同學(xué)所學(xué)相同的同學(xué)的學(xué)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(xué)過學(xué)號為“001”同學(xué)所有一門課的其他同學(xué)學(xué)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(xué)學(xué)習(xí)的課程完全相同的其他同學(xué)學(xué)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(xué)習(xí)“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(xué)學(xué)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(xué)生的“數(shù)據(jù)庫”、“企業(yè)管理”、“英語”三門的課程成績，按如下形式顯示： 學(xué)生ID,,數(shù)據(jù)庫,企業(yè)管理,英語,有效課程數(shù),有效平均分
SELECT S# as 學(xué)生ID
,(SELECT score FROM SC WHERE SC.S#=t.S# AND C#='004') AS 數(shù)據(jù)庫
,(SELECT score FROM SC WHERE SC.S#=t.S# AND C#='001') AS 企業(yè)管理
,(SELECT score FROM SC WHERE SC.S#=t.S# AND C#='006') AS 英語
,COUNT(*) AS 有效課程數(shù), 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?jǐn)?shù)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?jǐn)?shù)
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?nèi)缦抡n程平均成績和及格率的百分?jǐn)?shù)(用"1行"顯示): 企業(yè)管理(001)，馬克思(002)，OO&UML (003)，數(shù)據(jù)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(yè)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(yè)管理及格百分?jǐn)?shù)
,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?jǐn)?shù)
,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?jǐn)?shù)
,SUM(CASE WHEN C# = '004' THEN score ELSE 0 END)/SUM(CASE C# WHEN '004' THEN 1 ELSE 0 END) AS 數(shù)據(jù)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 數(shù)據(jù)庫及格百分?jǐn)?shù)
FROM SC
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