SQL查询面试题与答案
SQL语言是一种数据库查询和程序设计语言，用于存取数据以及查询、更新和管理关系数据库系统;同时也是数据库脚本文件的扩展名。下面是小编搜集的SQL查询面试题与答案，欢迎大家阅读。
SQL查询面试题与答案一
1.一道SQL语句面试题，关于group by表内容：
2005-05-09 胜
2005-05-09 胜
2005-05-09 负
2005-05-09 负
2005-05-10 胜
2005-05-10 负
2005-05-10 负
如果要生成下列结果, 该如何写sql语句?
胜 负
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','胜')
insert into #tmp values('2005-05-09','胜')
insert into #tmp values('2005-05-09','负')
insert into #tmp values('2005-05-09','负')
insert into #tmp values('2005-05-10','胜')
insert into #tmp values('2005-05-10','负')
insert into #tmp values('2005-05-10','负')
1)select rq, sum(case when shengfu='胜' then 1 else 0 end)'胜',sum(case when shengfu='负' then 1 else 0 end)'负' from #tmp group by rq
2) select N.rq,N.勝,M.負 from (
select rq,勝=count(*) from #tmp where shengfu='胜'group by rq)N inner join
(select rq,負=count(*) from #tmp where shengfu='负'group by rq)M on N.rq=M.rq
3)select a.col001,a.a1 胜,b.b1 负 from
(select col001,count(col001) a1 from temp1 where col002='胜' group by col001) a,
(select col001,count(col001) b1 from temp1 where col002='负' group by col001) b
where a.col001=b.col001
2.请教一个面试中遇到的SQL语句的查询问题
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 esle c end)
from table_name
3.面试题：一个日期判断的sql语句?
请取出tb_send表中日期(SendTime字段)为当天的所有记录?(SendTime字段为datetime型，包含日期与时间)
------------------------------------------
select * from tb where datediff(dd,SendTime,getdate())=0
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：
语文 数学 英语
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.在sqlserver2000中请用sql创建一张用户临时表和系统临时表，里面包含两个字段ID和IDValues,类型都是int型，并解释下两者的区别?
------------------------------------------
用户临时表:create table #xx(ID int, IDValues int)
系统临时表:create table ##xx(ID int, IDValues int)
区别:
用户临时表只对创建这个表的用户的Session可见,对其他进程是不可见的.
当创建它的进程消失时这个临时表就自动删除.
全局临时表对整个SQL Server实例都可见,但是所有访问它的Session都消失的时候,它也自动删除.
6.sqlserver2000是一种大型数据库，他的`存储容量只受存储介质的限制，请问它是通过什么方式实现这种无限容量机制的。
------------------------------------------
它的所有数据都存储在数据文件中(*.dbf),所以只要文件够大,SQL Server的存储容量是可以扩大的.
SQL Server 2000 数据库有三种类型的文件：
主要数据文件
主要数据文件是数据库的起点，指向数据库中文件的其它部分。每个数据库都有一个主要数据文件。主要数据文件的推荐文件扩展名是 .mdf。
次要数据文件
次要数据文件包含除主要数据文件外的所有数据文件。有些数据库可能没有次要数据文件，而有些数据库则有多个次要数据文件。次要数据文件的推荐文件扩展名是 .ndf。
日志文件
日志文件包含恢复数据库所需的所有日志信息。每个数据库必须至少有一个日志文件，但可以不止一个。日志文件的推荐文件扩展名是 .ldf。
7.请用一个sql语句得出结果
从table1,table2中取出如table3所列格式数据，注意提供的数据及结果不准确，只是作为一个格式向大家请教。
如使用存储过程也可以。
table1
月份mon 部门dep 业绩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
部门dep 部门名称dname
--------------------------------
01 国内业务一部
02 国内业务二部
03 国内业务三部
04 国际业务部
table3 (result)
部门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.部门名称dname,b.业绩yj as '一月份',c.业绩yj as '二月份',d.业绩yj as '三月份'
from table1 a,table2 b,table2 c,table2 d
where a.部门dep = b.部门dep and b.月份mon = '一月份' and
a.部门dep = c.部门dep and c.月份mon = '二月份' and
a.部门dep = d.部门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.华为一道面试题
一个表中的Id有多个记录，把所有这个id的记录查出来，并显示共有多少条记录数。
------------------------------------------
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查询面试题与答案二
1、查询不同老师所教不同课程平均分从高到低显示
SELECT max(Z.T#) AS 教师ID,MAX(Z.Tname) AS 教师姓名,C.C# AS 课程ID,MAX(C.Cname) AS 课程名称,AVG(Score) AS 平均成绩
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、查询如下课程成绩第 3 名到第 6 名的学生成绩单：企业管理(001)，马克思(002)，UML (003)，数据库(004)
[学生ID],[学生姓名],企业管理,马克思,UML,数据库,平均成绩
SELECT DISTINCT top 3
SC.S# As 学生学号,
Student.Sname AS 学生姓名 ,
T1.score AS 企业管理,
T2.score AS 马克思,
T3.score AS UML,
T4.score AS 数据库,
ISNULL(T1.score,0) + ISNULL(T2.score,0) + ISNULL(T3.score,0) + ISNULL(T4.score,0) as 总分
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、统计列印各科成绩,各分数段人数:课程ID,课程名称,[100-85],[85-70],[70-60],[ <60]
SELECT SC.C# as 课程ID, Cname as 课程名称
,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、查询学生平均成绩及其名次
SELECT 1+(SELECT COUNT( distinct 平均成绩)
FROM (SELECT S#,AVG(score) AS 平均成绩
FROM SC
GROUP BY S#
) AS T1
WHERE 平均成绩 > T2.平均成绩) as 名次,
S# as 学生学号,平均成绩
FROM (SELECT S#,AVG(score) 平均成绩
FROM SC
GROUP BY S#
) AS T2
ORDER BY 平均成绩 desc;
5、查询各科成绩前三名的记录:(不考虑成绩并列情况)
SELECT t1.S# as 学生ID,t1.C# as 课程ID,Score as 分数
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、查询每门课程被选修的学生数
select c#,count(S#) from sc group by C#;
7、查询出只选修了一门课程的全部学生的学号和姓名
select SC.S#,Student.Sname,count(C#) AS 选课数
from SC ,Student
where SC.S#=Student.S# group by SC.S# ,Student.Sname having count(C#)=1;
8、查询课程编号“002”的成绩比课程编号“001”课程低的所有同学的学号、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、查询所有课程成绩小于60分的同学的学号、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、查询没有学全所有课的同学的学号、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、查询至少有一门课与学号为“1001”的同学所学相同的同学的学号和姓名;
select S#,Sname from Student,SC where Student.S#=SC.S# and C# in select C# from SC where S#='1001';
12、查询至少学过学号为“001”同学所有一门课的其他同学学号和姓名;
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“叶平”老师教的课的成绩都更改为此课程的平均成绩;
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='叶平');
14、查询和“1002”号的同学学习的课程完全相同的其他同学学号和姓名;
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、删除学习“叶平”老师课的SC表记录;
Delect SC
from course ,Teacher
where Course.C#=SC.C# and Course.T#= Teacher.T# and Tname='叶平';
16、向SC表中插入一些记录，这些记录要求符合以下条件：没有上过编号“003”课程的同学学号、2、
号课的平均成绩;
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、按平均成绩从高到低显示所有学生的“数据库”、“企业管理”、“英语”三门的课程成绩，按如下形式显示： 学生ID,,数据库,企业管理,英语,有效课程数,有效平均分
SELECT S# as 学生ID
,(SELECT score FROM SC WHERE SC.S#=t.S# AND C#='004') AS 数据库
,(SELECT score FROM SC WHERE SC.S#=t.S# AND C#='001') AS 企业管理
,(SELECT score FROM SC WHERE SC.S#=t.S# AND C#='006') AS 英语
,COUNT(*) AS 有效课程数, AVG(t.score) AS 平均成绩
FROM SC AS t
GROUP BY S#
ORDER BY avg(t.score)
18、查询各科成绩最高和最低的分：以如下形式显示：课程ID，最高分，最低分
SELECT L.C# As 课程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、按各科平均成绩从低到高和及格率的百分数从高到低顺序
SELECT t.C# AS 课程号,max(course.Cname)AS 课程名,isnull(AVG(score),0) AS 平均成绩
,100 * SUM(CASE WHEN isnull(score,0)>=60 THEN 1 ELSE 0 END)/COUNT(*) AS 及格百分数
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、查询如下课程平均成绩和及格率的百分数(用"1行"显示): 企业管理(001)，马克思(002)，OO&UML (003)，数据库(004)
SELECT SUM(CASE WHEN C# ='001' THEN score ELSE 0 END)/SUM(CASE C# WHEN '001' THEN 1 ELSE 0 END) AS 企业管理平均分
,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 企业管理及格百分数
,SUM(CASE WHEN C# = '002' THEN score ELSE 0 END)/SUM(CASE C# WHEN '002' THEN 1 ELSE 0 END) AS 马克思平均分
,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 马克思及格百分数
,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及格百分数
,SUM(CASE WHEN C# = '004' THEN score ELSE 0 END)/SUM(CASE C# WHEN '004' THEN 1 ELSE 0 END) AS 数据库平均分
,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 数据库及格百分数
FROM SC
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