SQL查询面试题与答案

请教一下,SQL查询面试题与答案
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书生剑主

2024-05-19 01:56:50

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。下面是我搜集的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.胜,M.负 from (

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

  (select rq,负=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

;