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  • 比较经典的SQL面试题

    最近在收集SQL每日一题时,找到这套比较经典的SQL面试题。

    我根据题目重新梳理了一遍,包括表结构,表之间的关系,测试数据,题目,参考答案等。其中大部分参考答案在各种数据库平台上通用。

    由于题目数量较多(足足50道题),小伙伴们可能不容易消化理解,于是将内容分为上下两篇,希望对你有所帮助。

    一、表结构
    1、学生表
    Student(Sid,Sname,Sage,Ssex)
    学生编号,学生姓名,出生年月,学生性别

    2、课程表
    Course(Cid,Cname,Tid)
    课程编号,课程名称,教师编号

    3、教师表
    Teacher(Tid,Tname)
    教师编号,教师姓名

    4、成绩表
    SC(Sid,Cid,Score)
    学生编号,课程编号,分数

    二、表之间的关系
    四张表之间的关系如下图:

    我们来解读一下上面的关系:
    1、课程表Course的课程编号(Cid)作为主键,在成绩表(SC)中可以看到一个或多个学生的课程分数,两表之间是属于1:n的关系。同理学生表(Student)与成绩表(SC)也是1:n的关系
    2、教师表Teacher的教师编号(Tid)作为主键,在课程表(Course)中可以带一门或多门课程,两表之间也是属于1:n的关系。

    三、测试数据
    1、学生表
    --建表语句

    CREATE TABLE Student (
    SID VARCHAR (10),
    Sname nvarchar (10),
    Sage datetime,
    Ssex nvarchar (10)
    )

    --插入测试数据

    INSERT INTO Student VALUES('01' , N'赵雷' , '1990-01-01' , N'')
    INSERT INTO Student VALUES('02' , N'钱电' , '1990-12-21' , N'')
    INSERT INTO Student VALUES('03' , N'孙风' , '1990-05-20' , N'')
    INSERT INTO Student VALUES('04' , N'李云' , '1990-08-06' , N'')
    INSERT INTO Student VALUES('05' , N'周梅' , '1991-12-01' , N'')
    INSERT INTO Student VALUES('06' , N'吴兰' , '1992-03-01' , N'')
    INSERT INTO Student VALUES('07' , N'郑竹' , '1989-07-01' , N'')
    INSERT INTO Student VALUES('08' , N'王菊' , '1990-01-20' , N'')

    结果如下:

     2、课程表

    --建表语句

    CREATE TABLE Course (
    CID VARCHAR (10),
    Cname nvarchar (10),
    TID VARCHAR (10)
    )

     --插入测试数据

    INSERT INTO Course VALUES('01' , N'语文' , '02')
    INSERT INTO Course VALUES('02' , N'数学' , '01')
    INSERT INTO Course VALUES('03' , N'英语' , '03')

     结果如下:

     3、教师表

    --建表语句
     CREATE TABLE Teacher ( TID VARCHAR (10), Tname nvarchar (10) ) 
    --插入测试数据

    INSERT INTO Teacher VALUES('01' , N'张三')
    INSERT INTO Teacher VALUES('02' , N'李四')
    INSERT INTO Teacher VALUES('03' , N'王五')

     结果如下:

    4、成绩表
    --建表语句

    CREATE TABLE SC (
    SID VARCHAR (10),
    CID VARCHAR (10),
    score DECIMAL (18, 1)
    )

    --插入测试数据

    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)

    结果如下:

     四、面试题及参考答案

    1、查询" 01 "课程比" 02"课程成绩高的学生的信息及课程分数
    --方法一

    SELECT a.*, b.score
    FROM Student a
    JOIN SC b ON a.SID = b.SID
    JOIN sc c ON b.SID = c.SID
    WHERE b.Cid = '01'
    AND c.Cid = '02'
    AND b.Score > c.Score

     --方法二

    SELECT A.*, B.score
    FROM Student A
    JOIN
    (SELECT * FROM SC WHERE CID = '01') B ON A.SID = B.SID
    JOIN 
    (SELECT * FROM SC WHERE CID = '02') C ON C.SID = B.SID
    WHERE B.score > C.score

    2、查询同时存在" 01 "课程和"02 "课程的情况
    --方法一

    SELECT
    A.*
    FROM Student A
    JOIN SC B ON A.SID=B.SID
    JOIN SC C ON C.SID=B.SID
    WHERE B.CID='01'
    AND C.CID='02'

    --方法二

    SELECT
    A.*
    FROM Student A
    JOIN (SELECT * FROM SC WHERE CID = '01') B ON A.SID=B.SID
    JOIN (SELECT * FROM SC WHERE CID = '02') C ON B.SID = C.SID

    3、查询存在" 01 "课程但可能不存在"02 "课程的情况(不存在时显示为 null )

    SELECT
    *
    FROM
    (SELECT * FROM SC WHERE CID = '01') A
    LEFT JOIN 
    (SELECT * FROM SC WHERE CID = '02') B ON A.SID = B.SID

    4、查询不存在" 01 "课程但存在"02 "课程的情况

    SELECT *
    FROM SC
    WHERE CID = '02'
    AND SID NOT IN (
    SELECT SID FROM SC
    WHERE CID = '01'
    )

    5、查询平均成绩大于等于 60 分的同学的学生编号和学生姓名和平均成绩

    SELECT
    A.SID,
    B.Sname,
    A.dc
    FROM
    (
    SELECT SID,AVG (score) dc
    FROM SC GROUP BY SID
    HAVING AVG(score)>=60
    ) A
    JOIN Student B ON A.SID = B.SID

    6、查询在 SC 表存在成绩的学生信息
     SELECT * FROM Student WHERE SID IN (SELECT DISTINCT SID FROM SC) 

    7、查询所有同学的学生编号、学生姓名、选课总数、所有课程的总成绩(没成绩的显示为 null )

    SELECT
    A.SID,
    A.Sname,
    B.Cnt,
    B.Total
    FROM Student A
    LEFT JOIN 
    (
    SELECT
    SID,
    COUNT (CID) Cnt,
    SUM (score) Total
    FROM SC
    GROUP BY SID
    ) B ON A.SID = B.SID

    8、查有成绩的学生信息

    SELECT
    A.SID,
    A.Sname,
    B.Cnt,
    B.Total
    FROM Student A
    RIGHT JOIN 
    (
    SELECT
    SID,
    COUNT (CID) Cnt,
    SUM (score) Total
    FROM SC
    GROUP BY SID
    ) B ON A.SID = B.SID

    9、查询「李」姓老师的数量

    SELECT
    COUNT (*) 李姓老师数量
    FROM Teacher
    WHERE Tname LIKE '李%'

    10、查询学过「张三」老师授课的同学的信息

    SELECT * FROM Student
    WHERE SID IN 
    (select DISTINCT SID FROM SC a
    JOIN Course b ON a.cid=b.cid
    JOIN Teacher c ON b.Tid=c.Tid
    WHERE c.Tname='张三')

    11. 查询没有学全所有课程的同学的信息

    SELECT * FROM Student
    WHERE SID IN 
    (
    SELECT SID FROM SC
    GROUP BY SID
    HAVING COUNT (CID) < 3
    )

    12. 查询至少有一门课与学号为" 01 "的同学所学相同的同学的信息

    SELECT *
    FROM Student
    WHERE SID IN (
    SELECT DISTINCT SID
    FROM SC
    WHERE CID IN 
    (
    SELECT CID
    FROM SC
    WHERE SID = '01'
    )
    )

    13. 查询和" 01 "号的同学学习的课程完全相同的其他同学的信息

    SELECT * FROM Student
    WHERE SID in (
    SELECT SID FROM SC WHERE CID in 
    (SELECT DISTINCT CID FROM SC WHERE SID='01') and SID<>'01'
    GROUP BY SID
    having COUNT(CID) =3)

    14. 查询没学过"张三"老师讲授的任一门课程的学生姓名

    SELECT * FROM Student
    WHERE SID NOT IN 
    (select DISTINCT SID FROM SC a
    JOIN Course b ON a.cid=b.cid
    JOIN Teacher c ON b.Tid=c.Tid
    WHERE c.Tname='张三')

     

    15. 查询两门及其以上不及格课程的同学的学号,姓名及其平均成绩

    SELECT A.SID,A.Sname,B.平均成绩
    FROM Student A
    RIGHT JOIN
    (SELECT SID,AVG(score)平均成绩 FROM SC
    WHERE score<60 
    GROUP BY SID 
    HAVING COUNT(score)>=2
    )B
    on A.SID=B.SID

    16. 检索" 01 "课程分数小于 60,按分数降序排列的学生信息

    SELECT A.*,B.score FROM 
    Student A
    JOIN SC B ON A.SID=B.SID
    WHERE CID='01' AND Score<60 
    ORDER BY score DESC

    17. 按平均成绩从高到低显示所有学生的所有课程的成绩以及平均成绩

    SELECT SID,
    MAX(case CID when '01' then score else 0 end) '01',
    MAX(case CID when '02' then score else 0 end)'02',
    MAX(case CID when '03' then score else 0 end)'03',
    AVG(score)平均分 FROM SC
    GROUP BY SID ORDER BY 平均分 DESC

    18. 查询各科成绩最高分、最低分和平均分:

    以如下形式显示:课程 ID,课程 name,最高分,最低分,平均分,及格率,中等率,优良率,优秀率
    及格为>=60,中等为:70-80,优良为:80-90,优秀为:>=90
    --SQL Server的解法

    SELECT DISTINCT A.CID,Cname,最高分,最低分,平均分,及格率,中等率,优良率,优秀率 FROM SC A
    LEFT JOIN Course on A.CID=Course.CID
    LEFT JOIN (SELECT CID,MAX(score)最高分,MIN(score)最低分,AVG(score)平均分 FROM SC GROUP BY CID)B on A.CID=B.CID
    LEFT JOIN (SELECT CID,(convert(decimal(5,2),(sum(case when score>=60 then 1 else 0 end)*1.00)/COUNT(*))*100)及格率 FROM SC GROUP BY CID)C on A.CID=C.CID
    LEFT JOIN (SELECT CID,(convert(decimal(5,2),(sum(case when score >=70 and score<80 then 1 else 0 end)*1.00)/COUNT(*))*100)中等率 FROM SC GROUP BY CID)D on A.CID=D.CID
    LEFT JOIN (SELECT CID,(convert(decimal(5,2),(sum(case when score >=80 and score<90 then 1 else 0 end)*1.00)/COUNT(*))*100)优良率 FROM SC GROUP BY CID)E on A.CID=E.CID
    LEFT JOIN (SELECT CID,(convert(decimal(5,2),(sum(case when score >=90 then 1 else 0 end)*1.00)/COUNT(*))*100)优秀率
    FROM SC GROUP BY CID)F on A.CID=F.CID

    (提示:可以左右滑动代码)

    19. 按各科成绩进行排序,并显示排名, Score 重复时保留名次空缺
     SELECT *,RANK()over(order by score desc) 排名 FROM SC 

    20 按各科成绩进行排序,并显示排名, Score 重复时合并名次
     SELECT *,DENSE_RANK()over(order by score desc) 排名 FROM SC 

    21. 查询学生的总成绩,并进行排名,总分重复时保留名次空缺

    SELECT *,RANK()over(order by 总成绩 desc) 排名
    FROM(
    SELECT SID,SUM(score) 总成绩 FROM SC GROUP BY SID
    )A

    22 查询学生的总成绩,并进行排名,总分重复时不保留名次空缺

    SELECT *,DENSE_RANK()over(order by 总成绩 desc) 排名
    FROM(
    SELECT SID,SUM(score)总成绩 FROM SC GROUP BY SID
    )A

    23. 统计各科成绩各分数段人数:课程编号,课程名称,[100-85],[85-70],[70-60],[60-0] 及所占百分比

    SELECT DISTINCT A.CID,B.Cname,C.[100-85],C.所占百分比,D.[85-70],D.所占百分比,E.[70-60],E.所占百分比,F.[60-0],F.所占百分比
    FROM SC A
    LEFT JOIN Course B ON A.CID=B.CID
    LEFT JOIN (SELECT CID,sum(case when score>85 and score<=100 then 1 else null end) [100-85],
    convert(decimal(5,2),(sum(case when score>85 and score<100 then 1 else null end))*1.00/COUNT(*))*100 所占百分比 FROM SC GROUP BY CID)C on A.CID=C.CID
    LEFT JOIN (SELECT CID,sum(case when score>70 and score<=85 then 1 else null end)[85-70],
    convert(decimal(5,2),(sum(case when score>70 and score<=85 then 1 else null end))*1.00/COUNT(*))*100 所占百分比 FROM SC GROUP BY CID)D on A.CID=D.CID
    LEFT JOIN (SELECT CID,sum(case when score>60 and score<=70 then 1 else null end)[70-60],
    convert(decimal(5,2),(sum(case when score>60 and score<=70 then 1 else null end))*1.00/COUNT(*))*100 所占百分比 FROM SC GROUP BY CID)E on A.CID=E.CID
    LEFT JOIN (SELECT CID,sum(case when score>0 and score<=60 then 1 else null end)[60-0],
    convert(decimal(5,2),(sum(case when score>0 and score<=60 then 1 else null end))*1.00/COUNT(*))*100 所占百分比 FROM SC GROUP BY CID)F on A.CID=F.CID

     24. 查询各科成绩前三名的记录

    SELECT * FROM 
    (SELECT *,rank()over (partition by CID order by score desc) A
    FROM SC)B
    WHERE B.A<=3

     25. 查询每门课程被选修的学生数

    SELECT CID,COUNT(SID)学生数 FROM SC GROUP BY CID

    以上就是这次分享内容,如有不明白的地方,欢迎在底下留言讨论。

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  • 原文地址:https://www.cnblogs.com/jameswohu/p/13161541.html
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