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