50道SQL经典面试题(上)

50道SQL经典面试题(上)

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

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