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近在收集SQL每日一题时，找到这套比较经典的SQL面试题。
下面是我根据题目重新整理的内容，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），可能不容易消化理解，于是将内容分为上下两篇，希望对大家有所帮助。
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
--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.score2、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.SID3、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.SID4、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.SID6、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.SID8、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.SID9、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.SID16. 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 DESC17. 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 DESC18. 查询各科成绩高分、低分和平均分：
以如下形式显示：课程 ID，课程 name，高分，低分，平均分，及格率，中等率，优良率，率
及格为>=60，中等为：70-80，优良为：80-90，为：>=90
--SQL Server的解法
SELECT DISTINCT A.CID,Cname,高分,低分,平均分,及格率,中等率,优良率,率 FROM SC A
LEFT JOIN Course on A.CID=Course.CID
LEFT JOIN (SELECT CID,MAX(score)高分,MIN(score)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)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 SC20 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 SC21. 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
)A22 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
)A23. 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.CID24. 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<=325. 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
