2019秋招复习笔记--数据库基本操作

转自知乎，链接：https://zhuanlan.zhihu.com/p/80039369

数据表介绍

1、学生表 Student(SId,Sname,Sage,Ssex) SId 学生编号,Sname 学生姓名,Sage 出生年月,Ssex 学生性别 2、课程表 Course(CId,Cname,TId) CId 课程编号,Cname 课程名称,TId 教师编号 3、教师表 Teacher(TId,Tname) TId 教师编号,Tname 教师姓名 4、成绩表 SC(SId,CId,score) SId 学生编号,CId 课程编号,score 分数

学生表 Student

# 创建学生表格与插入学生student信息
create table Student(
                        Sid nvarchar(10) NOT NULL,
                        Sname nvarchar(20) NOT NULL default '',
                        Sage nvarchar(10) NOT NUll default '',
                        Ssex nvarchar(10) NOT NULL default '',
                        PRIMARY KEY (Sid)
                    );
insert into Student values('01' , '赵雷' , '19900101' , '男');
insert into Student values('02' , '钱电' , '19901221' , '男');
insert into Student values('03' , '孙风' , '19901220' , '男');
insert into Student values('04' , '李云' , '19901206' , '男');
insert into Student values('05' , '周梅' , '19911201' , '女');
insert into Student values('06' , '吴兰' , '19920101' , '女');
insert into Student values('07' , '郑竹' , '19890101' , '女');
insert into Student values('09' , '张三' , '20171220' , '女');
insert into Student values('10' , '李四' , '20171225' , '女');
insert into Student values('11' , '李四' , '20120606' , '女');
insert into Student values('12' , '赵六' , '20130613' , '女');
insert into Student values('13' , '孙七' , '20140601' , '女');

科目表 Course

# 创建科目信息表格与插入科目信息
create table Course(
                      Cid nvarchar(10) NOT NULL,
                      Cname nvarchar(20) NOT NULL default '',
                      Tid nvarchar(10) NOT NULL default '',
                      primary key(Cid)
                    );
insert into Course values('01' , '语文' , '02');
insert into Course values('02' , '数学' , '01');
insert into Course values('03' , '英语' , '03');

教师表 Teacher

# 创建教师表格与插入教师信息
create table Teacher(
                      Tid nvarchar(10) NOT NULL,
                      Tname nvarchar(20) NOT NULL default '',
                      primary key(Tid)
                    );
insert into Teacher values('01' , '张三');
insert into Teacher values('02' , '李四');
insert into Teacher values('03' , '王五');

成绩表 SC

# 创建成绩表格与插入成绩信息
create table SC(
                    Sid nvarchar(10) NOT NULL,
                    Cid nvarchar(10) NOT NULL default '',
                    score decimal(3,1) NOT NULL default 0,
                    primary key(Sid, Cid),
                    foreign key(Sid) references student(Sid),
                    foreign key(Cid) references course(Cid)
                );
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);

问题及答案

1-5

1、查询" 01 "课程比" 02 "课程成绩高的学生的信息及课程分数

SELECT s.*, a.cid, a.score, b.cid, b.score
FROM student s JOIN sc a ON s.sid = a.sid and a.cid = '01'
               JOIN sc b ON s.sid = b.sid and b.cid = '02'
WHERE a.score > b.score;

​ 1.1 查询同时存在" 01 "课程和" 02 "课程的情况

SELECT *
FROM student s JOIN sc a ON s.sid = a.sid and a.cid = '01'
           JOIN sc b ON s.sid = b.sid and b.cid = '02';

​ 1.2 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 sc a LEFT JOIN sc b ON (a.sid = b.sid and b.cid = '02')
WHERE a.cid = '01';

​ 1.3 查询不存在" 01 "课程但存在" 02 "课程的情况

SELECT a.sid sid, a.cid a_cid, a.score a_score, b.cid b_cid, b.score b_score
FROM sc a LEFT JOIN sc b ON a.sid = b.sid and b.cid = '01'
WHERE a.cid = '02' 
and b.cid is null;

2、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 s.*, sc.cid, avg(sc.score) mean
FROM student s JOIN sc ON s.sid = sc.sid
GROUP BY s.sid 
HAVING mean >= 60;

3、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);

4、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 s.sid, s.sname, count(sc.cid), sum(sc.score)
FROM student s LEFT JOIN sc ON s.sid = sc.sid
GROUP BY s.sid, s.sname;

​ 4.1 查有成绩的学生信息

SELECT DISTINCT s.* 
FROM student s JOIN sc ON s.sid = sc.sid;


#     4.1延伸-查没有成绩的学生信息
SELECT DISTINCT s.*
FROM student s LEFT JOIN sc ON s.sid = sc.sid
WHERE sc.score IS NULL;

5、 查询「李」姓老师的数量

SELECT COUNT(*) FROM teacher
WHERE tname LIKE '李%';

6-10

6、 查询学过「张三」老师授课的同学的信息

SELECT s.* 
FROM student s JOIN sc ON s.sid = sc.sid
           JOIN course c ON sc.cid = c.cid
               JOIN teacher t ON c.tid = t.tid
WHERE t.tname = '张三';

7、 查询没有学全所有课程的同学的信息

SELECT s.*
FROM student s LEFT JOIN sc ON s.sid = sc.sid
           LEFT JOIN course c ON sc.cid = c.cid
GROUP BY s.sid
HAVING COUNT(s.sid) < (SELECT count(*) FROM course);

8、 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 DISTINCT s.*
FROM student s JOIN sc ON s.sid = sc.sid
WHERE sc.cid IN (SELECT cid FROM sc
         WHERE sid = '01');

9、 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 s.*
FROM sc a JOIN sc b ON a.cid = b.cid and a.sid = '01' and a.sid != b.sid
          JOIN student s ON b.sid = s.sid
GROUP BY b.sid
HAVING COUNT(b.sid) = (SELECT count(*) FROM sc
               WHERE sid = '01');

10、 查询没学过"张三"老师讲授的任一门课程的学生姓名

SELECT DISTINCT s.sid, s.sname
FROM student s LEFT JOIN sc ON s.sid = sc.sid
WHERE s.sid NOT IN (SELECT sc.sid FROM course c JOIN teacher t ON c.Tid = t.Tid
                        JOIN sc ON sc.cid = c.Cid
            WHERE Tname = '张三');

11-15

11、 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 s.sid, s.sname, ROUND(avg(sc.score),2) mean_score
FROM student s JOIN sc ON s.sid = sc.sid
WHERE sc.score < 60
GROUP BY s.sid
HAVING COUNT(s.sid) >= 2;

12、 检索" 01 "课程分数小于 60，按分数降序排列的学生信息

SELECT s.*, sc.cid, sc.score 
FROM student s JOIN sc ON s.sid = sc.sid
WHERE sc.cid = '01' 
AND sc.score < 60
ORDER BY sc.score DESC;

13、 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 s.sid, s.sname, a.cid, a.score, b.cid, b.score, c.cid, c.score, d.mean_score
FROM student s LEFT JOIN sc a ON s.sid = a.sid AND a.cid = '01'
           LEFT JOIN sc b ON s.sid = b.sid AND b.cid = '02'
               LEFT JOIN sc c ON s.sid = c.sid AND c.cid = '03'
               LEFT JOIN (SELECT sid, ROUND(avg(score), 2) mean_score FROM sc
                          GROUP BY sid) d ON s.sid = d.sid
WHERE mean_score is NOT NULL
ORDER BY mean_score DESC;

14、 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
要求输出课程号和选修人数，查询结果按人数降序排列，若人数相同，按课程 号升序排列

SELECT c.cname, sc.cid, max(score) as max_score, min(score) as min_score, ROUND(avg(score), 2) as mean_score, 
        ROUND(100*sum(case WHEN score < 60 THEN 1 ELSE 0 END)/COUNT(*),2) as '不及格率(%)',
        ROUND(100*sum(case WHEN score >= 60 AND score < 70 THEN 1 ELSE 0 END)/COUNT(*),2) as '及格率(%)',
        ROUND(100*sum(case WHEN score >= 70 AND score < 80 THEN 1 ELSE 0 END)/COUNT(*),2) as '中等率(%)', 
        ROUND(100*sum(case WHEN score >= 80 AND score < 90 THEN 1 ELSE 0 END)/COUNT(*),2) as '优良率(%)', 
                ROUND(100*sum(case WHEN score >= 90 THEN 1 ELSE 0 END)/COUNT(*),2) as '优秀率(%)'
FROM sc JOIN course c ON sc.cid = c.cid
GROUP BY cid;

15、 按各科成绩进行排序，并显示排名， Score 重复时保留名次空缺

​ 15.1 按各科成绩进行排序，并显示排名， Score 重复时合并名次

#########################################################
# 1-依次排序（忽略分数相等情况，依次12345）
SELECT sc.*, 
CASE 
WHEN @curid = cid THEN @ranknum:=@ranknum+1 
WHEN @curid := cid THEN @ranknum:=1
END as ranknum,
FROM sc, (SELECT @ranknum := 0, @curid := 0) r
ORDER BY cid, score DESC;

#########################################################
# 2-出现并列情况，但排名依次排序（1223345）
SELECT sc.*,
CASE
WHEN (@curid = cid) and (@curscore = score) THEN @ranknum:=@ranknum
WHEN (@curid = cid) and (@curscore := score) THEN @ranknum:=@ranknum+1
WHEN (@curid := cid) and (@curscore := score) THEN @ranknum:=1
END as ranknum
FROM sc, (SELECT @ranknum:=0, @curid:=0, @curscore:=NULL) as r
ORDER BY cid, score DESC;

#########################################################
# 出现并列情况，但排名不连续（12245）
SELECT sc.*,
CASE 
WHEN (@curid = cid) and (@curscore = score) THEN @ranknum:=@ranknum+(@currow:=@currow+1)-@currow
WHEN (@curid = cid) and (@curscore := score) THEN @ranknum:=(@currow:=@currow+1)
WHEN (@curid := cid) and (@curscore := score) THEN @ranknum:=(@currow:=1) 
END as ranknum
FROM sc, (SELECT @ranknum:=0, @curid:=0, @curscore:=NULL, @currow:=0) as r
ORDER BY cid, score DESC;

# 另一种写法

SELECT a.*, count(b.score)+1 as ranknum
FROM sc a LEFT JOIN sc b ON a.cid = b.cid and a.score < b.score
GROUP BY a.sid, a.cid
ORDER BY a.cid, a.score DESC
;

16-20

16、 查询学生的总成绩，并进行排名，总分重复时保留名次空缺

16.1 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

# 创建一个view，后面直接引用，偷懒少写几次
CREATE VIEW total AS
SELECT sid, sum(score) as total_score 
FROM sc 
GROUP BY sid 
order by sum(score) DESC;

#########################################################
# 分别列出3种情况
# 1-没有并列【12345】
SELECT total.sid, total.total_score, @ranknum:=@ranknum+1 as ranknum
FROM (SELECT @ranknum:=0) as r, total;

#########################################################
# 2-并列【122345】
SELECT total.sid, total.total_score, 
CASE
WHEN @curscore = total.total_score THEN @ranknum
WHEN @curscore := total.total_score THEN @ranknum:=@ranknum+1
END as ranknum
FROM total, (SELECT @ranknum:=0, @curscore:=Null) as r;

#########################################################
# 3-并列【12245】
SELECT total.sid, total.total_score,
CASE
WHEN @curscore = total.total_score THEN @ranknum:=@ranknum+(@currow:=@currow+1)-@currow
WHEN @curscore := total.total_score THEN @ranknum:=(@currow:=@currow+1)
END as ranknum
FROM total, (SELECT @ranknum:=0, @curscore:=Null, @currow:=0) as r;

# 另一种写法，需要在view里增加一列辅助列
create view new_total as
SELECT * , count(sid) class FROM total GROUP BY sid;

SELECT *, count(t2.total_score)+1 as ranknum
FROM new_total t1 LEFT JOIN new_total t2 ON t1.class = t2.class and t1.total_score < t2.total_score
GROUP BY t1.sid
ORDER BY ranknum;

17、 统计各科成绩各分数段人数：课程编号，课程名称，[10085]，[8570]，[7060]，[600] 及所占百分比

SELECT sc.cid, c.Cname, 
       sum(CASE WHEN score >= 85 THEN 1 ELSE 0 END) AS  '[10085]',
       sum(CASE WHEN score >= 70 AND score < 85 THEN 1 ELSE 0 END) AS  '[8570]',
       sum(CASE WHEN score >= 60 AND score < 70 THEN 1 ELSE 0 END) AS  '[7060]',
       sum(CASE WHEN score < 60 THEN 1 ELSE 0 END) AS  '[600]',
       ROUND(100*sum(CASE WHEN score >= 85 THEN 1 ELSE 0 END)/count(*),2) AS  '[10085]%',
       ROUND(100*sum(CASE WHEN score >= 70 AND score < 85 THEN 1 ELSE 0 END)/count(*),2) AS  '[8570]%',
       ROUND(100*sum(CASE WHEN score >= 60 AND score < 70 THEN 1 ELSE 0 END)/count(*),2) AS  '[7060]%',
       ROUND(100*sum(CASE WHEN score < 60 THEN 1 ELSE 0 END)/count(*),2) AS  '[600]%'
FROM course c JOIN sc ON c.cid = sc.cid
GROUP BY c.cid;

18、 查询各科成绩前三名的记录

SELECT a.*, b.*, count(b.score)+1 as ranknum
FROM sc a LEFT JOIN sc b ON a.cid = b.cid and a.score < b.score
GROUP BY a.sid, a.cid
HAVING count(b.score)+1 <= 3
ORDER BY a.cid, a.score DESC;

19、 查询每门课程被选修的学生数

SELECT cid, count(*)
FROM sc
GROUP BY cid;

20、 查询出只选修两门课程的学生学号和姓名

SELECT s.sid, s.sname, sc.*
FROM student s JOIN sc ON s.sid = sc.sid
GROUP BY s.sid
HAVING count(s.sid) = 2;

21-25

21、 查询男生、女生人数

SELECT ssex, COUNT(sid)
 FROM student
 GROUP BY ssex;

22、 查询名字中含有「风」字的学生信息

SELECT * FROM student
 WHERE sname LIKE '%风%';

23、 查询同名同性学生名单，并统计同名人数

 SELECT sname, COUNT(sid)
 FROM student
 GROUP BY sname
 HAVING COUNT(sid) > 1;

24、 查询 1990 年出生的学生名单

SELECT sname FROM student
 WHERE year(sage) = 1990;
 
# 另一种写法
 SELECT sname FROM student
 WHERE sage LIKE '%1990%';

25、 查询每门课程的平均成绩，结果按平均成绩降序排列，平均成绩相同时，按 课程编号升序排列

SELECT cid, ROUND(avg(score), 2) as mean_score
FROM sc
GROUP BY cid
ORDER BY mean_score DESC, cid;

26-30

26、 查询平均成绩大于等于 85 的所有学生的学号、姓名和平均成绩

SELECT *, ROUND(avg(score), 2) as mean_score
FROM student s JOIN sc ON s.sid = sc.sid
GROUP BY s.sid
HAVING ROUND(avg(score), 2) >= 85;

27、 查询课程名称为「数学」，且分数低于 60 的学生姓名和分数

SELECT s.*, sc.cid, c.Cname, sc.score 
FROM student s JOIN sc ON s.sid = sc.sid
           JOIN course c ON sc.cid = c.cid
WHERE c.Cname = '数学' AND score < 60;

28、 查询所有学生的课程及分数情况

SELECT s.*, sc.cid, sc.score
FROM student s LEFT JOIN sc ON s.sid = sc.sid;

29、 查询任何一门课程成绩在 70 分以上的姓名、课程名称和分数

SELECT s.sid, s.sname, c.Cname, sc.score
FROM student s JOIN sc ON s.sid = sc.sid and sc.score > 70
               JOIN course c ON sc.cid = c.cid;

30、 查询不及格的课程

SELECT s.sid, s.sname, c.cname, sc.score
FROM student s JOIN sc ON s.sid = sc.sid and sc.score < 60
           JOIN course c ON sc.cid = c.cid;

31-35

31、 查询课程编号为 01 且课程成绩在 80 分以上的学生的学号和姓名

SELECT s.sid, s.sname
FROM student s JOIN sc ON s.sid = sc.sid and sc.score >= 80 and sc.cid = '01';

32、 求每门课程的学生人数

SELECT cid, COUNT(sid) FROM sc
GROUP BY cid;

33、 成绩不重复，查询选修「张三」老师所授课程的学生中，成绩最高的学生信 息及其成绩

SELECT s.*, sc.score
FROM student s JOIN sc ON s.sid = sc.sid
           JOIN course c ON sc.cid = c.cid
               JOIN teacher t ON c.tid = t.Tid
WHERE t.Tname = '张三'
ORDER BY sc.score DESC
LIMIT 1;

34、 成绩有重复的情况下，查询选修「张三」老师所授课程的学生中，成绩最高的学生信息及其成绩

SELECT s.*, a.*
FROM student s JOIN sc a ON s.sid = a.sid
           JOIN course c ON a.cid = c.cid
               JOIN teacher t ON c.tid = t.Tid and t.Tname = '张三'
               LEFT JOIN sc b ON a.cid = b.cid and a.score < b.score
GROUP BY s.sid
HAVING count(b.score)+1 = 1;

35、 查询不同课程成绩相同的学生的学生编号、课程编号、学生成绩

SELECT DISTINCT a.*
FROM sc a JOIN sc b ON a.score = b.score and a.cid != b.cid;

36-40

36、 查询每门功成绩最好的前两名

SELECT a.*
FROM sc a LEFT JOIN sc b ON a.cid = b.cid and a.score < b.score
GROUP BY a.sid, a.cid
HAVING count(b.score)+1 < 3
ORDER BY a.cid, a.sid;

37、 统计每门课程的学生选修人数（超过 5 人的课程才统计）

SELECT cid, COUNT(sid) FROM sc
GROUP BY cid
HAVING COUNT(sid) > 5;

38、 检索至少选修两门课程的学生学号

SELECT sid , count(*)
FROM sc
GROUP BY sid
HAVING count(*) >= 2;

39、 查询选修了全部课程的学生信息

SELECT s.*,count(*)
FROM student s JOIN sc ON s.sid = sc.sid
GROUP BY s.sid
HAVING count(*) = (select COUNT(*) FROM course);

40、 查询各学生的年龄，只按年份来算

SELECT sname, YEAR(curdate())-YEAR(sage) age 
FROM student;

41-45

41、 按照出生日期来算，当前月日 < 出生年月的月日则，年龄减一

SELECT sname, sage, curdate(), 
CASE
WHEN ((month(sage) = month(curdate()) and day(sage) < day(curdate())) or month(sage) < month(curdate())) THEN year(curdate())-year(sage)
WHEN ((month(sage) = month(curdate()) and day(sage) > day(curdate())) or month(sage) > month(curdate())) THEN year(curdate())-year(sage)-1
END age
FROM student;

42、 查询本周过生日的学生

SELECT * FROM student
WHERE weekofyear(sage) = weekofyear(curdate()); 

43、 查询下周过生日的学生

SELECT * FROM student
WHERE weekofyear(sage) = weekofyear(curdate())+1; 

44、 查询本月过生日的学生

SELECT * FROM student 
WHERE month(sage) = month(curdate());

45、查询下月过生日的学生

SELECT * FROM student 
WHERE month(sage) = month(curdate())+1;