一、表关系
先创建如下表，并创建相关约束
1. 班级表class
create table class ( cid int primary key auto_increment, caption char(10), grade_id int );
insert into class values (1,'少一一班',1), (2,'少二一班',2), (3,'少三二班',3), (4,'少四一班',4), (5,'少五三班',5);
2. 学生表student
create table student ( sid int primary key auto_increment, sname char(10), gender enum('男','女') not null, class_id int );
insert into student values (1,'乔丹','女',1), (2,'艾弗森','女',1), (3,'科比','男',2), (4,'葫芦娃','男',3), (5,'张三丰','男',5), (6,'洞房不败','男',4), (7,'樱木花道','男',2), (8,'松岛菜菜子','女',3), (9,'洞房不败','女',5);
3. 老师表teacher
create table teacher ( tid int primary key auto_increment, tname char(10) );
一、表关系请创建如下表，并创建相关约束1. 班级表class【创建表语句】create table class(cid int primary key auto_increment,caption char(10),grade_id int);
【插入记录语句】insert into class values(1,'少一一班',1),(2,'少二一班',2),(3,'少三二班',3),(4,'少四一班',4),(5,'少五三班',5);
2. 学生表student【创建表语句】create table student(sid int primary key auto_increment,sname char(10),gender enum('男','女') not null,class_id int);
【插入记录语句】insert into student values(1,'乔丹','女',1),(2,'艾弗森','女',1),(3,'科比','男',2),(4,'葫芦娃','男',3),(5,'张三丰','男',5),(6,'洞房不败','男',4),(7,'樱木花道','男',2),(8,'松岛菜菜子','女',3),(9,'洞房不败','女',5);
3. 老师表teacher【创建表语句】create table teacher(tid int primary key auto_increment,tname char(10));
【插入记录语句】Insert into teacher values(1,'张三'),(2,'李四'),(3,'王五'),(4,'萧峰'),(5,'一休哥'),(6,'诸葛'),(7,'李四');
4. 课程表course【创建表语句】create table course(cid int primary key auto_increment,cname char(10),teacher_id int);
【插入记录语句】insert into course values(1,'生物',1),(2,'体育',1),(3,'物理',2),(4,'数学',3),(5,'语文',4),(6,'英语',2),(7,'土遁?沙地送葬',5),(8,'夏日喂蚊子大法',3),(9,'麻将牌九扑克千术',6);
5. 成绩表score【创建表语句】create table score(sid int primary key auto_increment,student_id int,course_id int,score int);
【插入记录语句】insert score values(1,1,1,60),(2,1,2,21),(3,2,2,99),(4,3,3,56),(5,4,1,56),(6,5,3,94),(7,5,4,40),(8,6,4,80),(9,7,3,37),(10,8,5,100),(11,8,6,89),(12,8,7,0),(13,3,8,45),(14,7,1,89),(15,2,7,89),(16,2,1,61);
6. 年级表class_grade【创建表语句】create table class_grade(gid int primary key auto_increment,gname char(10));
【插入记录语句】insert class_grade values(1,'一年级'),(2,'二年级'),(3,'三年级'),(4,'四年级'),(5,'五年级');
7. 班级任职表teach2cls【创建表语句】create table teach2cls(tcid int primary key auto_increment,tid int,cid int);
【插入记录语句】insert into teach2cls values(1,1,1),(2,1,2),(3,2,1),(4,3,2),(5,4,5),(6,5,3),(7,5,5),(8,6,2),(9,6,4),(10,6,3),(11,4,1),(12,1,4);
二、操作表★注：由于样本数量有限，为了能够得到足够的查询结果，所有题目中涉及到“超过”或“以上”字样的，均默认为包含该值（例如：查询教授课程超过2门的老师的id和姓名，视作教授课程数>=2）
1、自行创建测试数据；（创建语句见"一、表关系"）
2、查询学生总人数；select count(*) as 学生总人数 from student;
3、查询“生物”课程和“物理”课程成绩都及格的学生id和姓名；【查法1——子查询】select sid, snamefrom student where sid in(select student_idfrom scorewhere student_id in(select student_idfrom scorewhere course_id = (select cid from course where cname = '生物') and score >= 60)and course_id = (select cid from course where cname = '物理') and score >= 60);
【查法2——联表】select sid, sname from studentwhere sid in (select t1.student_id from (select student_id from scorewhere course_id = (select cid from course where cname = '生物') and score >= 60) as t1inner join (select student_id from score where course_id = (select cid from course where cname = '物理') and score >= 60) as t2on t1.student_id=t2.student_id);
4、查询每个年级的班级数，取出班级数最多的前三个年级select class.grade_id, class_grade.gname, count(class.cid) as 班级数from class inner join class_grade on class.grade_id=class_grade.gidgroup by class.grade_idorder by count(class.cid) desclimit 3;
5、查询平均成绩最高和最低的学生的id和姓名以及平均成绩select stu.sid, stu.sname, avg(score) as 平均成绩from student as stu inner join score as scoon stu.sid = sco.student_idgroup by stu.sidhaving avg(score) = (select avg(score) from score group by student_idorder by avg(score) desclimit 1) or avg(score) = (select avg(score) from score group by student_idorder by avg(score) asclimit 1);
6、查询每个年级的学生人数；select t1.gname, count(s.sid) as 学生人数from (select * from class as c inner join class_grade as g on c.grade_id = g.gid) as t1inner join student as s on t1.cid = s.class_idgroup by t1.gid;
7、查询每位学生的学号，姓名，选课数，平均成绩；select stu.sid as 学号,stu.sname as 姓名,count(sco.course_id) as 选课数,avg(sco.score) as 平均成绩from student as stu left join score as sco on stu.sid = sco.student_idgroup by sco.student_id;
8、查询学生编号为“2”的学生的姓名、该学生成绩最高的课程名、成绩最低的课程名及分数；select t1.sname as 姓名,t2.cname as 课程名,t1.score as 分数from (select stu.sid, stu.sname, sco.course_id, sco.score from student as stu inner join score as sco on stu.sid = sco.student_id where stu.sid=2) as t1inner joincourse as t2 on t1.course_id = t2.cidgroup by t2.cidhaving score in (max(score),min(score));
9、查询姓“李”的老师的个数和所带班级数；select count(te.tid) as 姓李老师个数,count(tc.cid) as 所带班级数from teacher as te inner join teach2cls as tcon te.tid = tc.tidwhere te.tname regexp "^李.*"group by te.tid;
10、查询班级数小于5的年级id和年级名；select c.grade_id as 年级id,g.gname as 年级名from class as c inner join class_grade as gon c.grade_id = g.gidgroup by c.grade_idhaving count(c.cid)<5;
11、查询班级信息，包括班级id、班级名称、年级、年级级别(12为低年级，34为中年级，56为高年级)，示例结果如下；select cid as 班级id,caption as 班级名称,gname as 年级,casewhen g.gid in (1,2) then '低年级'when g.gid in (3,4) then '中年级'when g.gid in (5,6) then '高年级'else '其他' end as 年级级别from class as c inner join class_grade as gon c.grade_id = g.gid;
12、查询学过“张三”老师2门课以上的同学的学号、姓名；select stu.sid as 学号,stu.sname as 姓名from student as stu inner join score as sco on stu.sid = sco.student_idwhere sco.course_id in (select c.cidfrom teacher as t inner join course as con t.tid = c.teacher_idwhere t.tname = '张三')group by stu.sidhaving count(sco.course_id) >= 2;
13、查询教授课程超过2门的老师的id和姓名；selecttid as id,tname as 姓名from teacher as t inner join course as c on t.tid = c.teacher_idgroup by c.teacher_idhaving count(c.cid) >= 2;
14、查询学过编号“1”课程和编号“2”课程的同学的学号、姓名；select sid as 学号,sname as 姓名from studentwhere sid in (select student_id from scorewhere student_id in (select student_id from scorewhere course_id = 1)and course_id = 2);
15、查询没有带过高年级的老师id和姓名；select tid as 老师id,tname as 姓名from teacherwhere tid not in (select tc.tidfrom class as c inner join teach2cls as tc on c.cid = tc.cidwhere c.grade_id in (5,6));
16、查询学过“张三”老师所教的所有课的同学的学号、姓名；select distinctstu.sid as 学号,stu.sname as 姓名from student as stu inner join score as sco on stu.sid = sco.student_idwhere sco.course_id in (select c.cid from teacher as t inner join course as c on t.tid = c.teacher_idwhere t.tname = "张三");
17、查询带过超过2个班级的老师的id和姓名；select tid as id,tname as 姓名from teacherwhere tid in (select tid from teach2clsgroup by tidhaving count(cid) >= 2);
18、查询课程编号“2”的成绩比课程编号“1”课程低的所有同学的学号、姓名；select sid as 学号,sname as 姓名from studentwhere sid in(select t1.student_idfrom (select * from scorewhere course_id = 1) as t1inner join (select * from score where course_id = 2) as t2on t1.student_id = t2.student_idwhere t1.score > t2.score);
19、查询所带班级数最多的老师id和姓名；select tid as id,tname as 姓名from teacher where tid in (select tidfrom teach2clsgroup by tidhaving count(cid) = (select count(cid)from teach2clsgroup by tidorder by count(cid) desclimit 1));
20、查询有课程成绩小于60分的同学的学号、姓名；select sid as 学号,sname as 姓名from studentwhere sid in (select student_idfrom score where score < 60);
21、查询没有学全所有课的同学的学号、姓名；select sid as 学号,sname as 姓名from studentwhere sid in (select student_idfrom scoregroup by student_idhaving count(course_id) != (select count(cid) from course));
22、查询至少有一门课与学号为“1”的同学所学相同的同学的学号和姓名；select sid as 学号,sname as 姓名from studentwhere sid in (select student_idfrom scorewhere course_id in (select course_id from scorewhere student_id = 1));
23、查询至少学过学号为“1”同学所选课程中任意一门课的其他同学学号和姓名；select sid as 学号,sname as 姓名from studentwhere sid in (select student_idfrom scorewhere course_id in (select course_id from scorewhere student_id = 1) and student_id != 1);
24、查询和“2”号同学学习的课程完全相同的其他同学的学号和姓名；select sid as 学号,sname as 姓名from studentwhere sid in (select student_idfrom scorewhere student_id != 2group by student_idhaving group_concat(course_id order by course_id asc) = (select group_concat(course_id order by course_id asc)from scorewhere student_id = 2group by student_id));
25、删除学习“张三”老师课的score表记录；delete from scorewhere course_id in (select c.cid from teacher as t inner join course as con t.tid = c.teacher_idwhere t.tname = '张三');
26、向score表中插入一些记录，这些记录要求符合以下条件：①没有上过编号“2”课程的同学学号；②插入“2”号课程的平均成绩；【插入第一条】insert into score(student_id, course_id, score) values((select sidfrom studentwhere sid not in(select s.student_idfrom score as swhere s.course_id = 2)order by sid desclimit 0,1),2,(select avg(s.score)from score as swhere s.course_id = 2));【插入第二条】insert into score(student_id, course_id, score) values((select sidfrom studentwhere sid not in(select s.student_idfrom score as swhere s.course_id = 2)order by sid desclimit 1,1),2,(select avg(s.score)from score as swhere s.course_id = 2));
【改limit后的第一个参数值，可继续插入第三、四、...条】
27、按平均成绩从低到高显示所有学生的“语文”、“数学”、“英语”三门的课程成绩，按如下形式显示： 学生ID,语文,数学,英语,有效课程数,有效平均分；【这里题目有歧义：有效课程数和有效平均分是仅以这3门课来统计，还是以学生所有科目来统计】【解一：仅以这3门课来统计】select t2.sid as 学生ID, sum(case when t1.cname = '语文' then t1.score else null end) as 语文,sum(case when t1.cname = '数学' then t1.score else null end) as 数学,sum(case when t1.cname = '英语' then t1.score else null end) as 英语,count(case when t1.cname in ('语文','数学','英语') then 1 else null end) as 有效课程数,avg(case when t1.cname in ('语文','数学','英语') then t1.score else null end) as 有效平均分from (select * from score as s inner join course as con s.course_id = c.cid) as t1right join student as t2on t1.student_id = t2.sidgroup by t2.sidorder by avg(case when t1.cname in ('语文','数学','英语') then t1.score else null end) asc;
【解二：以该学生所有科目来统计】select t2.sid as 学生ID, sum(case when t1.cname = '语文' then t1.score else null end) as 语文,sum(case when t1.cname = '数学' then t1.score else null end) as 数学,sum(case when t1.cname = '英语' then t1.score else null end) as 英语,count(t1.score) as 有效课程数,avg(t1.score) as 有效平均分from (select * from score as s inner join course as con s.course_id = c.cid) as t1right join student as t2on t1.student_id = t2.sidgroup by t2.sidorder by avg(t1.score) asc;
28、查询各科成绩最高和最低的分：以如下形式显示：课程ID，最高分，最低分；select course_id as 课程ID,max(score) as 最高分,min(score) as 最低分from scoregroup by course_id;
29、按各科平均成绩从低到高和及格率的百分数从高到低顺序；【这里优先按平均成绩从低到高排序，若遇到平均成绩相同的则按及格率百分数从高到低排序】select course_id as 课程ID,avg(score) as 平均成绩,concat(100*count(case when score>=60 then 1 else null end)/count(score),"%") as 及格率from scoregroup by course_idorder by avg(score) asc, count(case when score>=60 then 1 else null end)/count(score) desc;
30、课程平均分从高到低显示（显示任课老师）；select t1.cname as 课程名称,avg(t2.score) as 平均分,t1.tname as 任课老师from (select * from teacher as t inner join course as con t.tid = c.teacher_id) as t1 inner join score as t2on t1.cid = t2.course_idgroup by t2.course_idorder by avg(t2.score) desc;
31、查询各科成绩前三名的记录(不考虑成绩并列情况)【本题与44题类似，不会做，于是百度了下"如何在mysql中查询每个分组的前几名"，参照其中的一个方法，写出了答案】【注：这里仍然是按照score表默认的排序，即sid的排序】select*from scorewhere(selectcount(*)from score as swheres.course_id = score.course_idands.score <= score.score)<= 3;
32、查询每门课程被选修的学生数；select cname as 课程名,count(s.student_id) as 选修学生数from course as c left join score as s on c.cid = s.course_idgroup by c.cid;
33、查询选修了2门以上课程的全部学生的学号和姓名；select sid as 学号,sname as 姓名from studentwhere sid in (select student_id from scoregroup by student_idhaving count(course_id) >= 2);
34、查询男生、女生的人数，按倒序排列；select gender, count(sid)from studentgroup by genderorder by count(sid) desc;
35、查询姓“张”的学生名单；【查法1——正则】select sname from studentwhere sname regexp "^张.*";
【查法2——like】select snamefrom studentwhere sname like "张%";
36、查询同名同姓学生名单，并统计同名人数；select sname as 姓名,count(sid) as 同名人数 from studentgroup by snamehaving count(sid) > 1;
37、查询每门课程的平均成绩，结果按平均成绩升序排列，平均成绩相同时，按课程号降序排列；select avg(score),course_idfrom scoregroup by course_idorder by avg(score) asc, course_id desc;
38、查询课程名称为“数学”，且分数低于60的学生姓名和分数；select stu.sname as 学生姓名,sco.score as 分数from student as stu inner join score as scoon stu.sid = sco.student_idwhere sco.course_id = (select cid from course where cname = '数学')and sco.score < 60;
39、查询课程编号为“3”且课程成绩在80分以上的学生的学号和姓名；select sid as 学号,sname as 姓名from studentwhere sid in (select student_id from scorewhere course_id = 3 and score >= 80);
40、求选修了课程的学生人数select count(1) as 学生人数from(select distinct student_idfrom score) as t1;
41、查询选修“王五”老师所授课程的学生中，成绩最高和最低的学生姓名及其成绩；select stu.sname as 学生姓名,sco.score as 成绩from student as stu inner join score as scoon stu.sid = sco.student_idwhere score in ((select max(score)from scorewhere course_id in (select c.cidfrom teacher as t inner join course as con t.tid = c.teacher_id where t.tname = '王五')), (select min(score)from scorewhere course_id in (select c.cidfrom teacher as t inner join course as con t.tid = c.teacher_id where t.tname = '王五')));
42、查询各个课程及相应的选修人数；select cname as 课程名,count(s.student_id) as 选修学生数from course as c left join score as s on c.cid = s.course_idgroup by c.cid;
43、查询不同课程但成绩相同的学生的学号、课程号、学生成绩；select student_id as 学号,course_id as 课程号,score as 学生成绩from scoregroup by scorehaving count(student_id) > 1;
44、查询每门课程成绩最好的前两名学生id和姓名；【注：这里指定了前两名，所以若出现多名同分的学生也只取倒序排的默认前2名】【与31题类似…不会写，于是百度了下"如何在mysql中查询每个分组的前几名"，参照其中一种比较高端且高效的自定义变量的方法，写出了答案】
set @num := 0, @cname := '';selectt2.cid as 课程ID,t2.cname as 课程名,t1.sid as 学生ID,t1.sname as 学生名,t1.score as 成绩,@num := if(@cname = t2.cname, @num + 1, 1) as 排名,@cname := t2.cname as 课程名确认from (select stu.sid, stu.sname, sco.course_id, sco.score from student as stu inner join score as scoon stu.sid = sco.student_id) as t1right joincourse as t2on t1.course_id = t2.cidgroup byt2.cid, t1.score, t1.snamehaving排名 <= 2;
45、检索至少选修两门课程的学生学号；select sid as 学号from studentwhere sid in (select student_id from scoregroup by student_idhaving count(course_id) >= 2);
46、查询没有学生选修的课程的课程号和课程名；select cid as 课程号,cname as 课程名from coursewhere cid not in (select distinct course_idfrom score);
47、查询没带过任何班级的老师id和姓名；selecttid as 老师id,tname as 姓名from teacherwhere tid not in (select distinct tidfrom teach2cls);
48、查询有两门以上课程超过80分的学生id及其平均成绩；select student_id as 学生id,avg(score) as 平均成绩from scorewhere student_id in (select student_idfrom scorewhere score >= 80group by student_idhaving count(course_id) >= 2)group by student_id;
49、检索“3”课程分数小于60，按分数降序排列的同学学号；select distinctstudent_id as 学号from scorewhere course_id = 3 and score < 60order by score desc;
50、删除编号为“2”的同学的“1”课程的成绩；delete from scorewhere student_id = 2 and course_id = 1;
51、查询同时选修了物理课和生物课的学生id和姓名；select sid as 学生id,sname as 姓名from studentwhere sid in (select student_idfrom scorewhere course_id = (select cid from course where cname = '生物'))and sid in (select student_idfrom scorewhere course_id = (select cid from course where cname = '物理'));