MySQL多表查询综合练习答案

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一、综合练习

1.1 init.sql文件内容

/*
 数据导入:
 Navicat Premium Data Transfer

 Source Server         : localhost
 Source Server Type    : MysqL
 Source Server Version : 50624
 Source Host           : localhost
 Source Database       : sqlexam

 Target Server Type    : MysqL
 Target Server Version : 50624
 File Encoding         : utf-8

 Date: 10/21/2016 06:46:46 AM
*/

SET NAMES utf8;
SET FOREIGN_KEY_CHECKS = 0;

-- ----------------------------
--  Table structure for `class`
-- ----------------------------
DROP TABLE IF EXISTS `class`;
CREATE TABLE `class` (
  `cid` int(11) NOT NULL AUTO_INCREMENT,`caption` varchar(32) NOT NULL,PRIMARY KEY (`cid`)
) ENGINE=InnoDB AUTO_INCREMENT=5 DEFAULT CHARSET=utf8;

-- ----------------------------
--  Records of `class`
-- ----------------------------
BEGIN;
INSERT INTO `class` VALUES ('1','三年二班'),('2','三年三班'),('3','一年二班'),('4','二年九班');
COMMIT;

-- ----------------------------
--  Table structure for `course`
-- ----------------------------
DROP TABLE IF EXISTS `course`;
CREATE TABLE `course` (
  `cid` int(11) NOT NULL AUTO_INCREMENT,`cname` varchar(32) NOT NULL,`teacher_id` int(11) NOT NULL,PRIMARY KEY (`cid`),KEY `fk_course_teacher` (`teacher_id`),CONSTRAINT `fk_course_teacher` FOREIGN KEY (`teacher_id`) REFERENCES `teacher` (`tid`)
) ENGINE=InnoDB AUTO_INCREMENT=5 DEFAULT CHARSET=utf8;

-- ----------------------------
--  Records of `course`
-- ----------------------------
BEGIN;
INSERT INTO `course` VALUES ('1','生物','1'),'物理','2'),'体育','3'),'美术','2');
COMMIT;

-- ----------------------------
--  Table structure for `score`
-- ----------------------------
DROP TABLE IF EXISTS `score`;
CREATE TABLE `score` (
  `sid` int(11) NOT NULL AUTO_INCREMENT,`student_id` int(11) NOT NULL,`course_id` int(11) NOT NULL,`num` int(11) NOT NULL,PRIMARY KEY (`sid`),KEY `fk_score_student` (`student_id`),KEY `fk_score_course` (`course_id`),CONSTRAINT `fk_score_course` FOREIGN KEY (`course_id`) REFERENCES `course` (`cid`),CONSTRAINT `fk_score_student` FOREIGN KEY (`student_id`) REFERENCES `student` (`sid`)
) ENGINE=InnoDB AUTO_INCREMENT=53 DEFAULT CHARSET=utf8;

-- ----------------------------
--  Records of `score`
-- ----------------------------
BEGIN;
INSERT INTO `score` VALUES ('1','1','10'),'2','9'),('5','4','66'),('6','8'),('8','3','68'),('9','99'),('10','77'),('11',('12','87'),('13',('14','79'),('15','11'),('16','67'),('17','100'),('18','5',('19',('20',('21',('22','6',('23',('24',('25',('26','7',('27',('28',('29','88'),('30','8',('31',('32',('33',('34','9','91'),('35',('36',('37','22'),('38','10','90'),('39',('40','43'),('41',('42','11',('43',('44',('45',('46','12',('47',('48',('49',('52','13','87');
COMMIT;

-- ----------------------------
--  Table structure for `student`
-- ----------------------------
DROP TABLE IF EXISTS `student`;
CREATE TABLE `student` (
  `sid` int(11) NOT NULL AUTO_INCREMENT,`gender` char(1) NOT NULL,`class_id` int(11) NOT NULL,`sname` varchar(32) NOT NULL,KEY `fk_class` (`class_id`),CONSTRAINT `fk_class` FOREIGN KEY (`class_id`) REFERENCES `class` (`cid`)
) ENGINE=InnoDB AUTO_INCREMENT=17 DEFAULT CHARSET=utf8;

-- ----------------------------
--  Records of `student`
-- ----------------------------
BEGIN;
INSERT INTO `student` VALUES ('1','男','理解'),'女','钢蛋'),'张三'),'张一'),'张二'),'张四'),('7','铁锤'),'李三'),'李一'),'李二'),'李四'),'如花'),'刘三'),'刘一'),'刘二'),'刘四');
COMMIT;

-- ----------------------------
--  Table structure for `teacher`
-- ----------------------------
DROP TABLE IF EXISTS `teacher`;
CREATE TABLE `teacher` (
  `tid` int(11) NOT NULL AUTO_INCREMENT,`tname` varchar(32) NOT NULL,PRIMARY KEY (`tid`)
) ENGINE=InnoDB AUTO_INCREMENT=6 DEFAULT CHARSET=utf8;

-- ----------------------------
--  Records of `teacher`
-- ----------------------------
BEGIN;
INSERT INTO `teacher` VALUES ('1','张磊老师'),'李平老师'),'刘海燕老师'),'朱云海老师'),'李杰老师');
COMMIT;

SET FOREIGN_KEY_CHECKS = 1;

1.2 从init.sql文件中导入数据

# 准备表、记录
MysqL> create database db1;
MysqL> use db1;
MysqL> source /root/init.sql

202-MySQL多表查询-01.png

1.3 基础练习

  1. 查询男生、女生的人数;

  2. 查询姓“张”的学生名单;

  3. 课程平均分从高到低显示

  4. 查询有课程成绩小于60分的同学的学号、姓名;

  5. 查询至少有一门课与学号为1的同学所学课程相同的同学的学号和姓名;

  6. 查询出只选修了一门课程的全部学生的学号和姓名;

  7. 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;

  8. 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;

  9. 查询“生物”课程比“物理”课程成绩高的所有学生的学号;

  10. 查询平均成绩大于60分的同学的学号和平均成绩;

  11. 查询所有同学的学号、姓名、选课数、总成绩;

  12. 查询姓“李”的老师的个数;

  13. 查询没学过“张磊老师”课的同学的学号、姓名;

  14. 查询学过“1”并且也学过编号“2”课程的同学的学号、姓名;

  15. 查询学过“李平老师”所教的所有课的同学的学号、姓名;

1.4 进阶练习

  1. 查询没有学全所有课的同学的学号、姓名;
  2. 查询和“002”号的同学学习的课程完全相同的其他同学学号和姓名;
  3. 删除学习“叶平”老师课的SC表记录;
  4. 向SC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“002”课程的同学学号;②插入“002”号课程的平均成绩;
  5. 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;
  6. 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;
  7. 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;
  8. 查询各科成绩前三名的记录:(不考虑成绩并列情况)
  9. 查询每门课程被选修的学生数;
  10. 查询同名同姓学生名单,并统计同名人数;
  11. 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;
  12. 查询平均成绩大于85的所有学生的学号. 姓名和平均成绩;
  13. 查询课程名称为“数学”,且分数低于60的学生姓名和分数;
  14. 查询课程编号为003且课程成绩在80分以上的学生的学号和姓名;
  15. 求选了课程的学生人数
  16. 查询选修“杨艳”老师所授课程的学生中,成绩最高的学生姓名及其成绩;
  17. 查询各个课程及相应的选修人数;
  18. 查询不同课程但成绩相同的学生的学号、课程号、学生成绩;
  19. 查询每门课程成绩最好的前两名;
  20. 检索至少选修两门课程的学生学号;
  21. 查询全部学生都选修的课程的课程号和课程名;
  22. 查询没学过“叶平”老师讲授的任一门课程的学生姓名;
  23. 查询两门以上不及格课程的同学的学号及其平均成绩;
  24. 检索“004”课程分数小于60,按分数降序排列的同学学号;
  25. 删除“002”同学的“001”课程的成绩;

二、基础练习答案

1、查询“生物”课程比“物理”课程成绩高的所有学生的学号;

select * from( 

 (select * from score where course_id in (select cid from course where cname = '生物')) t1  

left join 

 (select * from score where course_id in (select cid from course where cname = '物理')) t2  

on  t1.student_id = t2.student_id) 

where t1.num > t2.num;

 

2、查询平均成绩大于60分的同学的学号和平均成绩;

# 先查看每个同学的平均分数

select student_id,avg(num) from score group by student_id;

# 在筛选成绩大于60分的同学的学号和平均成绩;

# select student_id,avg(num) from score group by student_id having avg(num) > 60;

 

3、查询所有同学的学号、姓名、选课数、总成绩;

# 先查看每个同学的总成绩

select student_id,sum(num) from score group by student_id;

# 学生和课程的关系只有成绩表中存在,因此要获取每个学生选择的课程,需要通过score表

select count(sid),student_id from score group by student_id;

# 将上面两步合并

select sum(num),count(sid),student_id from score group by student_id;

# 将学生的信息和成绩选课情况拼在一起

select sid,sname,sum_num,count_stu 

from student  

left join 

 (select sum(num) sum_num,count(sid) count_stu,student_id from score group by student_id) t2  

on  sid = student_id;

# 还可以更严谨,那些没有选课的同学选课数和总成绩应该是0

select sid,(

           CASE

           WHEN sum_num is  null THEN 0   

   ELSE sum_num

           END

         ) as sum_num,(

           CASE

           WHEN count_stu is  null THEN 0   

   ELSE count_stu

           END

         ) as count_stu 

from student  

left join 

 (select sum(num) sum_num,student_id from score group by student_id) t2  

on  sid = student_id;

 

4、查询姓“李”的老师的个数;

# 找到所有姓李的

 # 方法一

 # select * from teacher where tname like '李%';

 # 方法二

 # select * from teacher where tname regexp '^李';

# 统计个数

 select count(tid) from teacher where tname regexp '^李';

 或者

 select count(id) from teacher where tname like '李%';

 

5、查询没学过“张磊老师”课的同学的学号、姓名;

# 找到张磊老师的id 

select tid from teacher where tname == '张磊老师';

# 找到张磊老师所教课程

select cid from course where teacher_id = (select tid from teacher where tname = '张磊老师');

# 找到所有学习这门课的学生id

select student_id from score where course_id = (select cid from course where teacher_id = (select tid from teacher where tname = '张磊老师'));

# 找到没有学过这门课的学生对应的学生学号、姓名

select sid,sname from student where sid not in 

 (select student_id from score where course_id = (select cid from course where teacher_id = (select tid from teacher where tname = '张磊老师'))

);

 

6、查询学过“1”并且也学过编号“2”课程的同学的学号、姓名;

# 先查询学习课程id为1的所有学生

select * from score where course_id = 1;

# 先查询学习课程id为2的所有学生

select * from score where course_id = 2;

# 把这两张表按照学生的id 内连接起来 去掉只学习某一门课程的学生

select t1.student_id from

(select student_id from score where course_id = 1)  t1

inner join

(select student_id from score where course_id = 2) t2

on t1.student_id = t2.student_id

# 根据学号在学生表中找到对应的姓名

select sid,sname from student where sid in (select t1.student_id from (select student_id from score where course_id = 1)  t1 inner join (select student_id from score where course_id = 2) t2 on t1.student_id = t2.student_id);

 

7、查询学过“李平老师”所教的所有课的同学的学号、姓名;

#找到李平老师的tid

select tid from teacher where tname ='李平老师';

# 找到李平老师教的所有课程cid

 select cid from course where teacher_id in (select tid from teacher where tname ='李平老师');

# 找到李平老师教的所有课程数

 select count(cid) from course where teacher_id in (select tid from teacher where tname ='李平老师');

# 找到所有学习李平老师课程的学生

select * from score where course_id in ( select cid from course where teacher_id in (select tid from teacher where tname ='李平老师'));

# 查看所有学习李平老师课程的学生选课数

select student_id,count(course_id) from score where course_id in ( select cid from course where teacher_id in (select tid from teacher where tname ='李平老师')) group by student_id;

# 找到所有选择了李平老师所有课程的学生id

select  student_id from (

select student_id,count(course_id) course_count from score where course_id in ( select cid from course where teacher_id in (select tid from teacher where tname ='李平老师')) group by student_id) t1

where t1.course_count =

(select count(cid) from course where teacher_id in (select tid from teacher where tname ='李平老师'));

# 找到学生的其他信息

select sid,sname from student where sid in (

select  student_id from (

select student_id,count(course_id) course_count from score where course_id in ( select cid from course where teacher_id in (select tid from teacher where tname ='李平老师')) group by student_id) t1

where t1.course_count =

(select count(cid) from course where teacher_id in (select tid from teacher where tname ='李平老师'))

);

 

8、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;

# 先找到每个学生的课程编号“1”的和课程编号“2”的成绩组成一张表

select t1.student_id from (select num num2,student_id from score where course_id = 2) t2 inner join (select student_id,num num1 from score where course_id = 1) t1 on t1.student_id = t2.student_id

# 再找到课程编号“2”的成绩比课程编号“1”课程低的所有学生的学号

select t1.student_id from (select num num2,num num1 from score where course_id = 1) t1 on t1.student_id = t2.student_id where num2 < num1

# 再找到所有学生的学号、姓名

select sid,sname from student where sid in(select t1.student_id from (select num num2,num num1 from score where course_id = 1) t1 on t1.student_id = t2.student_id where num2 < num1);

 

9、查询有课程成绩小于60分的同学的学号、姓名;

# 先查询成绩小于60分的同学的学号

select distinct student_id from score where num < 60;

# 再查询有课程成绩小于60分的同学的学号、姓名

select sid,sname from student where sid in (select distinct student_id from score where num < 60);

 

10、查询至少有一门课与学号为1的同学所学课程相同的同学的学号和姓名;

# 先看看学号为1的同学都学了哪些课程

select course_id from score where student_id = 1

# 找到学习 学号为1的同学所学课程 的学号

select distinct student_id from score where course_id in (select course_id from score where student_id = 1);

#  找到学习 学号为1的同学所学课程 的学号\姓名

select sid,sname from student where sid in (select distinct student_id from score where course_id in (select course_id from score where student_id = 1));

 

11、课程平均分从高到低显示

select course_id,avg(num) avg_num from score group by course_id order by avg_num desc;

 

12、查询出只选修了一门课程的全部学生的学号和姓名;

# 查询出只选修了一门课程的全部学生的学号

select student_id,count(student_id) from score group by student_id having count(student_id) =1;

# 查询出只选修了一门课程的全部学生的学号和姓名;

select sid,sname from student where sid in (select student_id from score group by student_id having count(student_id) =1);

 

13、查询男生、女生的人数;

select gender,count(sid) from student group by gender;

 

14、查询姓“张”的学生名单;

select * from student where sname like '张%';

 

15、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;

# 查询成绩的最高分

select course_id c1,max(num) from score group by course_id

# 查询成绩的最低分

select course_id c1,min(num) from score group by course_id

# 查询成绩的最高分和最低分拼接

select * from ( (select course_id c1,max(num) from score group by course_id) t1 inner join (select course_id c2,min(num) from score group by course_id) t2 on t1.c1 = t2.c2 );

# 格式整理

select t1.c1,t1.max_num,t2.min_num from ( (select course_id c1,max(num) max_num from score group by course_id) t1 inner join (select course_id c2,min(num) min_num from score group by course_id) t2 on t1.c1 = t2.c2 );

三、进阶练习答案

1、查询没有学全所有课的同学的学号、姓名;

# 先统计一共有多少门课程

select count(cid) from course;

# 查看每个学生选择的课程书

select count(course_id) from score group by student_id;

# 查询所学课程数小于总课程数的学生学号

select student_id

from (select count(course_id) c_course_id,student_id from score group by student_id) t1 

where t1.c_course_id <  (select count(cid) from course) ;

# 查询没有学全所有课的同学的学号、姓名;

select sid,sname from student where sid in (

 select student_id from (select count(course_id) c_course_id,student_id from score group by student_id

 ) t1 where t1.c_course_id <  (select count(cid) from course)

) ;

 

2、查询和“002”号的同学学习的课程完全相同的其他同学学号和姓名;

# 先查询2号同学学了哪些课程

select * from score where student_id =2;

# 找到学习了2号同学没学习课程的所有同学(找到所有和2号同学学习的课程不一样的同学)

select student_id from score where course_id not in (select course_id from score where student_id=2)

# 找到score表中所有的学生并且把 2号同学 以及(和2号同学学习的课程不一样的同学)排除出去

select student_id from score where student_id not in (select student_id from score where course_id not in (select course_id from score where student_id=2)) and student_id !=2

# 对剩余的和2号同学所选课程没有不同的同学所选课程数进行统计,如果和2号同学的课程数相同,就是选择了相同的课程

select student_id from score where student_id not in (

 select student_id from score where course_id not in (select course_id from score where student_id=2)

 ) and student_id !=2

group by student_id 

having count(course_id)= (select count(course_id) from score where student_id=2);

 

3、删除学习“叶平”老师课的SC(score)表记录;

# 先查出李平老师的id

select tid from teacher where tname = '李平老师';

# 查看李平老师所教授的课程

select cid from course where teacher_id = (select tid from teacher where tname = '李平老师');

# 查看李平老师所教课程的成绩数据

select * from score where course_id in (select cid from course where teacher_id = (select tid from teacher where tname = '李平老师'));

# 执行删除命令

delete from score where course_id in (select cid from course where teacher_id = (select tid from teacher where tname = '李平老师'));

4、向SC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“002”课程的同学学号;②插入“002”号课程的平均成绩; 

#  先找寻上过2号课程的同学

select student_id from score where course_id = 2;

# 再找到没上过2号课程的所有同学

select * from student where sid not in (select student_id from score where course_id = 2);

#  计算出学习2号课程的同学的平均成绩

select avg(num) from score where course_id = 2 group by course_id;

# 用笛卡尔积将上述两个表拼起来

select * from (select sid from student where sid not in (select student_id from score where course_id = 2)) t1,(select avg(num) from score where course_id = 2 group by course_id) t2;

#  向SC表中插入记录

insert into score (course_id,student_id,num)   select 2,t1.sid,t2.avg_num from (select sid from student where sid not in (select student_id from score where course_id = 2)) t1,(select avg(num) avg_num from score where course_id = 2 group by course_id) t2;

 

5、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,有效平均分;

# 查看每个学生的数学成绩

select student_id,num from score where course_id = (select cid from course where cname = '数学');

#  查看每个学生的语文成绩

select student_id,num from score where course_id = (select cid from course where cname = '语文');

#  查看每个学生的英语成绩

select student_id,num from score where course_id = (select cid from course where cname = '英语');

# 查看每个学生的平均成绩

select student_id,avg(num),count(num) from score group by student_id;

# 将上面的几张表拼接起来,为了生成所有学生的信息,用student表作为左连接的第一张表

select sid 学生ID,t2.num 语文,t1.num 数学,t3.num 英语,t4.count_course 有效课程数,t4.avg_num 有效平均分 from student 

 left join (select student_id,num from score where course_id = (select cid from course where cname = '数学')) t1

 on student.sid = t1.student_id

 left join (select student_id,num from score where course_id = (select cid from course where cname = '语文')) t2

 on student.sid = t2.student_id

 left join (select student_id,num from score where course_id = (select cid from course where cname = '英语')) t3

 on student.sid = t3.student_id

 left join (select student_id,avg(num) avg_num,count(num) count_course from score group by student_id)  t4

 on student.sid = t4.student_id

 

6、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 课程ID,max(num) 最高分,min(num) 最低分 from score group by course_id;

 

7、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;

# 方法1:

# 先求平均成绩

select course_id,avg(num) from score group by course_id;

# 解决计算各科及格率的问题

所有及格的人/所有人数

select t1.course_id,t1.count1/t2.count2 from 

(select course_id,count(course_id) count1 from score where num>60 group by course_id) t1 

left join

(select course_id,count(course_id) count2 from score group by course_id) t2

on t1.course_id = t2.course_id;

# 根据上述内容进行表的拼接

select  t_out1.course_id,t_out1.avgnum,t_out2.pass_per from 

(select course_id,avg(num) avgnum from score group by course_id ) t_out1

left join 

(select t1.course_id,t1.count1/t2.count2 pass_per from 

(select course_id,count(course_id) count2 from score group by course_id) t2

on t1.course_id = t2.course_id) t_out2

on  t_out1.course_id = t_out2.course_id

# 加上排序

select  t_out1.course_id,t_out2.pass_per from  (select course_id,avg(num) avgnum from score group by course_id ) t_out1 left join  (select t1.course_id,t1.count1/t2.count2 pass_per from  (select course_id,count(course_id) count1 from score where num>60 group by course_id) t1  left join (select course_id,count(course_id) count2 from score group by course_id) t2 on t1.course_id = t2.course_id) t_out2 on  t_out1.course_id = t_out2.course_id order by avgnum,pass_per desc;

 

# 方法2 

# 使用case when直接计算合格率

select 

sum(case when num>60 then 1 else 0 end)/count(course_id)

from score group by course_id

# 加上课程id和平均值

select  course_id,sum(case when num>60 then 1 else 0 end)/count(course_id)

from score group by course_id

# 加上排序

select  course_id,avg(num) avgnum,sum(case when num>60 then 1 else 0 end)/count(course_id) pass_per 

from score group by course_id

 order by avgnum,pass_per desc;

 

 

8、查询各科成绩前三名的记录:(不考虑成绩并列情况) 

select
t1.sid,t1.student_id,t1.course_id,t1.num from score t1
left join
    (
    select sid,course_id,(select num from score as s2 where s2.course_id = s1.course_id order by num desc limit 0,1) as first_num,(select num from score as s2 where s2.course_id = s1.course_id order by num desc limit 1,1) as second_num,(select num from score as s2 where s2.course_id = s1.course_id order by num desc limit 2,1) as third_num
    from score as s1
    ) t2
on t1.sid = t2.sid
where t1.num = t2.first_num or t1.num = t2.second_num or t1.num = t2.third_num;
 


9、查询每门课程被选修的学生数;

select course_id,count(course_id) from score group by course_id;

 

10、查询同名同姓学生名单,并统计同名人数;

select sname,count(1) as count from student group by sname;

 

11、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 course_id,avg(if(isnull(num),num)) as avg from score group by course_id order by avg  asc,course_id desc;

 

12、查询平均成绩大于85的所有学生的学号、姓名和平均成绩;

select student_id,num)) from score left join student on score.student_id = student.sid group by student_id;

 

13、查询课程名称为“数学”,且分数低于60的学生姓名和分数;

select student.sname,score.num from score

left join course on score.course_id = course.cid

left join student on score.student_id = student.sid

where score.num < 60 and course.cname = '数学'

 

 

14、查询课程编号为003且课程成绩在80分以上的学生的学号和姓名; 

select * from score where score.student_id = 3 and score.num > 80

 

15、求选了课程的学生人数

select sid,sname from student where sid not in (select student_id from score group by student_id);

 

16、查询选修“杨艳”老师所授课程的学生中,成绩最高的学生姓名及其成绩;

# 先找到“杨艳”老师的教师id

select tid from teacher where tname = '杨艳';

# 再找到杨艳老师教的所有课程

select cid from course where teacher_id in (select tid from teacher where tname = '杨艳');

# 再找到杨艳老师教的所有课程的最高分

select max(num) from score where course_id in (select cid from course where teacher_id in (select tid from teacher where tname = '李平老师'));

# 再找到杨艳老师教的所有课程的最高分对应的学生

select distinct student_id,num from score 

where num = (select max(num) from score where course_id in (select cid from course where teacher_id in (select tid from teacher where tname = '李平老师'))) 

and course_id in   (select cid from course where teacher_id in (select tid from teacher where tname = '李平老师'));

# 找到学生的姓名

select student.sname,t1.num from(

select distinct student_id,num from score 

where num = (select max(num) from score where course_id in (select cid from course where teacher_id in (select tid from teacher where tname = '李平老师'))) 

and course_id in   (select cid from course where teacher_id in (select tid from teacher where tname = '李平老师'))

) t1

left join

student

on 

t1.student_id = student.sid;

 

17、查询各个课程及相应的选修人数;

select course.cname,count(1) from score

left join course on score.course_id = course.cid

group by course_id;

 

18、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 DISTINCT s1.course_id,s2.course_id,s1.num,s2.num from score as s1,score as s2 where s1.num = s2.num and s1.course_id != s2.course_id;

 

19、查询每门课程成绩最好的前两名;

   先查询每条数据对应学科成绩的第一名和第二名,这里必须要保留所有的s1,以便后续进行连表查询

select sid,1) as second_num
from score as s1

按照sid连表,把学生的成绩和对应的第一名、第二名成绩连起来
select
* from score t1
left join
    (
    select sid,1) as second_num
    from score as s1
    ) t2
on t1.sid = t2.sid

判断如果学生的成绩是第一名、第二名的成绩,那么就符合条件,显示学生的id、学科和成绩
select
t1.sid,1) as second_num
    from score as s1
    ) t2
on t1.sid = t2.sid
where t1.num = t2.first_num or t1.num = t2.second_num;

20、检索至少选修两门课程的学生学号;

select student_id from score group by student_id having count(student_id) > 1;

 

21、查询全部学生都选修的课程的课程号和课程名;

# 先查看一共有多少学生

select count(sid) from student;

#  查看哪一门课选秀的学生个数和学生的总个数相等

select course_id from score group by course_id having count(student_id) = (select count(sid) from student);


22、查询没学过“叶平”老师讲授的任一门课程的学生姓名;

# 先查看要查找老师的id

select tid from teacher where tname = '李平老师';

# 查看该老师交了哪些课程

select cid from course where teacher_id in (select tid from teacher where tname = '李平老师')

# 看看有多少学生学习了该老师的课程

select distinct student_id from score where course_id in (select cid from course where teacher_id in (select tid from teacher where tname = '李平老师'));

# 把不在上表中的学生姓名查出来

select sname from student where sid not in (select distinct student_id from score where course_id in (select cid from course where teacher_id in (select tid from teacher where tname = '李平老师')));


23、查询两门以上不及格课程的同学的学号及其平均成绩;

select student_id,avg(num) from score where num<60 group by student_id having count(num)>=2;

 
24、检索“004”课程分数小于60,按分数降序排列的同学学号;

select student_id from score where num< 60 and course_id = 4 order by num desc;


25、删除“002”同学的“001”课程的成绩;

delete from score where course_id = 1 and student_id = 2;
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