概览
MySQL数据操作： DML
在MySQL管理软件中，可以通过SQL语句中的DML语言来实现数据的操作，包括
- 使用INSERT实现数据的插入
- UPDATE实现数据的更新
- 使用DELETE实现数据的删除
- 使用SELECT查询数据以及。
插入数据insert
1. 插入完整数据（顺序插入）
语法一：
INSERT INTO 表名(字段1,字段2,字段3…字段n) VALUES(值1,值2,值3…值n);
语法二：
INSERT INTO 表名 VALUES (值1,值2,值3…值n);
2. 指定字段插入数据
语法：
INSERT INTO 表名(字段1,字段2,字段3…) VALUES (值1,值2,值3…);
3. 插入多条记录
语法：
INSERT INTO 表名 VALUES
(值1,值2,值3…值n),
(值1,值2,值3…值n),
(值1,值2,值3…值n);
4. 插入查询结果
语法：
INSERT INTO 表名(字段1,字段2,字段3…字段n)
SELECT (字段1,字段2,字段3…字段n) FROM 表2
WHERE …;
更新数据update
语法：
UPDATE 表名 SET
字段1=值1,
字段2=值2,
WHERE CONDITION;
示例：
UPDATE mysql.user SET password=password(‘123’)
where user=’root’ and host=’localhost’;
删除数据delete
语法：
DELETE FROM 表名
WHERE CONITION;
示例：
DELETE FROM mysql.user
WHERE password=’’;
操作表的SQL语句补充
1.修改表名
alter table 表名 rename 新表名;
2.新增字段
alter table 表名 add 字段名 字段类型(数字) 约束条件;
alter table 表名 add 字段名 字段类型(数字) 约束条件 after 已经存在的字段;
alter table 表名 add 字段名 字段类型(数字) 约束条件 first;
3.修改字段
alter table 表名 change 旧字段 新字段 字段类型(数字) 约束条件;
alter table 表名 modify 字段名 新的字段类型(数字) 约束条件;
4.删除字段
alter table 表名 drop 字段名;
表查询关键字
1.数据准备(直接拷贝)
create table emp(
id int not null unique auto_increment,
name varchar(20) not null,
sex enum('male','female') not null default 'male', #大部分是男的
age int(3) unsigned not null default 28,
hire_date date not null,
post varchar(50),
post_comment varchar(100),
salary double(15,2),
office int, #一个部门一个屋子
depart_id int
);
#三个部门：教学，销售，运营
insert into emp(name,sex,age,hire_date,post,salary,office,depart_id) values
('jason','male',18,'20170301','浦东第一帅形象代言',7300.33,401,1), #以下是教学部
('tom','male',78,'20150302','teacher',1000000.31,401,1),
('kevin','male',81,'20130305','teacher',8300,401,1),
('tony','male',73,'20140701','teacher',3500,401,1),
('owen','male',28,'20121101','teacher',2100,401,1),
('jack','female',18,'20110211','teacher',9000,401,1),
('jenny','male',18,'19000301','teacher',30000,401,1),
('sank','male',48,'20101111','teacher',10000,401,1),
('哈哈','female',48,'20150311','sale',3000.13,402,2),#以下是销售部门
('呵呵','female',38,'20101101','sale',2000.35,402,2),
('西西','female',18,'20110312','sale',1000.37,402,2),
('乐乐','female',18,'20160513','sale',3000.29,402,2),
('拉拉','female',28,'20170127','sale',4000.33,402,2),
('僧龙','male',28,'20160311','operation',10000.13,403,3), #以下是运营部门
('程咬金','male',18,'19970312','operation',20000,403,3),
('程咬银','female',18,'20130311','operation',19000,403,3),
('程咬铜','male',18,'20150411','operation',18000,403,3),
('程咬铁','female',18,'20140512','operation',17000,403,3);
查询关键字之select与from
"""
SQL语句的关键字编写顺序与执行顺序是不一致的!!!
eg: select name from emp;
肯定是先支持from确定表 之后执行select确定字段
编写SQL语句针对select和from可以先写个固定模板
select * from 表名 其他操作
select后的字段可能是实际的 也可能是通过SQL动态产生的 所以可以先用*占位最后再修改
"""
select
自定义查询表中字段对应的数据
from
指定操作的对象(到底是哪张表 也可能是多张)
查询关键字之where筛选
# 1.查询id大于等于3小于等于6的数据
select id,name from emp where id >= 3 and id <= 6;
select * from emp where id between 3 and 6;
# 2.查询薪资是20000或者18000或者17000的数据
select * from emp where salary = 20000 or salary = 18000 or salary = 17000;
select * from emp where salary in (20000,18000,17000); # 简写
# 3.查询员工姓名中包含o字母的员工姓名和薪资
# 在你刚开始接触mysql查询的时候，建议你按照查询的优先级顺序拼写出你的sql语句
"""
先是查哪张表 from emp
再是根据什么条件去查 where name like ‘%o%’
再是对查询出来的数据筛选展示部分 select name,salary
"""
select name,salary from emp where name like '%o%';
# 4.查询员工姓名是由四个字符组成的员工姓名与其薪资
select name,salary from emp where name like '____';
select name,salary from emp where char_length(name) = 4;
# 5.查询id小于3或者大于6的数据
select * from emp where id not between 3 and 6;
# 6.查询薪资不在20000，18000，17000范围的数据
select * from emp where salary not in (20000,18000,17000);
# 7.查询岗位描述为空的员工名与岗位名 针对null不能用等号，只能用is
select name,post from emp where post_comment = NULL; # 查询为空！
select name,post from emp where post_comment is NULL;
select name,post from emp where post_comment is not NULL;
查询关键字之group by分组
分组:按照一些指定的条件将单个单个的数据分为一个个整体
"""
分组之后我们研究的对象应该是以组为单位 不应该再直接获取单个数据项 如果获取了应该直接报错 select后面可以直接填写的字段名只能是分组的依据(其他字段需要借助于一些方法才可以获取)
set global sql_mode='strict_trans_tables,only_full_group_by';
"""
select post from emp group by post;
"""
我们写SQL是否需要使用分组 可以在题目中得到答案
每个、平均、最大、最小
配合分组常见使用的有聚合函数
max 最大值
min 最小值
sum 总和
count 计数
avg 平均
"""
获取每个部门的最高工资
# 以组为单位统计组内数据>>>聚合查询(聚集到一起合成为一个结果)
# 每个部门的最高工资
select post,max(salary) from emp group by post;
补充:在显示的时候还可以给字段取别名
select post as '部门',max(salary) as '最高工资' from emp group by post;
as也可以省略 但是不推荐省 因为寓意不明确
# 每个部门的最低工资
select post,min(salary) from emp group by post;
# 每个部门的平均工资
select post,avg(salary) from emp group by post;
# 每个部门的工资总和
select post,sum(salary) from emp group by post;
# 每个部门的人数
select post,count(id) from emp group by post;
查询分组之后的部门名称和每个部门下所有的学生姓名
# group_concat（分组之后用）不仅可以用来显示除分组外字段还有拼接字符串的作用
select post,group_concat(name) from emp group by post;
select post,group_concat(name,"_SB") from emp group by post;
select post,group_concat(name,": ",salary) from emp group by post;
select post,group_concat(salary) from emp group by post;
查询关键字之having过滤
where与having的功能其实是一样的 都是用来筛选数据
只不过where用于分组之前的筛选 而having用于分组之后的筛选
为了人为的区分 所以叫where是筛选 having是过滤
1、统计各部门年龄在30岁以上的员工平均工资，并且保留平均工资大于10000的部门
select post,avg(salary) from emp
where age >= 30
group by post
having avg(salary) > 10000;
查询关键字之distinct去重
去重的前提是数据必须一模一样
select distinct age from emp;
查询关键字之order by排序
select * from emp order by salary asc; #默认升序排
select * from emp order by salary desc; #降序排
select * from emp order by age desc; #降序排
#先按照age降序排，在年轻相同的情况下再按照薪资升序排
select * from emp order by age desc,salary asc;
# 统计各部门年龄在10岁以上的员工平均工资，并且保留平均工资大于1000的部门，然后对平均工资进行排序
select post,avg(salary) from emp
where age > 10
group by post
having avg(salary) > 1000
order by avg(salary)
;
查询关键字之limit分页
# 限制展示条数
select * from emp limit 3;
# 查询工资最高的人的详细信息
select * from emp order by salary desc limit 1;
# 分页显示
select * from emp limit 0,5; # 第一个参数表示起始位置，第二个参数表示的是条数，不是索引位置
select * from emp limit 5,5;
查询关键字之regexp正则
select * from emp where name regexp '^j.*(n|y)$';
多表查询思路
子查询
将一张表的查询结果括号括起来当做另外一条SQL语句的条件
eg:类似以日常生活中解决问题的方式
第一步干什么
第二步基于第一步的结果在做操作 ...
连表操作
先将所有涉及到结果的表全部拼接到一起形成一张大表 然后从大表中查询数据
#建表
create table dep1(
id int primary key auto_increment,
name varchar(20)
);
create table emp1(
id int primary key auto_increment,
name varchar(20),
gender enum('male','female') not null default 'male',
age int,
dep_id int
);
#插入数据
insert into dep1 values
(200,'技术'),
(201,'人力资源'),
(202,'销售'),
(203,'运营'),
(205,'安保')
;
insert into emp1(name,gender,age,dep_id) values
('jason','male',18,200),
('dragon','female',48,201),
('kevin','male',18,201),
('nick','male',28,202),
('owen','male',18,203),
('jerry','female',18,204);
子查询
# 查询jason的部门名称
1.先获取jason的部门编号
select dep_id from emp1 where name = 'jason'; # 200
2.根据部门编号获取部门名称
select name from dep1 where id = 200;
子查询
select name from dep1 where id = (select dep_id from emp1 where name = 'jason');
连表操作
select * from emp1,dep1; # 笛卡尔积
'''我们不会使用笛卡尔积来求数据 效率太低 连表有专门的语法'''
inner join 内连接
只拼接两边都有的字段数据
left join 左连接
以左表为基准 展示所有的数据 没有对应则NULL填充
right join 右连接
以右表为基准 展示所有的数据 没有对应则NULL填充
union 全连接
select * from emp1 left join dep1 on emp1.dep_id = dep1.id
union
select * from emp1 right join dep1 on emp1.dep_id = dep1.id;
多表查询练习题
/*
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'), ('2', '物理', '2'), ('3', '体育', '3'), ('4', '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', '1', '10'), ('2', '1', '2', '9'), ('5', '1', '4', '66'), ('6', '2', '1', '8'), ('8', '2', '3', '68'), ('9', '2', '4', '99'), ('10', '3', '1', '77'), ('11', '3', '2', '66'), ('12', '3', '3', '87'), ('13', '3', '4', '99'), ('14', '4', '1', '79'), ('15', '4', '2', '11'), ('16', '4', '3', '67'), ('17', '4', '4', '100'), ('18', '5', '1', '79'), ('19', '5', '2', '11'), ('20', '5', '3', '67'), ('21', '5', '4', '100'), ('22', '6', '1', '9'), ('23', '6', '2', '100'), ('24', '6', '3', '67'), ('25', '6', '4', '100'), ('26', '7', '1', '9'), ('27', '7', '2', '100'), ('28', '7', '3', '67'), ('29', '7', '4', '88'), ('30', '8', '1', '9'), ('31', '8', '2', '100'), ('32', '8', '3', '67'), ('33', '8', '4', '88'), ('34', '9', '1', '91'), ('35', '9', '2', '88'), ('36', '9', '3', '67'), ('37', '9', '4', '22'), ('38', '10', '1', '90'), ('39', '10', '2', '77'), ('40', '10', '3', '43'), ('41', '10', '4', '87'), ('42', '11', '1', '90'), ('43', '11', '2', '77'), ('44', '11', '3', '43'), ('45', '11', '4', '87'), ('46', '12', '1', '90'), ('47', '12', '2', '77'), ('48', '12', '3', '43'), ('49', '12', '4', '87'), ('52', '13', '3', '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,
PRIMARY KEY (`sid`),
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', '1', '理解'), ('2', '女', '1', '钢蛋'), ('3', '男', '1', '张三'), ('4', '男', '1', '张一'), ('5', '女', '1', '张二'), ('6', '男', '1', '张四'), ('7', '女', '2', '铁锤'), ('8', '男', '2', '李三'), ('9', '男', '2', '李一'), ('10', '女', '2', '李二'), ('11', '男', '2', '李四'), ('12', '女', '3', '如花'), ('13', '男', '3', '刘三'), ('14', '男', '3', '刘一'), ('15', '女', '3', '刘二'), ('16', '男', '3', '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'), ('2', '李平老师'), ('3', '刘海燕老师'), ('4', '朱云海老师'), ('5', '李杰老师');
COMMIT;
SET FOREIGN_KEY_CHECKS = 1;
#1、查询所有的课程的名称以及对应的任课老师姓名
SELECT
course.cname,
teacher.tname
FROM
course
INNER JOIN teacher ON course.teacher_id = teacher.tid;
#2、查询学生表中男女生各有多少人
SELECT
gender 性别,
count(1) 人数
FROM
student
GROUP BY
gender;
#3、查询物理成绩等于100的学生的姓名
SELECT
student.sname
FROM
student
WHERE
sid IN (
SELECT
student_id
FROM
score
INNER JOIN course ON score.course_id = course.cid
WHERE
course.cname = '物理'
AND score.num = 100
);
#4、查询平均成绩大于八十分的同学的姓名和平均成绩
SELECT
student.sname,
t1.avg_num
FROM
student
INNER JOIN (
SELECT
student_id,
avg(num) AS avg_num
FROM
score
GROUP BY
student_id
HAVING
avg(num) > 80
) AS t1 ON student.sid = t1.student_id;
#5、查询所有学生的学号，姓名，选课数，总成绩(注意：对于那些没有选修任何课程的学生也算在内)
SELECT
student.sid,
student.sname,
t1.course_num,
t1.total_num
FROM
student
LEFT JOIN (
SELECT
student_id,
COUNT(course_id) course_num,
sum(num) total_num
FROM
score
GROUP BY
student_id
) AS t1 ON student.sid = t1.student_id;
#6、 查询姓李老师的个数
SELECT
count(tid)
FROM
teacher
WHERE
tname LIKE '李%';
#7、 查询没有报李平老师课的学生姓名(找出报名李平老师课程的学生，然后取反就可以)
SELECT
student.sname
FROM
student
WHERE
sid NOT IN (
SELECT DISTINCT
student_id
FROM
score
WHERE
course_id IN (
SELECT
course.cid
FROM
course
INNER JOIN teacher ON course.teacher_id = teacher.tid
WHERE
teacher.tname = '李平老师'
)
);
#8、 查询物理课程比生物课程高的学生的学号(分别得到物理成绩表与生物成绩表，然后连表即可)
SELECT
t1.student_id
FROM
(
SELECT
student_id,
num
FROM
score
WHERE
course_id = (
SELECT
cid
FROM
course
WHERE
cname = '物理'
)
) AS t1
INNER JOIN (
SELECT
student_id,
num
FROM
score
WHERE
course_id = (
SELECT
cid
FROM
course
WHERE
cname = '生物'
)
) AS t2 ON t1.student_id = t2.student_id
WHERE
t1.num > t2.num;
#9、 查询没有同时选修物理课程和体育课程的学生姓名(没有同时选修指的是选修了一门的，思路是得到物理+体育课程的学生信息表，然后基于学生分组，统计count(课程)=1)
SELECT
student.sname
FROM
student
WHERE
sid IN (
SELECT
student_id
FROM
score
WHERE
course_id IN (
SELECT
cid
FROM
course
WHERE
cname = '物理'
OR cname = '体育'
)
GROUP BY
student_id
HAVING
COUNT(course_id) = 1
);
#10、查询挂科超过两门(包括两门)的学生姓名和班级(求出<60的表，然后对学生进行分组，统计课程数目>=2)
SELECT
student.sname,
class.caption
FROM
student
INNER JOIN (
SELECT
student_id
FROM
score
WHERE
num < 60
GROUP BY
student_id
HAVING
count(course_id) >= 2
) AS t1
INNER JOIN class ON student.sid = t1.student_id
AND student.class_id = class.cid;
#11、查询选修了所有课程的学生姓名(先从course表统计课程的总数，然后基于score表按照student_id分组，统计课程数据等于课程总数即可)
SELECT
student.sname
FROM
student
WHERE
sid IN (
SELECT
student_id
FROM
score
GROUP BY
student_id
HAVING
COUNT(course_id) = (SELECT count(cid) FROM course)
);
#12、查询李平老师教的课程的所有成绩记录
SELECT
*
FROM
score
WHERE
course_id IN (
SELECT
cid
FROM
course
INNER JOIN teacher ON course.teacher_id = teacher.tid
WHERE
teacher.tname = '李平老师'
);
#13、查询全部学生都选修了的课程号和课程名(取所有学生数，然后基于score表的课程分组，找出count(student_id)等于学生数即可)
SELECT
cid,
cname
FROM
course
WHERE
cid IN (
SELECT
course_id
FROM
score
GROUP BY
course_id
HAVING
COUNT(student_id) = (
SELECT
COUNT(sid)
FROM
student
)
);
#14、查询每门课程被选修的次数
SELECT
course_id,
COUNT(student_id)
FROM
score
GROUP BY
course_id;
#15、查询之选修了一门课程的学生姓名和学号
SELECT
sid,
sname
FROM
student
WHERE
sid IN (
SELECT
student_id
FROM
score
GROUP BY
student_id
HAVING
COUNT(course_id) = 1
);
#16、查询所有学生考出的成绩并按从高到低排序（成绩去重）
SELECT DISTINCT
num
FROM
score
ORDER BY
num DESC;
#17、查询平均成绩大于85的学生姓名和平均成绩
SELECT
sname,
t1.avg_num
FROM
student
INNER JOIN (
SELECT
student_id,
avg(num) avg_num
FROM
score
GROUP BY
student_id
HAVING
AVG(num) > 85
) t1 ON student.sid = t1.student_id;
#18、查询生物成绩不及格的学生姓名和对应生物分数
SELECT
sname 姓名,
num 生物成绩
FROM
score
LEFT JOIN course ON score.course_id = course.cid
LEFT JOIN student ON score.student_id = student.sid
WHERE
course.cname = '生物'
AND score.num < 60;
#19、查询在所有选修了李平老师课程的学生中，这些课程(李平老师的课程，不是所有课程)平均成绩最高的学生姓名
SELECT
sname
FROM
student
WHERE
sid = (
SELECT
student_id
FROM
score
WHERE
course_id IN (
SELECT
course.cid
FROM
course
INNER JOIN teacher ON course.teacher_id = teacher.tid
WHERE
teacher.tname = '李平老师'
)
GROUP BY
student_id
ORDER BY
AVG(num) DESC
LIMIT 1
);
#20、查询每门课程成绩最好的前两名学生姓名
#查看每门课程按照分数排序的信息，为下列查找正确与否提供依据
SELECT
*
FROM
score
ORDER BY
course_id,
num DESC;
#表1：求出每门课程的课程course_id,与最高分数first_num
SELECT
course_id,
max(num) first_num
FROM
score
GROUP BY
course_id;
#表2：去掉最高分，再按照课程分组，取得的最高分，就是第二高的分数second_num
SELECT
score.course_id,
max(num) second_num
FROM
score
INNER JOIN (
SELECT
course_id,
max(num) first_num
FROM
score
GROUP BY
course_id
) AS t ON score.course_id = t.course_id
WHERE
score.num < t.first_num
GROUP BY
course_id;
#将表1和表2联合到一起，得到一张表t3，包含课程course_id与该们课程的first_num与second_num
SELECT
t1.course_id,
t1.first_num,
t2.second_num
FROM
(
SELECT
course_id,
max(num) first_num
FROM
score
GROUP BY
course_id
) AS t1
INNER JOIN (
SELECT
score.course_id,
max(num) second_num
FROM
score
INNER JOIN (
SELECT
course_id,
max(num) first_num
FROM
score
GROUP BY
course_id
) AS t ON score.course_id = t.course_id
WHERE
score.num < t.first_num
GROUP BY
course_id
) AS t2 ON t1.course_id = t2.course_id;
#查询前两名的学生（有可能出现并列第一或者并列第二的情况）
SELECT
score.student_id,
t3.course_id,
t3.first_num,
t3.second_num
FROM
score
INNER JOIN (
SELECT
t1.course_id,
t1.first_num,
t2.second_num
FROM
(
SELECT
course_id,
max(num) first_num
FROM
score
GROUP BY
course_id
) AS t1
INNER JOIN (
SELECT
score.course_id,
max(num) second_num
FROM
score
INNER JOIN (
SELECT
course_id,
max(num) first_num
FROM
score
GROUP BY
course_id
) AS t ON score.course_id = t.course_id
WHERE
score.num < t.first_num
GROUP BY
course_id
) AS t2 ON t1.course_id = t2.course_id
) AS t3 ON score.course_id = t3.course_id
WHERE
score.num >= t3.second_num
AND score.num <= t3.first_num;
#排序后可以看的明显点
SELECT
score.student_id,
t3.course_id,
t3.first_num,
t3.second_num
FROM
score
INNER JOIN (
SELECT
t1.course_id,
t1.first_num,
t2.second_num
FROM
(
SELECT
course_id,
max(num) first_num
FROM
score
GROUP BY
course_id
) AS t1
INNER JOIN (
SELECT
score.course_id,
max(num) second_num
FROM
score
INNER JOIN (
SELECT
course_id,
max(num) first_num
FROM
score
GROUP BY
course_id
) AS t ON score.course_id = t.course_id
WHERE
score.num < t.first_num
GROUP BY
course_id
) AS t2 ON t1.course_id = t2.course_id
) AS t3 ON score.course_id = t3.course_id
WHERE
score.num >= t3.second_num
AND score.num <= t3.first_num
ORDER BY
course_id;
#可以用以下命令验证上述查询的正确性
SELECT
*
FROM
score
ORDER BY
course_id,
num DESC;
-- 21、查询不同课程但成绩相同的学号，课程号，成绩
-- 22、查询没学过“叶平”老师课程的学生姓名以及选修的课程名称；
-- 23、查询所有选修了学号为1的同学选修过的一门或者多门课程的同学学号和姓名；
-- 24、任课最多的老师中学生单科成绩最高的学生姓名