mysql表查询练习

准备数据

create table book(
    id int primary key auto_increment,
    name varchar(16),
    price int
);
create table author(
    id int primary key auto_increment,
    name varchar(16)
);
create table book_author(
    id int primary key auto_increment,
    book_id int,
    author_id int
);
id name pirce
1 葵花宝典 88
2 九阴真经 78
3 辟邪剑谱 99
id name
1 张全蛋
2 李诞
3 寡人
id book_id author_id
1 1 1
2 1 2
3 1 3
4 2 1
5 2 3
6 3 2
7 3 2
2、查看每本书的作者
SELECT
    book.NAME,
    GROUP_CONCAT( author.`name` )
FROM
    book
    INNER JOIN book_to_author ON book.id = book_to_author.book_id
    INNER JOIN author ON author.id = author_id
GROUP BY
    book.NAME;
3、查看每位作者都有哪些作品
SELECT
    author.NAME,
    group_concat( book.NAME )
FROM
    author
    INNER JOIN book_to_author ON author.id = book_to_author.author_id
    INNER JOIN book ON book.id = book_to_author.book_id
GROUP BY
     author.NAME;
4、利用pymysql操作数据库实现sql注入效果
# -*- coding: utf-8 -*-

import pymysql

# 实例化产生一个 Connect 连接对象
conn = pymysql.Connect(
    host='127.0.0.1',
    port=3306,
    user='root',
    password='root',
    database='day41',
    charset='utf8'
)
# 游标对象
name = input('>>>name:').strip()
pwd = input('>>>>pwd:').strip()
cursor = conn.cursor(pymysql.cursors.DictCursor)

sql = "select * from userinfo where name='%s' and password=%s" % (name, pwd)
print(sql)
res = cursor.execute(sql)
print(res)
if res:
    print('登录成功')
else:
    print('登录失败')

# >>>name:qzk' -- dasad
# >>>>pwd:
# select * from userinfo where name='qzk' -- dasad' and password=
# 1
# 登录成功

# >>>name:qzk' or 1=1 -- asd
# >>>>pwd:''
# select * from userinfo where name='qzk' or 1=1 -- asd' and password=''
# 5
# 登录成功

5、利用pymysql实现登陆注册功能

注册:用户名不能重复

登陆:展示对应的错误信息(用户名错误，密码错误)

# 实例化产生一个 Connect 连接对象
conn = pymysql.Connect(
    host='127.0.0.1',
    port=3306,
    user='root',
    password='root',
    database='day41',
    charset='utf8'
)
# 游标对象
name = input('>>>name:').strip()
pwd = input('>>>>pwd:').strip()
cursor = conn.cursor(pymysql.cursors.DictCursor)

sql = 'select * from userinfo where name=%s'
print(sql)
res = cursor.execute(sql,(name,))
print(res)
dic_user = cursor.fetchone()
print(dic_user)
if res:
    if int(pwd) == dic_user['password']:
        print('登录成功')
    else:
        print('密码错误')
else:
    print('用户名不存在')

内容回顾

select distinct 字段1，字段2，。。。 from 表名
            where   分组之前的过滤条件
            group by 分组条件
            having  分组之后过滤条件
            order by 排序字段1 asc，排序字段2 desc
            limit 5,5

as语法中给某个查询结果起别名的时候需要把查询语句中的分号去除
(select name,salary*12 as '年薪' from emp) as t1;

# 一个字段展示用户名和年龄
select concat(name,':',age) as info from emp;

# 字段为NAME和AGE，值为‘NAME:jason’,'AGE:18'
select concat("NAME:",name) as NAME,concat("AGE:",age) as AGE from emp;

# 如果拼接的符号是统一的可以用
select concat_ws(':',name,age,sex) as info from emp;

# 1.子查询相关
# 查询平均年轻在25岁以上的部门名
select name from dep
            where id in
            (select dep_id from emp group by dep_id having avg(age)>25);

select dep.name from emp inner join dep on emp.dep_id = dep.id
            group by dep.name
            having avg(age) > 25;

# exist(了解)
EXISTS关字键字表示存在。在使用EXISTS关键字时，内层查询语句不返回查询的记录，
而是返回一个真假值，True或False。
当返回True时，外层查询语句将进行查询
当返回值为False时，外层查询语句不进行查询。
select * from employee
    where exists
    (select id from department where id > 3);

select * from employee
    where exists
    (select id from department where id > 250);

Navicat使用

下载地址:https://pan.baidu.com/s/1bpo5mqj

掌握：
#1. 测试+链接数据库
#2. 新建库
#3. 新建表，新增字段+类型+约束
#4. 设计表：外键
#5. 新建查询
#6. 建立表模型

#注意：
批量加注释：ctrl+？键
批量去注释：ctrl+shift+？键

练习题

导出的sql语句代码

/*
 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'), ('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;

拷贝上述代码，新建一个.sql文件，保存到桌面

打开navicat新建数据库day41，选中新建的数据库鼠标右键选择运行SQL文件

弹出文件框，选中刚刚保存到桌面的.sql文件即可

快速建表

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

# navicat建表

练习题

1、查询所有的课程的名称以及对应的任课老师姓名
4、查询平均成绩大于八十分的同学的姓名和平均成绩
7、 查询没有报李平老师课的学生姓名
8、 查询没有同时选修物理课程和体育课程的学生姓名
9、 查询挂科超过两门(包括两门)的学生姓名和班级
# 参考答案
#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;
#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;
#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
    );

# 9、 查询挂科超过两门(包括两门)的学生姓名和班级
select student.sname,class.caption from class INNER JOIN student
    on class.cid = student.class_id
    WHERE student.sid in
    (select student_id from score where num < 60
    GROUP BY student_id
    HAVING COUNT(course_id) >=2)
    ;

pymysql模块

# 1.安装:pip3 insatll pymysql

# 2.代码链接
import pymysql
#链接
conn=pymysql.connect(
  host='localhost',
  user='root',
  password='123',
  database='egon',
  charset='utf8')
#游标
cursor=conn.cursor() #执行完毕返回的结果集默认以元组显示
#cursor=conn.cursor(cursor=pymysql.cursors.DictCursor)  # 以字典的方式显示数据

# 3.pymysql操作数据库
#执行sql语句
user = input(">>>:").strip()
pwd = input(">>>:").strip()
sql='select * from userinfo where name="%s" and password="%s"' %(user,pwd) #注意%s需要加引号

rows=cursor.execute(sql) #执行sql语句，返回sql查询成功的记录数目
# 获取真实数据cursor.fetchone(),cursor.fetchall(),cursor.fetchmany(),类似管道取值，获取一条，所有，多条

cursor.scroll(1,'relative')  # 相对移动
cursor.scroll(3,'absolute')  # 绝对移动

cursor.close()
conn.close()

sql注入问题

# 不要手动去拼接查询的sql语句
username = input(">>>:").strip()
password = input(">>>:").strip()
sql = "select * from user where username='%s' and password='%s'"%(username,password)

# 用户名正确
username >>>: jason' -- jjsakfjjdkjjkjs
# 用户名密码都不对的情况
username >>>: xxx' or 1=1 --asdjkdklqwjdjkjasdljad
password >>>: ''

增删改

# 增
sql = "insert into user(username,password) values(%s,%s)"
rows = cursor.excute(sql,('jason','123'))

# 修改
sql = "update user set username='jasonDSB' where id=1"
rows = cursor.excute(sql)

"""
增和改单单执行excute并不会真正影响到数据，需要再执行conn.commit()才可以完成真正的增改
"""

# 一次插入多行记录
res = cursor,excutemany(sql,[(),(),()]

1、查询所有的课程的名称以及对应的任课老师姓名

SELECT
    cname,
    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, count( gender )
FROM
    student
GROUP BY
    gender
HAVING
    count( gender );

3、查询物理成绩等于100的学生的姓名

SELECT
    student.sname,
    course.cname,
    num
FROM
    student
    INNER JOIN score ON student.sid = score.student_id
    INNER JOIN course ON course_id = course.cid
WHERE
    cname = '物理'
    AND 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,
    avg( num )
FROM
    student
    INNER JOIN score ON student.sid = score.student_id
GROUP BY
    student.sname
HAVING
    avg( num ) > 80;

5、查询所有学生的学号，姓名，选课数，总成绩

SELECT
    student.sid,
    sname,
    count( course_id ),
    sum( num )
FROM
    student
    INNER JOIN score ON student.sid = score.student_id
    INNER JOIN course ON score.course_id = course.cid
GROUP BY
    sname
ORDER BY
    student.sid;

6、 查询姓李老师的个数

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

7、 查询没有报李平老师课的学生姓名

SELECT
    sname
FROM
    student
WHERE
    student.sid not IN ( SELECT student_id FROM score WHERE course_id IN ( SELECT cid FROM course INNER JOIN teacher ON teacher_id = tid WHERE tname like '%李平%' ) );

8、 查询物理课程比生物课程高的学生的学号

# 方法一、
SELECT
    t1.student_id
FROM
    ( SELECT student_id, num FROM score INNER JOIN course ON course_id = cid WHERE cname = '物理' ) AS t1
    INNER JOIN ( SELECT student_id, num FROM score INNER JOIN course ON course_id = cid WHERE cname = '生物' ) AS t2 ON t1.student_id = t2.student_id
WHERE
    t1.num > t2.num;

# 方法二、
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

# 方法一、
SELECT
    student.sname
FROM
    student
    INNER JOIN score ON student.sid = student_id
    INNER JOIN course ON course_id = cid
WHERE
    cname = '物理'
    OR cname = '体育'
GROUP BY
    student.sname
HAVING
    count( sname ) = 1;
# 方法三、
SELECT
    student.sname
FROM
    student
    INNER JOIN score ON student.sid = student_id
    INNER JOIN course ON course_id = cid
WHERE
    cname = '物理'
    OR cname = '体育'
GROUP BY
    student.sname
HAVING
    count( course_id ) = 1;
# 方法二、
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、查询选修了所有课程的学生姓名

# 方法一、
SELECT
    student.sname,
    class.caption
FROM
    class
    INNER JOIN student ON class.cid = student.class_id
    INNER JOIN score ON score.student_id = student.sid
WHERE
    num < 60
GROUP BY
    student.sname
HAVING
    count( course_id ) >= 2;
# 方法二、
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;

12、查询李平老师教的课程的所有成绩记录

SELECT
    num
FROM
    score
WHERE
    course_id IN ( SELECT cid FROM course INNER JOIN teacher ON teacher.tid = course.teacher_id WHERE tname = '李平老师' );

13、查询全部学生都选修了的课程号和课程名（count（sid）=13

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
    cid,cname,
    count( student_id )
FROM
    course
    INNER JOIN score ON course.cid = score.course_id
GROUP BY
    cid;

# 方法二
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
    student.sname,
    student.sid
FROM
    student
WHERE
    student.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）

17、查询平均成绩大于85的学生姓名和平均成绩

SELECT
    student.sname,
    avg( num )
FROM
    student
    INNER JOIN score ON student.sid = score.student_id
GROUP BY
    student.sname
HAVING
    avg( num ) > 80;

18、查询生物成绩不及格的学生姓名和对应生物分数

SELECT
    student.sname,
    course.cname,
    score.num
FROM
    student
    INNER JOIN score ON student.sid = student_id
    INNER JOIN course ON course_id = course.cid
WHERE
    course.cname = '生物'
    AND 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

20、查询每门课程成绩最好的前两名学生姓名

#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;

原文地址：https://www.cnblogs.com/qianzhengkai/p/10939766.html

时间： 2024-08-08 03:22:38

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