MySQL多表查询
目录
MySQL多表查询
一、数据准备
建表与数据准备
# 建表
create table department(
id int,
name varchar(20)
);
create table employee(
id int primary key auto_increment,
name varchar(20),
sex enum('male','female') not null default 'male',
age int,
dep_id int
);
# 插入数据
insert into department values
(200,'技术'),
(201,'人力资源'),
(202,'销售'),
(203,'运营');
insert into employee(name,sex,age,dep_id) values
('nick','male',18,200),
('jason','female',48,201),
('sean','male',38,201),
('tank','female',28,202),
('oscar','male',18,200),
('mac','female',18,204)
;
# 查看表结构和数据
mysql> desc department;
+-------+-------------+------+-----+---------+-------+
| Field | Type | Null | Key | Default | Extra |
+-------+-------------+------+-----+---------+-------+
| id | int(11) | YES | | NULL | |
| name | varchar(20) | YES | | NULL | |
+-------+-------------+------+-----+---------+-------+
mysql> desc employee;
+--------+-----------------------+------+-----+---------+----------------+
| Field | Type | Null | Key | Default | Extra |
+--------+-----------------------+------+-----+---------+----------------+
| id | int(11) | NO | PRI | NULL | auto_increment |
| name | varchar(20) | YES | | NULL | |
| sex | enum('male','female') | NO | | male | |
| age | int(11) | YES | | NULL | |
| dep_id | int(11) | YES | | NULL | |
+--------+-----------------------+------+-----+---------+----------------+
mysql> select * from department;
+------+--------------+
| id | name |
+------+--------------+
| 200 | 技术 |
| 201 | 人力资源 |
| 202 | 销售 |
| 203 | 运营 |
+------+--------------+
mysql> select * from employee;
+----+------------+--------+------+--------+
| id | name | sex | age | dep_id |
+----+------------+--------+------+--------+
| 1 | nick | male | 18 | 200 |
| 2 | jason | female | 48 | 201 |
| 3 | sean | male | 38 | 201 |
| 4 | tank | female | 28 | 202 |
| 5 | oscar | male | 18 | 200 |
| 6 | mac | female | 18 | 204 |
+----+------------+--------+------+--------+
表department与employee
二、多表链接查询
重点：外链接语法
SELECT 字段列表
FROM 表1 INNER|LEFT|RIGHT JOIN 表2
ON 表1.字段 = 表2.字段;
2.1 交叉连接
不适用任何匹配条件。生成笛卡尔积。
mysql> select * from employee,department;
+----+------------+--------+------+--------+------+--------------+
| id | name | sex | age | dep_id | id | name |
+----+------------+--------+------+--------+------+--------------+
| 1 | nick | male | 18 | 200 | 200 | 技术 |
| 1 | nick | male | 18 | 200 | 201 | 人力资源 |
| 1 | nick | male | 18 | 200 | 202 | 销售 |
| 1 | nick | male | 18 | 200 | 203 | 运营 |
| 2 | jason | female | 48 | 201 | 200 | 技术 |
| 2 | jason | female | 48 | 201 | 201 | 人力资源 |
| 2 | jason | female | 48 | 201 | 202 | 销售 |
| 2 | jason | female | 48 | 201 | 203 | 运营 |
| 3 | sean | male | 38 | 201 | 200 | 技术 |
| 3 | sean | male | 38 | 201 | 201 | 人力资源 |
| 3 | sean | male | 38 | 201 | 202 | 销售 |
| 3 | sean | male | 38 | 201 | 203 | 运营 |
| 4 | tank | female | 28 | 202 | 200 | 技术 |
| 4 | tank | female | 28 | 202 | 201 | 人力资源 |
| 4 | tank | female | 28 | 202 | 202 | 销售 |
| 4 | tank | female | 28 | 202 | 203 | 运营 |
| 5 | oscar | male | 18 | 200 | 200 | 技术 |
| 5 | oscar | male | 18 | 200 | 201 | 人力资源 |
| 5 | oscar | male | 18 | 200 | 202 | 销售 |
| 5 | oscar | male | 18 | 200 | 203 | 运营 |
| 6 | mac | female | 18 | 204 | 200 | 技术 |
| 6 | mac | female | 18 | 204 | 201 | 人力资源 |
| 6 | mac | female | 18 | 204 | 202 | 销售 |
| 6 | mac | female | 18 | 204 | 203 | 运营 |
+----+------------+--------+------+--------+------+--------------+
2.2 内连接
只连接有匹配关系的数据
# 关键字：inner join on
# 语法：from A表 inner join B表 on A表.关联字段=B表.关联字段
# 找两张表共有的部分，相当于利用条件从笛卡尔积结果中筛选出了正确的结果
# department没有204这个部门，因而employee表中关于204这条员工信息没有匹配出来
mysql> select employee.id,employee.name,employee.age,employee.sex,department.name from employee inner join department on employee.dep_id=department.id;
+----+-----------+------+--------+--------------+
| id | name | age | sex | name |
+----+-----------+------+--------+--------------+
| 1 | nick | 18 | male | 技术 |
| 2 | jason | 48 | female | 人力资源 |
| 3 | sean | 38 | male | 人力资源 |
| 4 | tank | 28 | female | 销售 |
| 5 | oscar | 18 | male | 技术 |
+----+-----------+------+--------+--------------+
# 上述sql等同于
mysql> select employee.id,employee.name,employee.age,employee.sex,department.name from employee,department where employee.dep_id=department.id;
2.3 外连接之左连接
保留左表的全部数据，右表有对应数据直接连表显示，没有对应关系空填充
# 关键字：left join on
# 语法：from 左表 left join 右表 on 左表.关联字段=右表.关联字段
# 以左表为准，即找出所有员工信息，当然包括没有部门的员工
# 本质就是：在内连接的基础上增加左边有右边没有的结果
mysql> select employee.id,employee.name,department.name as depart_name from employee left join department on employee.dep_id=department.id;
+----+------------+--------------+
| id | name | depart_name |
+----+------------+--------------+
| 1 | nick | 技术 |
| 5 | oscar | 技术 |
| 2 | jason | 人力资源 |
| 3 | sean | 人力资源 |
| 4 | tank | 销售 |
| 6 | mac | NULL |
+----+------------+--------------+
2.4外连接之右连接
保留右表的全部数据，左表有对应数据直接连表显示，没有对应关系空填充
# 关键字：right join on
# 语法：from A表 right join B表 on A表.关联字段=B表关联字段
# 以右表为准，即找出所有部门信息，包括没有员工的部门
# 本质就是：在内连接的基础上增加右边有左边没有的结果
mysql> select employee.id,employee.name,department.name as depart_name from employee right join department on employee.dep_id=department.id;
+------+-----------+--------------+
| id | name | depart_name |
+------+-----------+--------------+
| 1 | nick | 技术 |
| 2 | jason | 人力资源 |
| 3 | sean | 人力资源 |
| 4 | tank | 销售 |
| 5 | oscar | 技术 |
| NULL | NULL | 运营 |
+------+-----------+--------------+
2.5 外连接之左右连接互换
更换一下左右表的位置，相对应更换左右连接关键字，结果相同
# 左连接
mysql> select employee.id,employee.name,department.name as depart_name from employee left join department on employee.dep_id=department.id;
# 右连接
mysql> select employee.id,employee.name,department.name as depart_name from department right join employee on employee.dep_id=department.id;
2.6 全连接
左表右表数据都被保留，彼此有对应关系正常显示，彼此没有对应关系均空填充对方
全外连接：在内连接的基础上增加左边有右边没有的和右边有左边没有的结果
# 注意：mysql不支持全外连接 full JOIN
# 强调：mysql可以使用此种方式间接实现全外连接
select * from employee left join department on employee.dep_id = department.id
union
select * from employee right join department on employee.dep_id = department.id
;
# 查看结果
+------+------------+--------+------+--------+------+--------------+
| id | name | sex | age | dep_id | id | name |
+------+------------+--------+------+--------+------+--------------+
| 1 | nick | male | 18 | 200 | 200 | 技术 |
| 5 | oscar | male | 18 | 200 | 200 | 技术 |
| 2 | jason | female | 48 | 201 | 201 | 人力资源 |
| 3 | sean | male | 38 | 201 | 201 | 人力资源 |
| 4 | tank | female | 28 | 202 | 202 | 销售 |
| 6 | mac | female | 18 | 204 | NULL | NULL |
| NULL | NULL | NULL | NULL | NULL | 203 | 运营 |
+------+------------+--------+------+--------+------+--------------+
# 注意 union与union all的区别：union会去掉相同的纪录
2.7 符合条件连接查询
# 示例1：以内连接的方式查询employee和department表，并且employee表中的age字段值必须大于25,即找出年龄大于25岁的员工以及员工所在的部门
select employee.name,department.name from employee inner join department
on employee.dep_id = department.id
where age > 25;
# 示例2：以内连接的方式查询employee和department表，并且以age字段的升序方式显示
select employee.id,employee.name,employee.age,department.name from employee,department
where employee.dep_id = department.id
and age > 25
order by age asc;
三、子查询
- 子查询是将一个查询语句嵌套在另一个查询语句中
- 内层查询语句的查询结果,可以为外层查询语句提供查询条件
- 子查询中可以包含 IN、NOT IN、ANY、ALL、EXISTS和NOT EXISTS等关键字
- 还可以包含比较运算符 =、!=、>、<等
3.1带IN关键字的子查询
# 查询平均年龄在25岁以上的部门名
select id,name from department
where id in
(select dep_id from employee group by dep_id having avg(age) > 25);
# 查看技术部员工姓名
select name from employee
where dep_id in
(select id from department where name='技术');
# 查看不足1人的部门名(子查询得到的是有人的部门id)
select name from department where id not in (select distinct dep_id from employee);
3.2 带比较运算符的子查询
# 比较运算符：=、!=、>、>=、<、<=、<>
# 查询大于所有人平均年龄的员工名与年龄
mysql> select name,age from emp where age > (select avg(age) from emp);
+---------+------+
| name | age |
+---------+------+
| jason | 48 |
| sean | 38 |
+---------+------+
rows in set (0.00 sec)
# 查询大于部门内平均年龄的员工名、年龄
select t1.name,t1.age from emp t1
inner join
(select dep_id,avg(age) avg_age from emp group by dep_id) t2
on t1.dep_id = t2.dep_id
where t1.age > t2.avg_age;
3.3 带EXISTS关键字的子查询
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。
# department表中存在dept_id=203，Ture
mysql> select * from employee
-> where exists
-> (select id from department where id=200);
+----+------------+--------+------+--------+
| id | name | sex | age | dep_id |
+----+------------+--------+------+--------+
| 1 | nick | male | 18 | 200 |
| 2 | jason | female | 48 | 201 |
| 3 | sean | male | 38 | 201 |
| 4 | tank | female | 28 | 202 |
| 5 | oscar | male | 18 | 200 |
| 6 | mac | female | 18 | 204 |
+----+------------+--------+------+--------+
# department表中存在dept_id=205，False
mysql> select * from employee
-> where exists
-> (select id from department where id=204);
Empty set (0.00 sec)
3.4 all与any：区间修饰条件
# 语法规则
# where id in (1, 2, 3) => id是1或2或3
# where id not in (1, 2, 3) => id不是1，2，3
# where salary < all(3, 6, 9) => salary必须小于所有情况（小于最小）
# where salary > all(3, 6, 9) => salary必须大于所有情况（大于最大）
# where salary < any(3, 6, 9) => salary只要小于一种情况（小于最大）
# where salary > any(3, 6, 9) => salary只要大于一种情况（大于最小）
in < > ()
# 案例
select * from emp where salary < all(select salary from emp where id>11);
四、练习:查询每个部门最新入职的那位员工
4.1 表与数据准备
company.employee
员工id id int
姓名 emp_name varchar
性别 sex enum
年龄 age int
入职日期 hire_date date
岗位 post varchar
职位描述 post_comment varchar
薪水 salary double
办公室 office int
部门编号 depart_id int
# 创建表
create table employee(
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
);
# 查看表结构
mysql> desc employee;
+--------------+-----------------------+------+-----+---------+----------------+
| Field | Type | Null | Key | Default | Extra |
+--------------+-----------------------+------+-----+---------+----------------+
| id | int(11) | NO | PRI | NULL | auto_increment |
| name | varchar(20) | NO | | NULL | |
| sex | enum('male','female') | NO | | male | |
| age | int(3) unsigned | NO | | 28 | |
| hire_date | date | NO | | NULL | |
| post | varchar(50) | YES | | NULL | |
| post_comment | varchar(100) | YES | | NULL | |
| salary | double(15,2) | YES | | NULL | |
| office | int(11) | YES | | NULL | |
| depart_id | int(11) | YES | | NULL | |
+--------------+-----------------------+------+-----+---------+----------------+
# 插入记录
# 三个部门：教学，销售，运营
insert into employee(name,sex,age,hire_date,post,salary,office,depart_id) values
('nick','male',18,'20170301','老男孩上海虹桥最帅',7300.33,401,1), # 以下是教学部
('jason','male',78,'20150302','teacher',1000000.31,401,1),
('sean','male',81,'20130305','teacher',8300,401,1),
('tank','male',73,'20140701','teacher',3500,401,1),
('oscar','male',28,'20121101','teacher',2100,401,1),
('mac','female',18,'20110211','teacher',9000,401,1),
('rocky','male',18,'19000301','teacher',30000,401,1),
('成龙','male',48,'20101111','teacher',10000,401,1),
('歪歪','female',48,'20150311','sale',3000.13,402,2),# 以下是销售部门
('丫丫','female',38,'20101101','sale',2000.35,402,2),
('丁丁','female',18,'20110312','sale',1000.37,402,2),
('星星','female',18,'20160513','sale',3000.29,402,2),
('格格','female',28,'20170127','sale',4000.33,402,2),
('张野','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)
;
# ps：如果在windows系统中，插入中文字符，select的结果为空白，可以将所有字符编码统一设置成gbk
4.2 答案一(连表查询)
SELECT
*
FROM
emp AS t1
INNER JOIN (
SELECT
post,
max(hire_date) max_date
FROM
emp
GROUP BY
post
) AS t2 ON t1.post = t2.post
WHERE
t1.hire_date = t2.max_date;
4.3 答案二(子查询)
mysql> select (select t2.name from emp as t2 where t2.post=t1.post order by hire_date desc limit 1) from emp as t1 group by post;
+---------------------------------------------------------------------------------------+
| (select t2.name from emp as t2 where t2.post=t1.post order by hire_date desc limit 1) |
+---------------------------------------------------------------------------------------+
| 张野 |
| 格格 |
| jason |
| nick |
+---------------------------------------------------------------------------------------+
rows in set (0.00 sec)
mysql> select (select t2.id from emp as t2 where t2.post=t1.post order by hire_date desc limit 1) from emp as t1 group by post;
+-------------------------------------------------------------------------------------+
| (select t2.id from emp as t2 where t2.post=t1.post order by hire_date desc limit 1) |
+-------------------------------------------------------------------------------------+
| 14 |
| 13 |
| 2 |
| 1 |
+-------------------------------------------------------------------------------------+
rows in set (0.00 sec)
# 正确答案
mysql> select t3.name,t3.post,t3.hire_date from emp as t3 where id in (select (select id from emp as t2 where t2.post=t1.post order by hire_date desc limit 1) from emp as t1 group by post);
+--------+-----------------------------------------+------------+
| name | post | hire_date |
+--------+-----------------------------------------+------------+
| nick | 老男孩上海虹桥最帅 | 2017-03-01 |
| jason | teacher | 2015-03-02 |
| 格格 | sale | 2017-01-27 |
| 张野 | operation | 2016-03-11 |
+--------+-----------------------------------------+------------+
rows in set (0.00 sec)
答案一为正确答案，答案二中的limit 1有问题（每个部门可能有>1个为同一时间入职的新员工），我只是想用该例子来说明可以在select后使用子查询。
可以基于上述方法解决：比如某网站在全国各个市都有站点，每个站点一条数据，想取每个省下最新的那一条市的网站质量信息。
五、综合练习
5.1 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'), ('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;
5.2 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
5.3基础练习
- 查询男生、女生的人数；
- 查询姓“张”的学生名单；
- 课程平均分从高到低显示
- 查询有课程成绩小于60分的同学的学号、姓名；
- 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；
- 查询出只选修了一门课程的全部学生的学号和姓名；
- 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；
- 查询“生物”课程比“物理”课程成绩高的所有学生的学号；
- 查询平均成绩大于60分的同学的学号和平均成绩;
- 查询所有同学的学号、姓名、选课数、总成绩；
- 查询姓“李”的老师的个数；
- 查询没学过“张磊老师”课的同学的学号、姓名；
- 查询学过“1”并且也学过编号“2”课程的同学的学号、姓名；
- 查询学过“李平老师”所教的所有课的同学的学号、姓名；
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,sname,
(
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,count(sid) count_stu,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,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 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,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 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 );
5.4 进阶练习
- 查询没有学全所有课的同学的学号、姓名；
- 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；
- 删除学习“叶平”老师课的SC表记录；
- 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；
- 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；
- 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；
- 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；
- 查询各科成绩前三名的记录:(不考虑成绩并列情况)
- 查询每门课程被选修的学生数；
- 查询同名同姓学生名单，并统计同名人数；
- 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；
- 查询平均成绩大于85的所有学生的学号. 姓名和平均成绩；
- 查询课程名称为“数学”，且分数低于60的学生姓名和分数；
- 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；
- 求选了课程的学生人数
- 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；
- 查询各个课程及相应的选修人数；
- 查询不同课程但成绩相同的学生的学号、课程号、学生成绩；
- 查询每门课程成绩最好的前两名；
- 检索至少选修两门课程的学生学号；
- 查询全部学生都选修的课程的课程号和课程名；
- 查询没学过“叶平”老师讲授的任一门课程的学生姓名；
- 查询两门以上不及格课程的同学的学号及其平均成绩；
- 检索“004”课程分数小于60，按分数降序排列的同学学号；
- 删除“002”同学的“001”课程的成绩；
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,语文,数学,英语,有效课程数,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) 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
# 加上排序
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) 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,avg(num),
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), 0 ,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,sname, avg(if(isnull(num), 0 ,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,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
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,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
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,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
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；
