MySQL之多表查询
MySQL之多表查询
阅读目录
一 介绍
本节主题
- 多表连接查询
- 复合条件连接查询
- 子查询
首先说一下，我们写项目一般都会建一个数据库，那数据库里面是不是存了好多张表啊，不可能把所有的数据都放到一张表里面，肯定要分表来存数据，这样节省空间，数据的组织结构更清晰，解耦和程度更高，但是这些表本质上是不是还是一个整体啊，是一个项目所有的数据，那既然分表存了，就要涉及到多个表连接查询了，比如说员工信息一张表，部门信息一张表，那如果我想让你帮我查一下技术部门有哪些员工的姓名，你怎么办，单独找员工表能实现吗，不能，单独找部门表也无法实现，因为部门表里面没有员工的信息，对不对，所以就涉及到部门表和员工表来关联到一起进行查询了，好，那我们来建立这么两张表：
#建表#部门表
create table department(
id int,
name varchar(20)
);
#员工表，之前我们学过foreign key，强行加上约束关联，但是我下面这个表并没有直接加foreign key，这两个表我只是让它们在逻辑意义上有关系，并没有加foreign key来强制两表建立关系，为什么要这样搞，是有些效果要给大家演示一下#所以，这两个表是不是先建立哪个表都行啊，如果有foreign key的话，是不是就需要注意表建立的顺序了。那我们来建表。
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
);
#给两个表插入一些数据
insert into department values
(200,'技术'),
(201,'人力资源'),
(202,'销售'),
(203,'运营'); #注意这一条数据，在下面的员工表里面没有对应这个部门的数据
insert into employee(name,sex,age,dep_id) values
('egon','male',18,200),
('alex','female',48,201),
('wupeiqi','male',38,201),
('yuanhao','female',28,202),
('liwenzhou','male',18,200),
('jingliyang','female',18,204) #注意这条数据的dep_id字段的值，这个204，在上面的部门表里面也没有对应的部门id。所以两者都含有一条双方没有涉及到的数据，这都是为了演示一下效果设计的昂
;
#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 | egon | male | 18 | 200 |
| 2 | alex | female | 48 | 201 |
| 3 | wupeiqi | male | 38 | 201 |
| 4 | yuanhao | female | 28 | 202 |
| 5 | liwenzhou | male | 18 | 200 |
| 6 | jingliyang | female | 18 | 204 |
+----+------------+--------+------+--------+
二 多表连接查询
#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;
1、交叉连接：不适用任何匹配条件。生成笛卡尔积
补充一点：select 查询表的时候，后面可以跟多张表一起查询：
mysql> select * from department,employee; #表用逗号分隔，看我查询时表的顺序，先department后employee，所以你看结果表的这些字段，是不是就是我们两个表字段并且哪个表在前面，哪个表的字段就在前面
+------+--------------+----+------------+--------+------+--------+
| id | name | id | name | sex | age | dep_id |
+------+--------------+----+------------+--------+------+--------+
| 200 | 技术 | 1 | egon | male | 18 | 200 |
| 201 | 人力资源 | 1 | egon | male | 18 | 200 |
| 202 | 销售 | 1 | egon | male | 18 | 200 |
| 203 | 运营 | 1 | egon | male | 18 | 200 |
| 200 | 技术 | 2 | alex | female | 48 | 201 |
| 201 | 人力资源 | 2 | alex | female | 48 | 201 |
| 202 | 销售 | 2 | alex | female | 48 | 201 |
| 203 | 运营 | 2 | alex | female | 48 | 201 |
| 200 | 技术 | 3 | wupeiqi | male | 38 | 201 |
| 201 | 人力资源 | 3 | wupeiqi | male | 38 | 201 |
| 202 | 销售 | 3 | wupeiqi | male | 38 | 201 |
| 203 | 运营 | 3 | wupeiqi | male | 38 | 201 |
| 200 | 技术 | 4 | yuanhao | female | 28 | 202 |
| 201 | 人力资源 | 4 | yuanhao | female | 28 | 202 |
| 202 | 销售 | 4 | yuanhao | female | 28 | 202 |
| 203 | 运营 | 4 | yuanhao | female | 28 | 202 |
| 200 | 技术 | 5 | liwenzhou | male | 18 | 200 |
| 201 | 人力资源 | 5 | liwenzhou | male | 18 | 200 |
| 202 | 销售 | 5 | liwenzhou | male | 18 | 200 |
| 203 | 运营 | 5 | liwenzhou | male | 18 | 200 |
| 200 | 技术 | 6 | jingliyang | female | 18 | 204 |
| 201 | 人力资源 | 6 | jingliyang | female | 18 | 204 |
| 202 | 销售 | 6 | jingliyang | female | 18 | 204 |
| 203 | 运营 | 6 | jingliyang | female | 18 | 204 |
+------+--------------+----+------------+--------+------+--------+
24 rows in set (0.12 sec)我们让employee表在前面看看结果，注意看结果表的字段
mysql> select * from employee,department;
+----+------------+--------+------+--------+------+--------------+
| id | name | sex | age | dep_id | id | name |
+----+------------+--------+------+--------+------+--------------+
| 1 | egon | male | 18 | 200 | 200 | 技术 |
| 1 | egon | male | 18 | 200 | 201 | 人力资源 |
| 1 | egon | male | 18 | 200 | 202 | 销售 |
| 1 | egon | male | 18 | 200 | 203 | 运营 |
| 2 | alex | female | 48 | 201 | 200 | 技术 |
| 2 | alex | female | 48 | 201 | 201 | 人力资源 |
| 2 | alex | female | 48 | 201 | 202 | 销售 |
| 2 | alex | female | 48 | 201 | 203 | 运营 |
| 3 | wupeiqi | male | 38 | 201 | 200 | 技术 |
| 3 | wupeiqi | male | 38 | 201 | 201 | 人力资源 |
| 3 | wupeiqi | male | 38 | 201 | 202 | 销售 |
| 3 | wupeiqi | male | 38 | 201 | 203 | 运营 |
| 4 | yuanhao | female | 28 | 202 | 200 | 技术 |
| 4 | yuanhao | female | 28 | 202 | 201 | 人力资源 |
| 4 | yuanhao | female | 28 | 202 | 202 | 销售 |
| 4 | yuanhao | female | 28 | 202 | 203 | 运营 |
| 5 | liwenzhou | male | 18 | 200 | 200 | 技术 |
| 5 | liwenzhou | male | 18 | 200 | 201 | 人力资源 |
| 5 | liwenzhou | male | 18 | 200 | 202 | 销售 |
| 5 | liwenzhou | male | 18 | 200 | 203 | 运营 |
| 6 | jingliyang | female | 18 | 204 | 200 | 技术 |
| 6 | jingliyang | female | 18 | 204 | 201 | 人力资源 |
| 6 | jingliyang | female | 18 | 204 | 202 | 销售 |
| 6 | jingliyang | female | 18 | 204 | 203 | 运营 |
+----+------------+--------+------+--------+------+--------------+
24 rows in set (0.00 sec)
关于笛卡儿积：我们看一下上面的这些数据，有什么发现，首先看到这些字段都显示出来了，并且数据变得很多，我们来看一下，这么多条数据都是怎么来的，为什么会出现这么条数据，笛卡儿积这是一个数据名词，你可以去研究研究~~ 因为我们要进行连表查询，那么mysql并不知道你想要如何连接两个表的关系进行查询，那么mysql会将你两个表数据的所有组合关系都给你拼接成一条数据来显示，这样你就可以想查哪个关联关系的数据就查哪个了，如果还是不太理解看一下下面的图：
关于笛卡儿积现象的解释图：
咱们为了更好的管理数据，为了节省空间，为了数据组织结构更清晰，将数据拆分到了不同表里面，但是本质上是不是还是一份数据，一份重复内容很多的很大的数据，所以我们即便是分表了，但是咱们是不是还需要找到一个方案把两个本来分开的表能够合并到一起来进行查询，那你是不是就可以根据部门找员工，根据员工找部门了，对不对，但是我们合并两个表的时候，如何合并，根据什么来合并，通过笛卡儿积这种合并有没有浪费，我们其实想做的是不是说我们的员工表中dep_id这个字段中的数据和部门表里面的id能够对应上就可以了，因为我们知道我们设计表的时候，是通过这两个字段来给两个表建立关系的，对不对，看下图：
我们的目标就是将两个分散出去的表，按照两者之间有关系的字段，能对应上的字段，把两者合并成一张表，这就是多表查询的一个本质。那么笛卡儿积干了什么事儿，就是简单粗暴的将两个表的数据全部对应了一遍，用处就是什么呢，它肯定就能保证有一条是对应准的，你需要做的事情就是在笛卡儿积的基础上只过滤出我们需要的那些数据就行了，笛卡儿积不是咱们最终要得到的结果，只是给你提供了一个基础，它不管对应的对不对，全部给你对应一遍，然后你自己去筛选就可以了，然后基于笛卡儿积我们来找一下对应的数据，看看能不能找到：
2、内连接：只连接匹配的行
#我们要找的数据就是员工表里面dep_id字段的值和部门表里面id字段的值能对应上的那些数据啊，所以你看下面的写法：
mysql> select * from employee,department where employee.dep_id=department.id;
+----+-----------+--------+------+--------+------+--------------+
| id | name | sex | age | dep_id | id | name |
+----+-----------+--------+------+--------+------+--------------+
| 1 | egon | male | 18 | 200 | 200 | 技术 |
| 2 | alex | female | 48 | 201 | 201 | 人力资源 |
| 3 | wupeiqi | male | 38 | 201 | 201 | 人力资源 |
| 4 | yuanhao | female | 28 | 202 | 202 | 销售 |
| 5 | liwenzhou | male | 18 | 200 | 200 | 技术 |
+----+-----------+--------+------+--------+------+--------------+
5 rows in set (0.14 sec)
拿到了我们想要的结果。
但是你看，我们左表employee表中的dep_id为204的那个数据没有了，右表department表的id为203的数据没有了，因为我们现在要的就是两表能对应上的数据一起查出来，那个204和203双方对应不上。
#再看一个需求，我要查出技术部的员工的名字
mysql> select name from employee,department where employee.dep_id=department.id and department.name='技术';
ERROR 1052 (23000): Column 'name' in field list is ambiguous
#上面直接就报错了，因为select后面直接写的name，在两个表合并起来的表中，是有两个name字段的，直接写name是不行的，要加上表名，再看：
mysql> select employee.name from employee,department where employee.dep_id=department.id and department.name='技术';
+-----------+
| name |
+-----------+
| egon |
| liwenzhou |
+-----------+
2 rows in set (0.09 sec)
结果就没问题了
但是你看上面的代码有没有什么不太好的地方，虽然我们能够完成我们的事情，但是代码可读性不好，所以以后不要这么写，但是看图：
所以mysql为我们提供了一些专门做连表操作的方法，这些方法语义更加的明确，你一看就知道那些代码是连表的，那些代码是查询的，其实上面的连表也是个查询操作，但是我们为了区分明确，连表专门用连表的方法，查询就专门用查询的方法。那这些专门的方法都是什么呢，看后面的内容：
3 、外链接之左连接：优先显示左表全部记录
#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 #注意语法：
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 | egon | 技术 |
| 5 | liwenzhou | 技术 |
| 2 | alex | 人力资源 |
| 3 | wupeiqi | 人力资源 |
| 4 | yuanhao | 销售 |
| 6 | jingliyang | NULL |
+----+------------+--------------+
4 、外链接之右连接：优先显示右表全部记录
#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 | egon | 技术 |
| 2 | alex | 人力资源 |
| 3 | wupeiqi | 人力资源 |
| 4 | yuanhao | 销售 |
| 5 | liwenzhou | 技术 |
| NULL | NULL | 运营 |
+------+-----------+--------------+
5 、全外连接：显示左右两个表全部记录
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 | egon | male | 18 | 200 | 200 | 技术 |
| 5 | liwenzhou | male | 18 | 200 | 200 | 技术 |
| 2 | alex | female | 48 | 201 | 201 | 人力资源 |
| 3 | wupeiqi | male | 38 | 201 | 201 | 人力资源 |
| 4 | yuanhao | female | 28 | 202 | 202 | 销售 |
| 6 | jingliyang | 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，因为union all是left join 和right join合并，所以有重复的记录，通过union就将重复的记录去重了。
三 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;
四 子查询
子查询其实就是将你的一个查询结果用括号括起来，这个结果也是一张表，就可以将它交给另外一个sql语句，作为它的一个查询依据来进行操作。
来，我们简单来个需求：技术部都有哪些员工的姓名，都显示出来： 1、看一下和哪个表有关，然后from找到两个表 2、进行一个连表操作 3、基于连表的结果来一个过滤就可以了
#我们之前的做法是：先连表
mysql> select * from employee inner join department on employee.dep_id = department.id;
+----+-----------+--------+------+--------+------+--------------+
| id | name | sex | age | dep_id | id | name |
+----+-----------+--------+------+--------+------+--------------+
| 1 | egon | male | 18 | 200 | 200 | 技术 |
| 2 | alex | female | 48 | 201 | 201 | 人力资源 |
| 3 | wupeiqi | male | 38 | 201 | 201 | 人力资源 |
| 4 | yuanhao | female | 28 | 202 | 202 | 销售 |
| 5 | liwenzhou | male | 18 | 200 | 200 | 技术 |
+----+-----------+--------+------+--------+------+--------------+
5 rows in set (0.10 sec)
#然后根据连表的结果进行where过滤，将select*改为select employee.namemysql> select employee.name from employee inner join department on employee.dep_id = department.id where department.name='技术';
+-----------+
| name |
+-----------+
| egon |
| liwenzhou |
+-----------+
2 rows in set (0.09 sec)
然后看一下子查询这种方式的写法：它的做法就是解决完一个问题，再解决下一个问题，针对我们上面的需求，你想，我们的需求是不是说找技术部门下面有哪些员工对不对，如果你直接找员工表，你能确定哪个dep_id的数值表示的是技术部门吗，不能，所以咱们是不是应该先确定一个技术部门对应的id号是多少，然后根据部门的id号，再去员工表里面查询一下dep_id为技术部门对应的部门表的那个id号的所有的员工表里面的记录：好，那我们看一下下面的操作
#首先从部门表里面找到技术部门对应的id
mysql> select id from department where name='技术';
+------+
| id |
+------+
| 200 |
+------+
1 row in set (0.00 sec)
#那我们把上面的查询结果用括号括起来，它就表示一条id=200的数据，然后我们通过员工表来查询dep_id=这条数据作为条件来查询员工的name
mysql> select name from employee where dep_id = (select id from department where name='技术');
+-----------+
| name |
+-----------+
| egon |
| liwenzhou |
+-----------+
2 rows in set (0.00 sec)上面这些就是子查询的一个思路，解决一个问题，再解决另外一个问题，你子查询里面可不可以是多个表的查询结果，当然可以，然后再通过这个结果作为依据来进行过滤，然后我们学一下子查询里面其他的内容，往下学。
子查询：#1：子查询是将一个查询语句嵌套在另一个查询语句中。
#2：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。
#3：子查询中可以包含：IN、NOT IN、ANY、ALL、EXISTS 和 NOT EXISTS等关键字
#4：还可以包含比较运算符：= 、 !=、> 、<等
1、带IN关键字的子查询
#查询员工平均年龄在25岁以上的部门名，可以用连表，也可以用子查询，我们用子查询来搞一下
select id,name from department
where id in
(select dep_id from employee group by dep_id having avg(age) > 25);
#连表来搞一下上面这个需求
select department.name from department inner join employee on department.id=employee.dep_id
group by department.name
having avg(age)>25;
总结：子查询的思路和解决问题一样，先解决一个然后拿着这个的结果再去解决另外一个问题，连表的思路是先将两个表关联在一起，然后在进行group by啊过滤啊等等操作，两者的思路是不一样的
#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);
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 |
+---------+------+
| alex | 48 |
| wupeiqi | 38 |
+---------+------+
2 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、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。还可以写not exists，和exists的效果就是反的
#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 | egon | male | 18 | 200 |
| 2 | alex | female | 48 | 201 |
| 3 | wupeiqi | male | 38 | 201 |
| 4 | yuanhao | female | 28 | 202 |
| 5 | liwenzhou | male | 18 | 200 |
| 6 | jingliyang | 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)
练习：通过连表的方式来查询每个部门最新入职的那位员工
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，只需要创建这一张表
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
('egon','male',18,'20170301','老男孩驻沙河办事处外交大使',7300.33,401,1), #以下是教学部
('alex','male',78,'20150302','teacher',1000000.31,401,1),
('wupeiqi','male',81,'20130305','teacher',8300,401,1),
('yuanhao','male',73,'20140701','teacher',3500,401,1),
('liwenzhou','male',28,'20121101','teacher',2100,401,1),
('jingliyang','female',18,'20110211','teacher',9000,401,1),
('jinxin','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
答案：
SELECT
*
FROM
emp AS t1
INNER JOIN ( #和虚拟表进行连表
SELECT
post,
max(hire_date) as max_date #给这个最大的日期取个别名叫做max_date，先将每个部门最近入职的最大的日期的信息筛选出来，通过这个表来和我们上面的总表进行关联
FROM
emp
GROUP BY
post
) AS t2 ON t1.post = t2.post #给虚拟表取个别名叫做t2
WHERE
t1.hire_date = t2.max_date; #然后再通过where来过滤出，入职日期和最大日期相等的记录，就是我们要的内容
五 综合练习
表结构为
#创建表及插入记录
CREATE TABLE class (
cid int(11) NOT NULL AUTO_INCREMENT,
caption varchar(32) NOT NULL,
PRIMARY KEY (cid)
) ENGINE=InnoDB CHARSET=utf8;
INSERT INTO class VALUES
(1, '三年二班'),
(2, '三年三班'),
(3, '一年二班'),
(4, '二年九班');
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 DEFAULT CHARSET=utf8;
INSERT INTO course VALUES
(1, '生物', 1),
(2, '物理', 2),
(3, '体育', 3),
(4, '美术', 2);
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 DEFAULT CHARSET=utf8;
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);
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 DEFAULT CHARSET=utf8;
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');
CREATE TABLE teacher(
tid int(11) NOT NULL AUTO_INCREMENT,
tname varchar(32) NOT NULL,
PRIMARY KEY (tid)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
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');
1、查询所有的课程的名称以及对应的任课老师姓名
2、查询学生表中男女生各有多少人
3、查询物理成绩等于100的学生的姓名
4、查询平均成绩大于八十分的同学的姓名和平均成绩
5、查询所有学生的学号，姓名，选课数，总成绩
6、 查询姓李老师的个数
7、 查询没有报李平老师课的学生姓名
8、 查询物理课程比生物课程高的学生的学号
9、 查询没有同时选修物理课程和体育课程的学生姓名
10、查询挂科超过两门(包括两门)的学生姓名和班级
11 、查询选修了所有课程的学生姓名
12、查询李平老师教的课程的所有成绩记录
13、查询全部学生都选修了的课程号和课程名
14、查询每门课程被选修的次数
15、查询之选修了一门课程的学生姓名和学号
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
18、查询生物成绩不及格的学生姓名和对应生物分数
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
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
#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
导出现有数据库数据：
- mysqldump -u用户名 -p密码 数据库名称 >导出文件路径 # 结构+数据
- mysqldump -u用户名 -p密码 -d 数据库名称 >导出文件路径 # 结构
导入现有数据库数据：
- mysqldump -uroot -p密码 数据库名称 < 文件路径
[
](javascript:void(0)😉
/*
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;
2、查询“生物”课程比“物理”课程成绩高的所有学生的学号；
思路：
获取所有有生物课程的人（学号，成绩） - 临时表
获取所有有物理课程的人（学号，成绩） - 临时表
根据【学号】连接两个临时表：
学号 物理成绩 生物成绩
然后再进行筛选
select A.student_id,sw,ty from
(select student_id,num as sw from score left join course on score.course_id = course.cid where course.cname = '生物') as A
left join
(select student_id,num as ty from score left join course on score.course_id = course.cid where course.cname = '体育') as B
on A.student_id = B.student_id where sw > if(isnull(ty),0,ty);
3、查询平均成绩大于60分的同学的学号和平均成绩；
思路：
根据学生分组，使用avg获取平均值，通过having对avg进行筛选
select student_id,avg(num) from score group by student_id having avg(num) > 60
4、查询所有同学的学号、姓名、选课数、总成绩；
select score.student_id,sum(score.num),count(score.student_id),student.sname
from
score left join student on score.student_id = student.sid
group by score.student_id
5、查询姓“李”的老师的个数；
select count(tid) from teacher where tname like '李%'
select count(1) from (select tid from teacher where tname like '李%') as B
6、查询没学过“叶平”老师课的同学的学号、姓名；
思路：
先查到“李平老师”老师教的所有课ID
获取选过课的所有学生ID
学生表中筛选
select * from student where sid not in (
select DISTINCT student_id from score where score.course_id in (
select cid from course left join teacher on course.teacher_id = teacher.tid where tname = '李平老师'
)
)
7、查询学过“001”并且也学过编号“002”课程的同学的学号、姓名；
思路：
先查到既选择001又选择002课程的所有同学
根据学生进行分组，如果学生数量等于2表示，两门均已选择
select student_id,sname from
(select student_id,course_id from score where course_id = 1 or course_id = 2) as B
left join student on B.student_id = student.sid group by student_id HAVING count(student_id) > 1
8、查询学过“叶平”老师所教的所有课的同学的学号、姓名；
同上，只不过将001和002变成 in (叶平老师的所有课)
9、查询课程编号“002”的成绩比课程编号“001”课程低的所有同学的学号、姓名；
同第1题
10、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
)
11、查询没有学全所有课的同学的学号、姓名；
思路：
在分数表中根据学生进行分组，获取每一个学生选课数量
如果数量 == 总课程数量，表示已经选择了所有课程
select student_id,sname
from score left join student on score.student_id = student.sid
group by student_id HAVING count(course_id) = (select count(1) from course)
12、查询至少有一门课与学号为“001”的同学所学相同的同学的学号和姓名；
思路：
获取 001 同学选择的所有课程
获取课程在其中的所有人以及所有课程
根据学生筛选，获取所有学生信息
再与学生表连接，获取姓名
select student_id,sname, count(course_id)
from score left join student on score.student_id = student.sid
where student_id != 1 and course_id in (select course_id from score where student_id = 1) group by student_id
13、查询至少学过学号为“001”同学所有课的其他同学学号和姓名；
先找到和001的学过的所有人
然后个数 ＝ 001所有学科 ＝＝》 其他人可能选择的更多
select student_id,sname, count(course_id)
from score left join student on score.student_id = student.sid
where student_id != 1 and course_id in (select course_id from score where student_id = 1) group by student_id having count(course_id) ＝ (select count(course_id) from score where student_id = 1)
14、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；
个数相同
002学过的也学过
select student_id,sname from score left join student on score.student_id = student.sid where student_id in (
select student_id from score where student_id != 1 group by student_id HAVING count(course_id) = (select count(1) from score where student_id = 1)
) and course_id in (select course_id from score where student_id = 1) group by student_id HAVING count(course_id) = (select count(1) from score where student_id = 1)
15、删除学习“叶平”老师课的score表记录；
delete from score where course_id in (
select cid from course left join teacher on course.teacher_id = teacher.tid where teacher.name = '叶平'
)
16、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；
思路：
由于insert 支持
inset into tb1(xx,xx) select x1,x2 from tb2;
所有，获取所有没上过002课的所有人，获取002的平均成绩
insert into score(student_id, course_id, num) select sid,2,(select avg(num) from score where course_id = 2)
from student where sid not in (
select student_id from score where course_id = 2
)
17、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；
select sc.student_id,
(select num from score left join course on score.course_id = course.cid where course.cname = "生物" and score.student_id=sc.student_id) as sy,
(select num from score left join course on score.course_id = course.cid where course.cname = "物理" and score.student_id=sc.student_id) as wl,
(select num from score left join course on score.course_id = course.cid where course.cname = "体育" and score.student_id=sc.student_id) as ty,
count(sc.course_id),
avg(sc.num)
from score as sc
group by student_id desc
18、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, max(num) as max_num, min(num) as min_num from score group by course_id;
19、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；
思路：case when .. then
select course_id, avg(num) as avgnum,sum(case when score.num > 60 then 1 else 0 END)/count(1)*100 as percent from score group by course_id order by avgnum asc,percent desc;
20、课程平均分从高到低显示（现实任课老师）；
select avg(if(isnull(score.num),0,score.num)),teacher.tname from course
left join score on course.cid = score.course_id
left join teacher on course.teacher_id = teacher.tid
group by score.course_id
21、查询各科成绩前三名的记录:(不考虑成绩并列情况)
select score.sid,score.course_id,score.num,T.first_num,T.second_num from score left join
(
select
sid,
(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 3,1) as second_num
from
score as s1
) as T
on score.sid =T.sid
where score.num <= T.first_num and score.num >= T.second_num
22、查询每门课程被选修的学生数；
select course_id, count(1) from score group by course_id;
23、查询出只选修了一门课程的全部学生的学号和姓名；
select student.sid, student.sname, count(1) from score
left join student on score.student_id = student.sid
group by course_id having count(1) = 1
24、查询男生、女生的人数；
select * from
(select count(1) as man from student where gender='男') as A ,
(select count(1) as feman from student where gender='女') as B
25、查询姓“张”的学生名单；
select sname from student where sname like '张%';
26、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;
27、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;
28、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;
29、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 = '生物'
30、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
31、求选了课程的学生人数
select count(distinct student_id) from score
select count(c) from (
select count(student_id) as c from score group by student_id) as A
32、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；
select sname,num from score
left join student on score.student_id = student.sid
where score.course_id in (select course.cid from course left join teacher on course.teacher_id = teacher.tid where tname='张磊老师') order by num desc limit 1;
33、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;
34、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;
35、查询每门课程成绩最好的前两名；
select score.sid,score.course_id,score.num,T.first_num,T.second_num from score left join
(
select
sid,
(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
) as T
on score.sid =T.sid
where score.num <= T.first_num and score.num >= T.second_num
36、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
37、查询全部学生都选修的课程的课程号和课程名；
select course_id,count(1) from score group by course_id having count(1) = (select count(1) from student);
38、查询没学过“叶平”老师讲授的任一门课程的学生姓名；
select student_id,student.sname from score
left join student on score.student_id = student.sid
where score.course_id not in (
select cid from course left join teacher on course.teacher_id = teacher.tid where tname = '张磊老师'
)
group by student_id
39、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,count(1) from score where num < 60 group by student_id having count(1) > 2
40、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;
41、删除“002”同学的“001”课程的成绩；
delete from score where course_id = 1 and student_id = 2
