MySQL连接查询
进阶6：多表连接查询
笛卡尔乘积：如果连接条件省略或无效则会出现
解决办法：添加上连接条件
一、传统模式下的连接 ：等值连接——非等值连接
1.等值连接的结果 = 多个表的交集
2.n表连接，至少需要n-1个连接条件
3.多个表不分主次，没有顺序要求
4.一般为表起别名，提高阅读性和性能
二、sql99语法：通过join关键字实现连接
内连接：等值连接，非等值连接，自连接
外连接：左外连接，右外连接，全外连接
交叉连接
含义：1999年推出的sql语法
支持：
等值连接、非等值连接 （内连接）
外连接
交叉连接
语法：
select 字段，...
from 表1
【inner|left outer|right outer|cross】join 表2 on 连接条件
【inner|left outer|right outer|cross】join 表3 on 连接条件
【where 筛选条件】
【group by 分组字段】
【having 分组后的筛选条件】
【order by 排序的字段或表达式】
好处：语句上，连接条件和筛选条件实现了分离，简洁明了！
mysql> select name,boyName from boys,beauty where boys.id= beauty.boyfriend_id;
查询员工名和对应的部门名
mysql> select last_name,department_name from employees,departments where employees.department_id=departments.department_id;
查询员工名、工种号、工种名
mysql> select last_name,department_name from employees,departments where employees.department_id=departments.department_id;
为表其别名
mysql> select last_name,e.job_id,job_title from employees e,jobs j where e.job_id =j.job_id;
起别名前面再用完整表名就不好用了
加筛选
查询城市名中第二个字母为o的部门名和城市名
三、自连接
案例：查询员工名和直接上级的名称
都在这个表中，建立某两个记录的关系，一张表找了两遍
mysql> select e.last_name,m.last_name,e.manager_id from employees e, employees m where e.manager_id=m.employee_id;
+-------------+-----------+------------+
| last_name | last_name | manager_id |
+-------------+-----------+------------+
| Kochhar | K_ing | 100 |
| De Haan | K_ing | 100 |
| Hunold | De Haan | 102 |
sql99
SELECT e.last_name,m.last_name
FROM employees e
JOIN employees m ON e.`manager_id`=m.`employee_id`;
sql92
SELECT e.last_name,m.last_name
FROM employees e,employees m
WHERE e.`manager_id`=m.`employee_id`;
**********92语法容易混淆，没有join
mysql> select department_name,city from departments d,locations l where d.location_id =l.location_id and city like'_o%';
+-----------------+---------------------+
| department_name | city |
+-----------------+---------------------+
| IT | Southlake |
| Shi | South San Francisco |
| Mar | Toronto |
| Hum | London |
+-----------------+---------------------+
加分组
查询每个城市的部门个数
mysql> select count(*) num,city from departments d,locations l where d.location_id=l.location_id group by city;
+-----+---------------------+
| num | city |
+-----+---------------------+
| 1 | London |
| 1 | Munich |
| 1 | Oxford |
| 21 | Seattle |
查询有奖金的每个部门名和部门的领导编号和该部门的最低工资
mysql> select department_name,d.manager_id,min(salary) from departments d,employees e where d.department_id=e.department_id and commission_pct is not null group by department_namee,d.manager_id;
+-----------------+------------+-------------+
| department_name | manager_id | min(salary) |
+-----------------+------------+-------------+
| Sal | 145 | 6100.00 |
+-----------------+------------+-------------+
加排序
查询每个工种的工种名和员工的个数，并按照员工个数降序
mysql> select job_title,count(*) from jobs j,employees e where e.job_id=j.job_id group by job_title order by count(*) desc;
+---------------------------------+----------+
| job_title | count(*) |
+---------------------------------+----------+
| Sales Representative | 30 |
| Shipping Clerk | 20 |
#非等值连接：
大于小于等等
查询员工的工资和工资级别
mysql> select employee_id,salary,grade_level from employees e,job_grades j where salary between j.lowest_sal and j.highest_sal;
+-------------+----------+-------------+
| employee_id | salary | grade_level |
+-------------+----------+-------------+
| 100 | 24000.00 | E |
| 101 | 17000.00 | E |
| 102 | 17000.00 | E |
mysql> select department_name,city from departments d inner join locations l on d.location_id =l.location_id where city like'_o%';
*****inner 可以省略
Inner join 对应了92语法中的等值连接
查询哪个部门没员工
mysql> select d.*,employee_id from departments d left outer join employees e on d.department_id=e.department_id where e.employee_id is null;
#外连接
用于查询一个表中有一个表中没有
查询不在男神表男朋友的女生
如果从表中有和它匹配的则显示匹配的值
如果没有则显示null
主表从表
左外连接，left join左边的是主表
右外连接，right join右边的是主表
左外和右外交换两个表的顺序实现同样的效果
mysql> select b.name,bo.* from beauty b left outer join boys bo on b.boyfriend_id=bo.id where bo.id is null; 最好选主键
全外连接，不能执行
mysql> select b.*,bo.* from breauty b full outer join boys bo on b.boyfriend_id=bo.id;
交叉连接，其实就是笛卡尔乘积
mysql> select b.*,bo.* from beauty b cross join boys bo;
mysql> select last_name,e.job_id,job_title from employees e inner join jobs j on e.job_id =j.job_id;
