多表查询练习题
MySQL所有笔记链接
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实验环境（表的状况）：
#看到是所有的员工，就要意识到可能是外连接
#注意多个表的共有元素要确切的指明查询哪一个表中的内容，否则会报错
#因为左表中的内容较多，因此是左外连接
SELECT last_name, e.department_id, department_name
FROM employees e LEFT OUTER JOIN departments d
ON e.department_id = d.department_id #连接条件
#两表连接的方式就是在ON后面声明的，尽管可能不查询连接的属性
SELECT e.job_id, d.location_id
FROM employees e JOIN departments d
ON e.department_id = d.department_id #连接条件
WHERE d.department_id = 90
#请注意本题目的实验环境：employees表中有一个员工是没有部门id的，但是这哥们有奖金
#因此如果是查询所有的员工，必须用外连接，否则这个哥们就被漏掉了（好坑）
#e 是 d表的左外连接， e d 就是 l表的左外连接
SELECT e.last_name, d.department_name, e.commission_pct, l.city
FROM employees e LEFT JOIN departments d
ON e.department_id = d.department_id
LEFT JOIN locations l
ON d.location_id = l.location_id
WHERE e.commission_pct IS NOT NULL #查询条件
SELECT e.last_name, e.department_id, d.department_name, e.job_id
FROM employees e LEFT JOIN departments d
ON e.department_id = d.department_id
JOIN locations l
ON d.location_id = l.location_id
WHERE l.city = 'Toronto'
#你可能会有这样的疑问：在连接的时候不是左外连接，那么那个没有部门的哥们就被排除在外了
#那么如果他的工作city就在Toronto怎么办呢?
#A:因为没有工作部门，所以一定没有city,有city依赖于有工作部门
SELECT d.department_name,l.street_address,e.last_name, e.job_id, e.salary
FROM employees e JOIN departments d
ON e.department_id = d.department_id
JOIN locations l
ON d.location_id = l.location_id
WHERE d.department_name = 'Executive'
SELECT d.department_id
FROM departments d LEFT JOIN employees e
ON d.department_id = e.department_id
WHERE e.department_id IS NULL
SELECT l.location_id,l.city
FROM locations l LEFT JOIN departments d
ON l.location_id = d.location_id
WHERE d.location_id IS NULL
SELECT e.last_name, e.employee_id, e.department_id
FROM employees e JOIN departments d
ON e.department_id = d.department_id
WHERE d.department_name = 'Sales' OR d.department_name = 'IT'
#WHERE d.department_name IN ('Sales','IT')
