【MySQL】连接查询

545 06-08
这套题还不错，感兴趣的猿可以试一试：前端开发工程师
/*
含义：又称多表查询，当查询的字段来自于多个表时，就会用到连接查询

笛卡尔乘积现象：表1 有m行，表2有n行，结果=m*n行

发生原因：没有有效的连接条件
如何避免：添加有效的连接条件	
	按功能分类：
		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
*/
#案例1：查询女神名和对应的男神名
SELECT NAME,boyName 
FROM boys,beauty
WHERE beauty.boyfriend_id= boys.id;

#案例2：查询员工名和对应的部门名

SELECT last_name,department_name
FROM employees,departments
WHERE employees.`department_id`=departments.`department_id`;

#2、为表起别名

#查询员工名、工种号、工种名

SELECT e.last_name,e.job_id,j.job_title
FROM employees  e,jobs j
WHERE e.`job_id`=j.`job_id`;
#4、可以加筛选

#案例：查询有奖金的员工名、部门名

SELECT last_name,department_name,commission_pct

FROM employees e,departments d
WHERE e.`department_id`=d.`department_id`
AND e.`commission_pct` IS NOT NULL;

#案例2：查询城市名中第二个字符为o的部门名和城市名

SELECT department_name,city
FROM departments d,locations l
WHERE d.`location_id` = l.`location_id`
AND city LIKE '_o%';

#5、可以加分组

#案例1：查询每个城市的部门个数

SELECT COUNT(*) 个数,city
FROM departments d,locations l
WHERE d.`location_id`=l.`location_id`
GROUP BY city;

#案例2：查询有奖金的每个部门的部门名和部门的领导编号和该部门的最低工资
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_name,d.`manager_id`;
#6、可以加排序

#案例：查询每个工种的工种名和员工的个数，并且按员工个数降序

SELECT job_title,COUNT(*)
FROM employees e,jobs j
WHERE e.`job_id`=j.`job_id`
GROUP BY job_title
ORDER BY COUNT(*) DESC;

#7、可以实现三表连接？

#案例：查询员工名、部门名和所在的城市

SELECT last_name,department_name,city
FROM employees e,departments d,locations l
WHERE e.`department_id`=d.`department_id`
AND d.`location_id`=l.`location_id`
AND city LIKE 's%'

ORDER BY department_name DESC;

#2、非等值连接

#案例1：查询员工的工资和工资级别
SELECT salary,grade_level
FROM employees e,job_grades g
WHERE salary BETWEEN g.`lowest_sal` AND g.`highest_sal`
AND g.`grade_level`='A';

#3、自连接

#案例：查询 员工名和上级的名称

SELECT e.employee_id,e.last_name,m.employee_id,m.last_name
FROM employees e,employees m
WHERE e.`manager_id`=m.`employee_id`;
#二、sql99语法

语法：
	select 查询列表
	from 表1 别名 【连接类型】
	join 表2 别名 
	on 连接条件
	【where 筛选条件】
	【group by 分组】
	【having 筛选条件】
	【order by 排序列表】
	

分类：
内连接（★）：inner
外连接
	左外(★):left 【outer】
	右外(★)：right 【outer】
	全外：full【outer】
交叉连接：cross 
#一）内连接

语法：

select 查询列表
from 表1 别名
inner join 表2 别名
on 连接条件;
特点：
①添加排序、分组、筛选
②inner可以省略

#案例1.查询员工名、部门名

SELECT last_name,department_name
FROM departments d
 JOIN  employees e
ON e.`department_id` = d.`department_id`;

#案例2.查询名字中包含e的员工名和工种名（添加筛选）
SELECT last_name,job_title
FROM employees e
INNER JOIN jobs j
ON e.`job_id`=  j.`job_id`
WHERE e.`last_name` LIKE '%e%';

#3. 查询部门个数>3的城市名和部门个数，（添加分组+筛选）

#①查询每个城市的部门个数
#②在①结果上筛选满足条件的
SELECT city,COUNT(*) 部门个数
FROM departments d
INNER JOIN locations l
ON d.`location_id`=l.`location_id`
GROUP BY city
HAVING COUNT(*)>3;

#案例4.查询哪个部门的员工个数>3的部门名和员工个数，并按个数降序（添加排序）

#①查询每个部门的员工个数
SELECT COUNT(*),department_name
FROM employees e
INNER JOIN departments d
ON e.`department_id`=d.`department_id`
GROUP BY department_name

#② 在①结果上筛选员工个数>3的记录，并排序

SELECT COUNT(*) 个数,department_name
FROM employees e
INNER JOIN departments d
ON e.`department_id`=d.`department_id`
GROUP BY department_name
HAVING COUNT(*)>3
ORDER BY COUNT(*) DESC;

#5.查询员工名、部门名、工种名，并按部门名降序（添加三表连接）

SELECT last_name,department_name,job_title
FROM employees e
INNER JOIN departments d ON e.`department_id`=d.`department_id`
INNER JOIN jobs j ON e.`job_id` = j.`job_id`

ORDER BY department_name DESC;

#查询员工的工资级别

SELECT salary,grade_level
FROM employees e
 JOIN job_grades g
 ON e.`salary` BETWEEN g.`lowest_sal` AND g.`highest_sal`;
 
 #查询工资级别的个数>20的个数，并且按工资级别降序
 SELECT COUNT(*),grade_level
FROM employees e
 JOIN job_grades g
 ON e.`salary` BETWEEN g.`lowest_sal` AND g.`highest_sal`
 GROUP BY grade_level
 HAVING COUNT(*)>20
 ORDER BY grade_level DESC;
 
 #三）自连接
 
 #查询员工的名字、上级的名字
 SELECT e.last_name,m.last_name
 FROM employees e
 JOIN employees m
 ON e.`manager_id`= m.`employee_id`;
 
  #查询姓名中包含字符k的员工的名字、上级的名字
 SELECT e.last_name,m.last_name
 FROM employees e
 JOIN employees m
 ON e.`manager_id`= m.`employee_id`
 WHERE e.`last_name` LIKE '%k%';
 
 #二、外连接 

 #左外连接
 SELECT b.*,bo.*
 FROM boys bo
 LEFT OUTER JOIN beauty b
 ON b.`boyfriend_id` = bo.`id`
 WHERE b.`id` IS NULL; 
 #案例1：查询哪个部门没有员工
 #左外
 SELECT d.*,e.employee_id
 FROM departments d
 LEFT OUTER JOIN employees e
 ON d.`department_id` = e.`department_id`
 WHERE e.`employee_id` IS NULL;  
 #右外 
  SELECT d.*,e.employee_id
 FROM employees e
 RIGHT OUTER JOIN departments d
 ON d.`department_id` = e.`department_id`
 WHERE e.`employee_id` IS NULL;
 #全外
 USE girls;
 SELECT b.*,bo.*
 FROM beauty b
 FULL OUTER JOIN boys bo
 ON b.`boyfriend_id` = bo.id;
标签： MYSQL