03、MySQL连接查询之内连接

在连接查询之前，同样导入一份sql文件。

连接查询

笛卡尔乘积：表一有m行，表二有n行，结果 = m*n 行。

内连接

#进阶七、连接查询、多表查询
/*
当我们要查询的字段涉及多张表时，这时我们就会用到连接查询。
*/
#笛卡尔乘积
SELECT NAME, boyname FROM beauty, boys;

#正确的写法(等值连接)
SELECT NAME, boyname
FROM beauty, boys
WHERE beauty.`boyfriend_id` = boys.`id`;

/*
连接查询的分类：
1、按年代分
	SQL92标准:在MySQL中仅仅支持内连接
	SQL99标准（推荐）：支持所有内连接、外连接（左外连接、右外连接）、交叉连接
2、按功能分
	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
*/
#一、---------------------------------SQL92语法----------------------------------
#内连接
-- 1、等值连接
/*
语法：
select 查询列表
from 表名1, 表名2, ...
where 等值连接条件;

1、因为会涉及多张表，容易造成字段名重复，为了解决这个问题，往往为表起别名;
2、from子句表的顺序无要求;
*/
#查询员工名和部门名
SELECT last_name, department_name
FROM employees e, departments d
WHERE e.`department_id` = d.`department_id`;

#查询部门编号>100的部门名和城市名
SELECT department_id, city
FROM departments d, locations l
WHERE d.`department_id` > 100 AND d.`location_id` = l.`location_id`;

#查询有奖金的员工名和部门名
SELECT last_name, department_name
FROM employees e, departments d
WHERE e.`commission_pct` IS NOT NULL AND e.`department_id` = d.`department_id`;

#查询城市名第二个字符为o的部门名和城市名
SELECT department_name, city
FROM departments d, locations l
WHERE l.`city` LIKE '_o%' AND d.`location_id` = l.`location_id`;

#查询每个城市的部门个数
SELECT COUNT(*) 部门个数, d.`location_id`, l.`city`
FROM departments d, locations l
WHERE d.`location_id` = l.`location_id`
GROUP BY d.`location_id`;

#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 d.department_name, d.manager_id, MIN(salary)
FROM employees e, departments d
WHERE e.`commission_pct` IS NOT NULL AND e.`department_id` = d.`department_id`
GROUP BY e.`department_id`;

#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(*) 员工个数
FROM employees e, jobs j
WHERE e.`job_id` = j.`job_id`
GROUP BY e.`job_id`
ORDER BY 员工个数 DESC;

#三表连接：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`;

-- 2、非等值连接
CREATE TABLE job_grades(
  grade_level VARCHAR(3),
  lowest_sal  INT,
  highest_sal INT
);

INSERT INTO job_grades VALUES ('A', 1000, 2999);
INSERT INTO job_grades VALUES ('B', 3000, 5999);
INSERT INTO job_grades VALUES('C', 6000, 9999);
INSERT INTO job_grades VALUES('D', 10000, 14999);
INSERT INTO job_grades VALUES('E', 15000, 24999);
INSERT INTO job_grades VALUES('F', 25000, 40000);

#查询出员工的工资和工资级别
SELECT salary, grade_level
FROM employees e, job_grades j
WHERE (e.`salary` BETWEEN j.`lowest_sal` AND j.`highest_sal`) AND grade_level IN ('A', 'B', 'C')
ORDER BY salary DESC;

-- 3、自连接
#相当于等值连接，只不过涉及到的表只有自己一张
#查询员工名以及他的领导的名称
SELECT e1.`last_name`, e1.`manager_id`, e2.`last_name`, e2.`employee_id`
FROM employees e1, employees e2
WHERE e1.`manager_id` = e2.`employee_id`
ORDER BY e1.`last_name`;

#二、---------------------------------SQL99语法----------------------------------
#内连接
/*
语法：
select 查询列表
from 表1 别名1
[连接类型] join 表2 别名2
on 连接条件
where 筛选条件;

内连：[inner]
左外：left [outer]
右外：right [outer]
全外：full [outer]
交叉：cross 
*/
-- 1、等值连接
#查询员工名、部门名
SELECT last_name, department_name
FROM employees e
INNER JOIN departments d
ON e.`department_id` = d.`department_id`;

#查询名字中包含e的员工名和工种名
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%';

#查询部门个数大于3的部门个数和城市名
SELECT COUNT(*) 部门个数, l.`city`
FROM departments d
INNER JOIN locations l
ON d.`location_id` = l.`location_id`
GROUP BY d.`location_id`
HAVING COUNT(*) > 3;

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

#查询员工名、部门名、工种名，并按部门名降序
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;

-- 2、非等值连接
#查询员工工资和工资级别
SELECT last_name, salary, grade_level
FROM employees e
JOIN job_grades j ON e.`salary` BETWEEN j.`lowest_sal` AND j.`highest_sal`;

#查询每个工资级别的个数,并按照级别降序
SELECT last_name, salary, grade_level, COUNT(*) 个数
FROM employees e
JOIN job_grades j ON e.`salary` BETWEEN j.`lowest_sal` AND j.`highest_sal`
GROUP BY grade_level
ORDER BY grade_level DESC;

-- 3、自连接
#查询员工的名字和领导的名字
SELECT e1.last_name, e1.`manager_id`, e2.`last_name`, e2.`employee_id`
FROM employees e1
JOIN employees e2
ON e1.`manager_id` = e2.`employee_id`;