2022-12-05：部门工资前三高的所有员工。编写一个SQL查询找出每个部门中收入前三高的员工 。 +------------+----------+----

DROP TABLE IF EXISTS `department`;
CREATE TABLE `department` (
  `id` int(11) NOT NULL,
  `name` varchar(255) NOT NULL,
  PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1;

INSERT INTO `department` VALUES ('1', 'IT');
INSERT INTO `department` VALUES ('2', 'Sales');

DROP TABLE IF EXISTS `employee`;
CREATE TABLE `employee` (
  `id` int(11) NOT NULL,
  `name` varchar(255) NOT NULL,
  `salary` int(11) NOT NULL,
  `department_id` int(11) DEFAULT NULL,
  PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1;

INSERT INTO `employee` VALUES ('1', 'Joe', '85000', '1');
INSERT INTO `employee` VALUES ('2', 'Henry', '80000', '2');
INSERT INTO `employee` VALUES ('3', 'Sam', '60000', '2');
INSERT INTO `employee` VALUES ('4', 'Max', '90000', '1');
INSERT INTO `employee` VALUES ('5', 'Janet', '69000', '1');
INSERT INTO `employee` VALUES ('6', 'Randy', '85000', '1');
INSERT INTO `employee` VALUES ('7', 'Will', '70000', '1');

SELECT
    d.Name AS 'department', e1.Name AS 'employee', e1.salary
FROM
    employee e1
        JOIN
    department d ON e1.department_id = d.id
WHERE
    3 > (SELECT
            COUNT(DISTINCT e2.salary)
        FROM
            employee e2
        WHERE
            e2.salary > e1.salary
                AND e1.department_id = e2.department_id
        )
;