Question

在 mysql 中，使用连接表 employees_positions 在员工和职位表之间建立了m-n关系。我找到了一种方法来传递position_id并检索具有该位置的员工。

但是我也会传递两个[或更多] position_id，并检索已经传递了两个[或所有] position_id的员工。

例如，寻找开发人员和经理将返回BOB和FRANK

数据库创建代码：

CREATE TABLE `employees` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `name` varchar(50) NOT NULL,
  PRIMARY KEY (`id`)
) ;

INSERT INTO `employees` (`id`, `name`) VALUES
    (1, 'bob'),
    (2, 'frank'),
    (3, 'jim'),
    (4, 'paul'),
    (5, 'mick');

CREATE TABLE `employees_positions` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `employe_id` int(11) NOT NULL,
  `position_id` int(11) NOT NULL,
  PRIMARY KEY (`id`),
  UNIQUE KEY `Index 2` (`position_id`,`employe_id`)
);

INSERT INTO `employees_positions` (`id`, `employe_id`, `position_id`) VALUES
    (1, 1, 1),
    (2, 2, 1),
    (3, 3, 3),
    (4, 4, 3),
    (5, 5, 3),
    (6, 1, 2),
    (7, 1, 3),
    (8, 2, 2);

CREATE TABLE `positions` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `name` varchar(50) NOT NULL,
  PRIMARY KEY (`id`)
) ;

INSERT INTO `positions` (`id`, `name`) VALUES
    (1, 'manager'),
    (2, 'developer'),
    (3, 'executive');

目前我正在使用此查询来检索返回员工的职位：

SELECT
  name,
  (SELECT
      GROUP_CONCAT(p.name)
    FROM positions AS p
      LEFT JOIN employees_positions AS e_p
        ON p.id = e_p.position_id
    WHERE e_p.employe_id = e.id) AS positionsHeld,
  (SELECT
      GROUP_CONCAT(p.id)
    FROM positions AS p
      LEFT JOIN employees_positions AS e_p
        ON p.id = e_p.position_id
    WHERE e_p.employe_id = e.id) AS positionIDs

FROM employees AS e;

但我想只返回包含所有请求的元素 positions_id

Answer 1

使用HAVING子句确保他们两个位置：

SELECT e.name
FROM employees e
JOIN employees_positions e_p ON e_p.employe_id = e.id
WHERE e_p.position_id in (1,2)
GROUP BY 1
HAVING count(*) = 2

请参阅SQLFiddle

请注意，您不需要group_concat()或加入position表，但如果确实需要：

SELECT e.name, group_concat(p.name order by 1) postitions
FROM employees e
JOIN employees_positions e_p ON e_p.employe_id = e.id
JOIN positions p ON p.id = e_p.position_id
WHERE p.id in (1,2)
GROUP BY 1
HAVING count(*) = 2

请参阅上面的相同SQLFiddle（它有两个查询）。

Answer 2

波西米亚人给出了一个很好的答案，只要一名员工只被分配到一个职位一次另一种方法是：

select emp.id, emp.name, 
       group_concat(p.name) as positions_held, 
       group_concat(p.id) as position_ids
  from employees emp
  join employees_positions epos on epos.employe_id = emp.id
  join positions p on p.id = epos.position_id
  join employees_positions epos1 on epos1.employe_id = emp.id and epos1.position_id = 1
  join employees_positions epos2 on epos2.employe_id = emp.id and epos2.position_id = 2
group by emp.id, emp.name;

当您添加更多职位时，您只需添加更多行：

  join employees_positions epos1 on epos1.employe_id = emp.id and epos1.position_id = 1

group_concat将为您提供员工所有职位的列表，而不仅仅是与提供的职位相匹配的职位。但是提供的位置数据是静态的，并且不会逐行更改，因此您可以使用应用程序插入该数据。

Link to SQL Fiddle

使用AND进行子查询的mysql - 过滤结果集

2 个答案: