MySQL - 選擇行的id不存在作爲另一個表中的外鍵

Question

在我的rails應用程序中，我有兩個表 - device_ports和circuits。我的目標是獲得device_ports的列表，其id未被用於circuits表的physical_port_id列中。

我在其他表格上做過類似的事情，但這裏我的查詢只返回一行，當它返回23行時 - 這個設備有24個設備端口，一個正在使用。

select id, name, device_id, multiuse 
from device_ports 
where (device_id = 6 and multiuse = 1) 
or device_ports.id not in (select physical_port_id from circuits) 

所以這個查詢獲取所有多用途端口（這樣即使ID是外鍵引用，該行仍然應該退還），並應該得到的所有行DEVICE_ID是6，但沒有被引用電路，但只有多用途行正在返回。

從查詢結果是

id | name | device_id | multiuse 
------------------------------------ 
268 | test-1 |  6  | 1 

我曾嘗試創建一個SQL小提琴，但在構建似乎只是超時。

CREATE TABLE `device_ports` (
    `id` int(11) NOT NULL AUTO_INCREMENT, 
    `device_id` int(11) DEFAULT NULL, 
    `name` tinytext, 
    `speed` tinytext, 
    `multiuse` tinyint(1) DEFAULT NULL, 
    `created_at` datetime DEFAULT NULL, 
    `updated_at` datetime DEFAULT NULL, 
    PRIMARY KEY (`id`), 
    KEY `id` (`id`) 
) ENGINE=MyISAM AUTO_INCREMENT=291 DEFAULT CHARSET=latin1; 

INSERT INTO `device_ports` (`id`, `device_id`, `name`, `speed`, `multiuse`, `created_at`, `updated_at`) 
*emphasized text*VALUES 
(1, 1, 'Test Device Port', '100', 0, NULL, NULL), 
(2, 1, 'Test Port 2', '300', 1, NULL, NULL), 
(289, 6, 'test-22', '100', 0, NULL, NULL), 
(290, 6, 'test-23', '100', 0, NULL, NULL), 
(288, 6, 'test-21', '100', 0, NULL, NULL), 
(287, 6, 'test-20', '100', 0, NULL, NULL), 
(286, 6, 'test-19', '100', 0, NULL, NULL), 
(284, 6, 'test-17', '100', 0, NULL, NULL), 
(285, 6, 'test-18', '100', 0, NULL, NULL), 
(283, 6, 'test-16', '100', 0, NULL, NULL), 
(282, 6, 'test-15', '100', 0, NULL, NULL), 
(281, 6, 'test-14', '100', 0, NULL, NULL), 
(280, 6, 'test-13', '100', 0, NULL, NULL), 
(279, 6, 'test-12', '100', 0, NULL, NULL), 
(278, 6, 'test-11', '100', 0, NULL, NULL), 
(277, 6, 'test-10', '100', 0, NULL, NULL), 
(276, 6, 'test-9', '100', 0, NULL, NULL), 
(275, 6, 'test-8', '100', 0, NULL, NULL), 
(274, 6, 'test-7', '100', 0, NULL, NULL), 
(273, 6, 'test-6', '100', 0, NULL, NULL), 
(272, 6, 'test-5', '100', 0, NULL, NULL), 
(271, 6, 'test-4', '100', 0, NULL, NULL), 
(270, 6, 'test-3', '100', 0, NULL, NULL), 
(269, 6, 'test-2', '100', 0, NULL, NULL), 
(268, 6, 'test-1', '100', 1, NULL, NULL), 
(267, 6, 'test-0', '100', 0, NULL, NULL); 


CREATE TABLE `circuits` (
    `id` int(11) NOT NULL AUTO_INCREMENT, 
    `organisation_id` int(11) DEFAULT NULL, 
    `physical_port_id` int(11) DEFAULT NULL, 
    PRIMARY KEY (`id`) 
) ENGINE=InnoDB AUTO_INCREMENT=248 DEFAULT CHARSET=latin1; 

INSERT INTO `circuits` (`id`, `organisation_id`, `physical_port_id`) 
VALUES (1, 125, 267); 

Answer 1

你可以嘗試使用LEFT OUTER JOIN：

SELECT DISTINCT d.id, d.name, d.device_id, d.multiuse 
FROM device_ports d 
LEFT OUTER JOIN circuits c ON c.physical_port_id = d.id 
WHERE 
(c.physical_port_id IS NULL AND d.device_id = 6) 
OR (d.multiuse = 1 AND d.device_id = 6) 
ORDER BY d.id 

這有幾種查詢技術，看看What's the difference between NOT EXISTS vs. NOT IN vs. LEFT JOIN WHERE IS NULL?。

Answer 2

SELECT p.* 
    FROM device_ports p 
    LEFT 
    JOIN circuits c 
    ON c.physical_port_id = p.id 
WHERE p.device_id = 6 
    AND multiuse = 1 
    AND c.id IS NULL;