php - 从 MySQL 连接中的第一个表中获取所有结果

Question

我有两张 table 。我需要从第一个表中选择所有行(只有一个条件 hotel_id=2)，并从第二个表中选择行(基于条件)。但是我使用左连接只从第二个表中获取数据。

查询

 SELECT R.name room_name,
        R.id room_id,
        UD.discount 
   FROM user_discounts UD 
   LEFT 
   JOIN rooms R 
     ON R.id = UD.room_id
  WHERE UD.user_id = 1482
    AND UD.hotel_id = 2

我需要显示所有的房间，但现在在两个表中显示共同点。

Answer 1

如果你想要所有房间，反转表关系

SELECT `R`.`name` as `room_name`, `R`.`id` as `room_id`, `UD`.`discount` as `discount`
FROM  `rooms` as `R`
LEFT JOIN `user_discounts` as `UD` ON `R`.`id`= `UD`.`room_id`
             AND`UD`.`user_id` = '1482'
             AND `UD`.`hotel_id` = '2'

但是您还需要更改 where 子句。它们可以代替原来的 where 子句用作连接条件的一部分。

where 子句的作用很容易被忽略，但如果您在 where 子句中引用左连接表，您还必须允许该表中的数据为 NULL。例如

SELECT `R`.`name` as `room_name`, `R`.`id` as `room_id`, `UD`.`discount` as `discount`
FROM  `rooms` as `R`
LEFT JOIN `user_discounts` as `UD` ON `R`.`id`= `UD`.`room_id`
WHERE (`UD`.`user_id` = '1482'
   AND `UD`.`hotel_id` = '2'
      )
   OR `UD`.`room_id` IS NULL

---

如果 table rooms 有一个 hotel_id 那么:

SELECT `R`.`name` as `room_name`, `R`.`id` as `room_id`, `UD`.`discount` as `discount`
FROM  `rooms` as `R`
LEFT JOIN `user_discounts` as `UD` ON `R`.`id`= `UD`.`room_id`
             AND`UD`.`user_id` = '1482'
             AND `UD`.`hotel_id` = `R`.`hotel_id`
WHERE `R`.`hotel_id` = 2