[英]Get all results from first table in a MySQL join
我有兩張桌子。 我需要從第一張表中選擇所有行(僅一個條件hotel_id = 2),並從第二張表中選擇所有行(基於條件)。 但是我使用的左聯接僅從第二張表中獲取數據。
詢問
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
我需要顯示所有房間,但現在在兩個表中顯示公共房間。
如果要所有房間,請反轉表關系
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,您還需要更改where子句。 它們可以代替您的原始where子句用作連接條件的一部分。
可以很容易地忽略where子句的影響,但是,如果您在where子句中引用左連接表,則還必須允許該表中的數據為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
如果餐桌房間有hotel_id,則:
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
據我了解,您需要第一張表中的所有數據,而僅第二張表中符合您的條件的數據。
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'
根據您的查詢, where
部分將左聯接變為內部聯接,因此您不會從第一張表中獲取所有行,要解決此問題,請在on
子句中移動第二張表的過濾器,因此只有符合條件的記錄才是從第二張表返回
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