用不同的值加入同一個表兩次

Question

我正在嘗試使用 ID 將“transfers.pickup_areas_group_id”和“transfers.drop_areas_group_id”的值替換為“areas_group”表中的值

我正在使用這個查詢：

SELECT 
transfers.id AS transfer_id, 
transfers.name AS transfer_name,  
transfers.pickup_areas_group_id AS transfer_pickup_areas_group_id,
transfers.drop_areas_group_id AS transfer_drop_areas_group_id, 
transfers_pricing.vehicle_id AS vehicle_id, 
transfers_pricing.date_start AS date_start, 
transfers_pricing.date_end AS date_end, 
transfers_pricing.price AS price 
FROM transfers
INNER JOIN transfers_pricing ON transfers_pricing.transfer_id = transfers.id

我嘗試了一個額外的 INNER JOIN 來替換第一個值“transfers.pickup_areas_group_id”，但我找不到替換第二個值“transfers.drop_areas_group_id”的方法

我試過這個查詢：

SELECT 
transfers.id AS transfer_id, 
transfers.name AS transfer_name,  
transfers.pickup_areas_group_id AS transfer_pickup_areas_group_id,
areas_group.area_id AS pickup_area_ids,
transfers.drop_areas_group_id AS transfer_drop_areas_group_id, 
transfers_pricing.vehicle_id AS vehicle_id, 
transfers_pricing.date_start AS date_start, 
transfers_pricing.date_end AS date_end, 
transfers_pricing.price AS price 
FROM transfers
INNER JOIN transfers_pricing ON transfers_pricing.transfer_id = transfers.id
INNER JOIN areas_group ON areas_group.id = transfers.pickup_areas_group_id

謝謝，

Answer 1

基本上，您需要再次加入areas_group ； 要消除對同一個表的兩個引用的歧義，您需要使用表別名。

實際上，對查詢中起作用的所有表使用表別名是一種很好的做法：這使得查詢的讀寫時間更短。

SELECT 
    t.id AS transfer_id, 
    t.name AS transfer_name,  
    t.pickup_areas_group_id AS transfer_pickup_areas_group_id,
    ag1.area_id AS pickup_area_ids,
    t.drop_areas_group_id AS transfer_drop_areas_group_id, 
    ag2.area_id AS drop_area_ids
    tp.vehicle_id AS vehicle_id, 
    tp.date_start AS date_start, 
    tp.date_end AS date_end, 
    tp.price AS price 
FROM transfers t
INNER JOIN transfers_pricing tp ON tp.transfer_id = t.id
INNER JOIN areas_group ag1 ON ag1.id = t.pickup_areas_group_id
INNER JOIN areas_group ag2 ON ag2.id = t.drop_areas_group_id