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php - 用不同的值加入同一个表两次

问题描述

我正在尝试使用 ID 将“transfers.pickup_areas_group_id”和“transfers.drop_areas_group_id”的值替换为“areas_group”表中的值

我正在使用这个查询：

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

我尝试了一个额外的 INNER JOIN 来替换第一个值“transfers.pickup_areas_group_id”，但我找不到替换第二个值“transfers.drop_areas_group_id”的方法

我试过这个查询：

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

谢谢，

标签： phpmysqlsqldatabasejoin

解决方案

基本上，你需要另一个加入areas_group；要消除对同一个表的两个引用的歧义，您需要使用表别名。

实际上，对查询中起作用的所有表使用表别名是一种很好的做法：这使得查询的读写时间更短。

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

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