我正在嘗試使用 ID 將“transfers.pickup_areas_group_id”和“transfers.drop_areas_group_id”的值替換為表“areas_group”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 transfersINNER 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 transfersINNER JOIN transfers_pricing ON transfers_pricing.transfer_id = transfers.idINNER JOIN areas_group ON areas_group.id = transfers.pickup_areas_group_id謝謝，
1 回答
PIPIONE
TA貢獻1829條經(jīng)驗 獲得超9個贊
基本上，您需要另一個加入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，對在查詢中起作用的所有表使用表別名是一個很好的做法：這使得查詢更易于讀寫。
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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