你好我試圖找出如何在口若懸河,你可以定義這種關係定義一個鏈接表的主鍵從表關係例如:用雄辯的Laravel
return $this->belongsToMany('App\Models\EquipmentType','equipment_types_manufacturers',
'manufacturer_id', 'equipment_type_id');
但是,試圖定義關係,我找不到你如何定義從equipment_model鏈接FK etml_id - > equipment_types_manufacturers PK
return $this->hasMany('App\Models\EquipmentType','equipment_types_manufacturers',
'equipment_types_manufacturers_id', 'equipment_type_id');
其中在SQL失敗
SQLSTATE [42S22]:列未找到:1054未知列在 'where子句'(SQL 'manufacturers.equipment_types_manufacturers_id':SELECT * FROM
equipment_models
其中存在(SELECT * FROMmanufacturers
哪裏equipment_models
。equipment_types_manufacturers_id
=manufacturers
。equipment_types_manufacturers_id
和manufacturer_id
= 1)和存在(來自equipment_types
其中equipment_models
選擇*。equipment_types_manufacturers_id
=equipment_types
。equipment_types_manufacturers_id
和equipment_type_id
= 1))
我感覺這可以通過創建的鏈接表的模型來實現,但是我」我不確定這是否正確?
更新:我現在想用這個 $車型查詢= EquipmentModel ::有( 'equipmentTypesManufacturer.manufacturer') - >使用( [ 'equipmentTypesManufacturer'=>功能($查詢){$ 查詢 - >其中([[ 'MANUFACTURER_ID',5],[ 'equipment_type_id',5]]);} , 'equipmentTypesManufacturer.manufacturer' , 'equipmentTypesManufacturer.equipmentType']) - > get() ; 這將工作,但你會永遠得到頂級模型? – jwtea