你好我试图找出如何在口若悬河,你可以定义这种关系定义一个链接表的主键从表关系例如:用雄辩的Laravel
return $this->belongsToMany('App\Models\EquipmentType','equipment_types_manufacturers',
'manufacturer_id', 'equipment_type_id');
但是,试图定义关系,我找不到你如何定义从equipment_model链接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失败
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
哪里equipment_models
。equipment_types_manufacturers_id
=manufacturers
。equipment_types_manufacturers_id
和manufacturer_id
= 1)和存在(来自equipment_types
其中equipment_models
选择*。equipment_types_manufacturers_id
=equipment_types
。equipment_types_manufacturers_id
和equipment_type_id
= 1))
我感觉这可以通过创建的链接表的模型来实现,但是我”我不确定这是否正确?
更新:我现在想用这个 $车型查询= EquipmentModel ::有( 'equipmentTypesManufacturer.manufacturer') - >使用( [ 'equipmentTypesManufacturer'=>功能($查询){$ 查询 - >其中([[ 'MANUFACTURER_ID',5],[ 'equipment_type_id',5]]);} , 'equipmentTypesManufacturer.manufacturer' , 'equipmentTypesManufacturer.equipmentType']) - > get() ; 这将工作,但你会永远得到顶级模型? – jwtea