【发布时间】:2018-01-18 11:39:36
【问题描述】:
如何在 $query 中传递 _user_id 列值。这是有很多关系。我无法弄清楚如何将 RFX 的 user_id 用户输入到 $query where 条件中。
public function response_pricings(){
return $this->hasMany('App\Models\Website\RFXRequestPricingResponse', ['rfx_request_id'=>'_rfx_request_id', 'user_id'=>'_user_id'])->selectRaw("*");
}
return RFXRequestSupplierResponded::select(
'id as _id',
'rfx_request_id as _rfx_request_id',
'user_id as _user_id',
'status',
'is_qualified',
DB::RAW('(SELECT name FROM users WHERE id=user_id) as user_name'),
DB::RAW('(SELECT note FROM rfx_request_response_notes WHERE rfx_request_id='.$rfx_request_id.' AND user_id=rfx_request_suppliers_responded.user_id LIMIT 1) as note')
)
->with(
[
'response_pricings' => function ($query) {
/*$query->where('user_id', $_user_id);*/
}
]
)
->where('rfx_request_id',$rfx_request_id)
->get();
【问题讨论】: