【发布时间】:2018-08-10 19:50:25
【问题描述】:
表user 和clinic 之间存在多对多关系,第三个表是user_clinics。所有三个表都完美地单独返回它们的值,但是当我调用 App\User::find(1)->clinics 或其逆时它返回 null。此外,user_clinic 具有 user_id 和 clinic_id 以及 previlage_id 作为外键。
public function users() {
return $this->belongsToMany(User::class,'user_clinics','user_id','clinic_id');
}
public function clinics() {
return $this->belongsToMany(Clinic::class,'user_clinics','clinic_id','user_id');
}
public function adminDashboard(Request $request) {
$clinic = new Clinic();
$User_clinic = new User_clinic();
$user = new User();
$clinic->name = $request->name;
$clinic->address = $request->address;
if($request->hasFile('logo')) {
$fileName = $request->logo->getClientOriginalName();
$request->logo->storeAs('public/logos',$fileName);
$clinic->logo = $request->logo;
}
$clinic->save();
$User_clinic->user_id = auth::user()->id;
$test=$User_clinic->clinic_id = $clinic->id;
//now hardcoded previlage_id but deal with it in future...
$User_clinic->previlage_id = 1;
$User_clinic->save();
$test= $clinic::find(2)->users;
dd($test);
//return view("admin.dashboard.dashboardFirstPage");
}
【问题讨论】:
-
我在数据库 user_clinics 中有一个表并创建它的模型 user_clinic 以使其与另一个优先表的关系。没事还是有什么问题?