Question

public function chat($id,$team1,$team2){

    $relation=Crelation::where('match_id',$id)->where('first_team_id',$team1)->where('second_team_id',$team2)->first();

    if($relation == null){

        $data=[
            'match_id'=>$id,
            'first_team_id'=>$team1,
            'second_team_id'=>$team2
        ];

        $rel=  Crelation::create($data);
       $whatRelation=$rel->id;
        $this->sendMessage($whatRelation);

    }else{

    $whatRelation=$relation->id;
        $this->sendMessage($whatRelation);
    }

    return view('chat',compact('whatRelation'));
}


public function sendMessage(Request $request,$whatRelation)
{

    $id=(int)$whatRelation;

$user = Auth::user();

$message = $user->messages()->create([
    'message' => $request->input('message'),
    'crelation_id'=>$id


]);

broadcast(new MessageSent($user, $message))->toOthers();

return ['status' => 'Message Sent!'];
}

我收到此错误：

参数1传递给 App \ Http \ Controllers \ ChatsController :: sendMessage（）必须是一个 Illuminate \ Http \ Request的实例，给定的整数，调用 C：\ XAMPP \ htdocs中\ ScrimWithMe \程序\ HTTP \控制器\ ChatsController.php 在第77行并定义

Answer 1

chat函数需要两个参数..但您只是传递一个参数..

您可以做的是在public function chat(Request $request, $id,$team1,$team2){ .... $this->sendMessage($request,$whatRelation); }功能聊天中添加另一个参数，例如

container

然后在所述函数中添加一个参数，应该这样做..

Answer 2

@Demonyowh解释了为什么会出现此错误

我认为factory design模式有两种方法可以解决您的问题

如果你的方法没有在任何其他地方使用，那么删除第一个参数并在下一行声明类
```
METHOD(){
  $request = new Request();
  // your code;
}
```
在__construct方法

我想从同一个控制器传递不同函数的变量 - Laravel

2 个答案: