Question

我的控制器：

 public function show($id){
    $user_id_1_connections = Connection::whereUser_id_1AndConnection_status($id, 1)->get();
    $user_id_2_connections = Connection::whereUser_id_2AndConnection_status($id, 1)->get();
    return view('connection.showConnection',['user_id_1_connections' => $user_id_1_connections, 'user_id_2_connections' => $user_id_2_connections]);
}

我的模特：

protected $table = 'connections';
protected $fillable = ['user_id_1','user_id_2','connection_status'];

public function user()
{
    return $this->belongsTo('App\User');
}

我的刀锋：

@foreach($user_id_1_connections as $user_id_1_connection)
{{ $user_id_1_connection->user->name }}
{{ $comment->user->name }}
@endforeach
@foreach($user_id_2_connections as $user_id_2_connection)
{{ $user_id_2_connection->user->name }}
@endforeach

我已将user_id_1和user_id_2的外键设为用户表。

$table->integer('user_id_1')->unsigned();
        $table->foreign('user_id_1')->references('id')->on('users')->onDelete('cascade');
        $table->integer('user_id_2')->unsigned();
        $table->foreign('user_id_2')->references('id')->on('users')->onDelete('cascade');

但是当我运行这段代码时。它显示错误：试图获得非对象的属性。

Answer 1

这里的问题可能是您没有为每个连接分配user，而不是：

{{ $user_id_1_connection->user->name }}

你应该写这样的东西：

{{ $user_id_1_connection->user ? $user_id_1_connection->user->name : 'unknown' }}

当您使用$x->y->z语法时，

在所有其他位置相同。要显示z，您应该确保$x->y不为空

Laravel 5，试图获得非对象的属性

1 个答案: