【发布时间】:2016-01-13 15:32:40
【问题描述】:
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 $this->belongsTo('App\User');
}
我的刀片:
@foreach($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
我已经为 user_id_1 和 user_id_2 创建了外键到 users 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: 试图获取非对象的属性。
【问题讨论】:
-
这可能发生在任何地方 - 您应该在错误消息中包含文件名和行号,请先发布该信息。
标签: laravel laravel-5 laravel-5.1 blade