php - 试图在laravel中获取非对象错误的属性

Question

我一直在尝试将我的用户的 User_id 与 Creator_id 进行匹配，以便在我的帖子中显示创建者的用户名但我正在尝试从非对象错误中获取属性，但我无法解决它

我 Controller 中的函数后 Controller .php:

public function index()
    {
    $allusers = users::getusers();

    $posts = DB::table('posts')->get();
    foreach ($posts as $posts) {
        foreach ($allusers as $allusers) {
            if ($posts->creator_id == $allusers->user_id) {
              array_push($posts,$allusers->username); 
            }
        }
    }
     return view('blog',['posts'=>$posts]);   
    }

我模型中的函数:

public static function getusers(){
    $allusers = users::all();
    return $allusers;
    }

Answer 1

您在 foreach 循环中使用了相同的变量名:

foreach ($posts as $post) {
    foreach ($allusers as $user) {
        if ($post->creator_id == $user->user_id) {
          array_push($post,$user->username); 
        }
    }
}

但是对于您的问题，您应该尝试将其添加到您的 Post 模型中

public function user(){
    return $this->belongsTo(User::class, 'creator_id', 'user_id);
}

那么在你看来:

@foreach($posts as $post)
    {{ $post->user->name }}
@endforeach

https://laravel.com/docs/5.4/eloquent-relationships#one-to-many-inverse