我試圖通過子表列通過故事表從列表中獲取數據。我已經設法與作者合作,但我認爲我必須錯過某些東西或遺忘某些東西,因爲它不再有用。我收到此錯誤此收集實例上不存在屬性[註釋]
Collection.php行1498中的異常:此集合實例上不存在屬性[註釋]。
我的控制器功能
public function slug($slug){
$menus_child = Menu::where('menu_id', 0)->with('menusP')->get();
$stories = Story::where('slug', $slug)->get();
$slug = $slug;
// I used this to get the author's email going through the slug from
// story table
$story = Story::with('author')->where('slug', $slug)->first();
$name = $story->author->first()->email;
// I would like to get the name in the comment table by going through
// the slug from the story table
$comment = Story::with('comment')->where('slug', $slug)->get();
$test = $comment->comment->first()->name;
return view('open::public.single-story', compact('menus_child', 'stories', 'slug'));
}
我Comment.php
namespace App\Modules\Authors\Models;
use Illuminate\Database\Eloquent\Model;
class Comment extends Model
{
protected $fillable = array('name', 'comment');
protected $guarded = array('id');
protected $table = 'comments';
//Validation rules
public static $rules = array(
'name' => '',
'comment' => ''
);
public function story(){
return $this->belongsToMany('App\Modules\Authors\Models\Story', 'comment_story', 'comment_id', 'story_id');
}
}
我Story.php
class Story extends Model
{
protected $fillable = array('title', 'content', 'type', 'slug');
protected $guarded = array('id');
protected $table = 'story';
//Validation rules
public static $rules = array(
'title' => '',
'content' => '',
'type' => ''
);
public function slug($title){
$this->attributes['title'] = $title;
$this->attributes['slug'] = Str::slug($title);
}
public function author(){
return $this->belongsToMany('App\Modules\Authors\Models\Author', 'author_story', 'story_id', 'author_id');
}
public function comment(){
return $this->belongsToMany('App\Modules\Authors\Models\Comment', 'comment_story', 'story_id', 'comment_id');
}
}
你的關係是否正確?他們都被設置爲'belongsToMany()'。我期望一個Story有一個作者('belongsTo()')和許多評論('hasMany()'),相反地一個評論有一個Story('belongsTo()')。 – Mei