从其他表中获取ID以供审阅

时间:2017-01-30 22:03:57

标签: php laravel model-view-controller

我创建的社交媒体允许用户查看他/她访问过的地方。我已经通过了评级,内容以及审核该地点的用户ID,但我似乎无法通过该地方的ID。这是我发布评论的控制器:

class PlaceController extends Controller
{
    public function postReview(Request $request)
        {

            $rating     = $request['rating'];
            $content    = $request['content'];

            $review     = new Review();
            $review     ->rating=$rating;
            $review     ->content=$content;

            $request->user()->reviews()->save($review);

            return redirect()->back();
        }
}

以下是我的评论模型:

class Review extends Model
{
    use SoftDeletes;

    protected $table = 'place_reviews';

    protected $dates = ['deleted_at'];

    public function place()
    {
        return $this->belongsTo('Keleus\Eloquent\Place\Place');
    }

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

    public function scopeOfPlace($query, $placeId)
    {
        return $query->where('place_id', $placeId)->whereNotNull('place_id');
    }
}

请注意,评论的输入仅为评分和内容,用户ID和地点ID。

0 个答案:

没有答案