2015-05-15 89 views
0

我使用由imgur API返回的对象链接: 1. Laravel 5 2. https://github.com/MarkRedeman/Imgur-Laravel如何获得

我的控制器:

public function store() 
    { 
     $input = Request::all(); 

     $upload = new Upload($input); 

     $upload->save(); 

     $upload_id = $upload->id; 

     $path = $this->resizeAndSaveTempImage($upload->raw_image_url); 

     $done = $this->uploadToImgur($path); 

     dd($done); 



    } 

public function resizeAndSaveTempImage($image_url) 
    { 


     $img = Image::make($image_url)->insert('http://i.imgur.com/Ned0D1ub.jpg', 'top', 130, 330); 

     $tmp_name = 'final-output-' . microtime(true) . '.jpg'; 

     $path = 'tmp/' . $tmp_name; 

     $img->save($path); 

     return $path; 



    } 

    public function uploadToImgur($image_path) 
    { 
     $imageData = array(
      'image' => $image_path, 
      'type' => 'file', 
      'name' => 'Lipsum', 
      'title' => 'Lorem Ipsum', 
      'description' => 'Lorem Ipsum Dolor Sit Amet' 
     ); 

     $basic = Imgur::api('image')->upload($imageData); 

     return $basic; 
    } 

模和转储的结果

Basic {#236 ▼ 
    -data: array:20 [▼ 
    "id" => "KWIhRsS" 
    "title" => "Lorem Ipsum" 
    "description" => "Lorem Ipsum Dolor Sit Amet" 
    "datetime" => 1431732751 
    "type" => "image/jpeg" 
    "animated" => false 
    "width" => 640 
    "height" => 640 
    "size" => 95601 
    "views" => 0 
    "bandwidth" => 0 
    "vote" => null 
    "favorite" => false 
    "nsfw" => null 
    "section" => null 
    "account_url" => null 
    "account_id" => 0 
    "deletehash" => "z8M21wNHdsFOBhJ" 
    "name" => "Lipsum" 
    "link" => "http://i.imgur.com/KWIhRsS.jpg" 
    ] 
    -success: true 
    -status: 200 
} 

但是,如果我尝试访问链接

dd($done->data['link']); 

它给我这个错误:

Cannot access private property Imgur\Api\Model\Basic::$data 

如何访问从响应对象的链接?

回答