从 Graph API 显示 [图像][url]

构建 Facebook 视频应用程序。用户可以通过在应用程序og.like中使用来收藏视频。

I use

$response = $facebook->api(
  'me/og.likes',
  'GET'

我会得到

"data": {
    "object": {
      "id": "1399918593560042", 
      "url": "http://some_url.com", 
      "type": "video.tv_show", 
      "title": "the_video_title"
    }

为了获取我使用的网址。

$response = $facebook->api(
 'me/og.likes?app_id_filter=381977341837631',
 'GET'
);

foreach ( $response['data'] as $data ); 
$Object = $data['data']['object'];

Then

 <li class="program"><a class="thumbnail" data-transition="slide" href="<?php echo $Object['url']; ?>">
  <img src="IMG_URL"></a></li> 

问题是显示图像。如果我点击图形 API 中的 ID，我会得到

{
  "id": "1399918593560042", 
  "url": "http://some_url.com", 
  "type": "video.tv_show", 
 "title": "the_video_title", 
  "image": [
{
  "url": "https://v.redli.se/p/102/sp/10200/thumbnail/entry_id/0_53lzx39w/width/1024/height/720", 
  "secure_url": "https://v.redli.se/p/102/sp/10200/thumbnail/entry_id/0_53lzx39w/width/1024/height/720", 
  "type": "image/jpg", 
  "width": 1024, 
  "height": 576
}

我的问题是。我如何显示图像？

当你得到$Object，使\GET对该对象的请求-

$resp = $facebook->api($Object['id']);

然后从响应中获取图像网址 -

if(isset($resp['image']))
{
   foreach ($resp['image'] as $image) 
   {
       echo $imag_url = $image['url'];
   }
}