php - 如何组合功能以获取多种类型的信息

Question

我使用Youtube Api来获取有关单个视频的信息。当我使用类似网址的请求时:

https://www.googleapis.com/youtube/v3/videos?part=status,snippet,contentdetails&id=$videoID&key=$apikey

我得到了我需要的所有信息。问题是我无法找到一种方法来一次性组合获得信息所需的功能。

我到目前为止所得到的:

function getDescription($videoID){
    $apikey = "<MYKEY>";

    $desc = file_get_contents("https://www.googleapis.com/youtube/v3/videos?part=snippet&id=$videoID&key=$apikey");
    $description =json_decode($desc, true);
    foreach ($description['items'] as $videodesc)
   {
       $description= $videodesc['snippet']['description'];
   }
   return $description;
}

function getPublishedAt($videoID){
    $apikey = "<MYKEY>";

    $pub = file_get_contents("https://www.googleapis.com/youtube/v3/videos?part=snippet&id=$videoID&key=$apikey");
    $publish =json_decode($pub, true);
    foreach ($publish['items'] as $published)
   {
       $publish= $published['snippet']['publishedAt'];
   }

   $publish = new DateTime($publish);
   $publish = $publish->format('Y-m-d H:i:s');

   return $publish;

}

echo "<br>Description: ";
echo getDescription("<VIDEO-ID>");
echo "<br>PublishedAt: ";
echo getPublishedAt("<VIDEO-ID>");

因此，此代码有效，但是我想向api发出一个请求(使用url中的多个部分)并获取信息。

有谁知道如何仅使用一个功能执行此操作？

Answer 1

好吧，我自己找到了答案:

$videoinfo = file_get_contents("https://www.googleapis.com/youtube/v3/videos?id=<video-id>&part=snippet,status,contentDetails,statistics&part=statistics&key=<api-key>");
   $videoinfo =json_decode($videoinfo, true);
   foreach ($videoinfo['items'] as $embed)
   {
       $emb= $embed['status']['embeddable'];
       $license= $embed['status']['license'];
       $duration= $embed['contentDetails']['duration'];
   }
echo $emb;   
echo $license;
echo $duration;

这不是一个函数，但这将执行公制:)