Question

我找不到Vine的API来获取页面内容的标题，描述和图像。 JSON位于脚本标记中的页面主体中：。如何使用PHP获取此脚本标记的内容（JSON）以便对其进行解析？

＆＃xA;＆＃xA;

Vine页面：

＆＃xA;＆＃ xA;

  https://vine.co/v/igO3EbIXDlI

＆＃xA;＆＃xA;

来自页面来源

＆＃ xA;＆＃xA;

 ＆lt; script type =“application / ld + json”＆gt;＆＃xA; {＆＃XA; “@context”：“http://schema.org”，＆＃xA; “@type”：“SocialMediaPosting”，＆＃xA; “url”：“https://vine.co/v/igO3EbIXDlI",
 “datePublished”：“2016-03-01T00：58：35”，＆＃xA; “作者”：{＆＃xA; “@type”：“人”，＆＃xA; “name”：“MotorAddicts \ u2122”，＆＃xA; “image”：“https://v.cdn.vine.co/r/avatars/39FEFED72B1242718633613316096_pic-r-1439261422661708f3e9755.jpg.jpg?versionId=LPjQUQ4KmTIPLu3iDbXw4FipgjEpC6fw",
 “url”：“https://vine.co/u/989736283540746240"
 }，＆＃XA; “articleBody”：“嗯...... Black black blaaaaack !! \ ud83d \ ude0d（Drift \ u53d1）”，＆＃xA; “image”：“https://v.cdn.vine.co/r/videos/98C3799A811316254965085667328_SW_WEBM_14567938452154dc600dbde.webm.jpg?versionId=wPuaQvDxnpwF7KjSGao21hoddooc3eCl",
 “interactionCount”：[{＆＃xA; “@type”：“UserInteraction”，＆＃xA; “userInteractionType”：“http://schema.org/UserLikes",
 “价值”：“1382”＆＃xA; }，{＆＃xA; “@type”：“UserInteraction”，＆＃xA; “userInteractionType”：“http://schema.org/UserShares",
 “价值”：“368”＆＃xA; }，{＆＃xA; “@type”：“UserInteraction”，＆＃xA; “userInteractionType”：“http://schema.org/UserComments",
 “价值”：“41”＆＃xA; }，{＆＃xA; “@type”：“UserInteraction”，＆＃xA; “userInteractionType”：“http://schema.org/UserViews",
 “价值”：“80575”＆＃xA; }]，＆＃XA;＆＃XA; “sharedContent”：{＆＃xA; “@type”：“VideoObject”，＆＃xA; “名字”：“嗯......黑色黑色blaaaaack !! \ ud83d \ ude0d（Drift \ u53d1）”，＆＃xA; “description”：“”，＆＃xA; “thumbnailUrl”：“https://v.cdn.vine.co/r/videos/98C3799A811316254965085667328_SW_WEBM_14567938452154dc600dbde.webm.jpg?versionId=wPuaQvDxnpwF7KjSGao21hoddooc3eCl",
 “uploadDate”：“2016-03-01T00：58：35”，＆＃xA; “contentUrl”：“https://v.cdn.vine.co/r/videos_h264high/98C3799A811316254965085667328_SW_WEBM_14567938452154dc600dbde.mp4?versionId=w7ugLPYtj5LWeVUsXaH1bt2VuK8QE0qv",
 “embedUrl”：“https://vine.co/v/igO3EbIXDlI/embed/simple",
 “interactionCount”：“82366”＆＃xA; }＆＃XA; }＆＃XA; ＆lt; / script＆gt;＆＃xA;

＆＃xA;＆＃xA;

此后该怎么办？

＆＃xA;＆＃xA;

 < code> $ html ='https://vine.co/v/igO3EbIXDlI';
$dom = new DOMDocument;＆＃xA; $ dom-＆gt; loadHTML（$ html）;＆＃xA;

＆＃xA;＆＃xA;

更新：

＆＃xA;＆＃xA;

我在这里找到了Vine API的说明：

＆＃ XA;＆＃XA;

 <代码> https://dev.twitter.com/web/vine/oembed

＆＃XA;＆＃XA;

要查询Vine API for JSON，从以下地址获取请求：

＆＃xA;＆＃xA;

  https://vine.co/oembed.json?url=https%3A%2F% 2Fvine.co％2FV％2F [VideoID的]＆＃XA;

＆＃XA;＆＃XA; 示例：
＆＃XA;＆＃XA;  <代码> https://vine.co/oembed.json?url=https%3A%2F%2Fvine.co%2Fv%2FMl16lZVTTxe
  ＆＃XA;

Answer 1

您可以使用DOMDocument和DOMXpath：

$html = file_get_contents( $url );
$dom  = new DOMDocument();
libxml_use_internal_errors( 1 );
$dom->loadHTML( $html );
$xpath = new DOMXpath( $dom );

$jsonScripts = $xpath->query( '//script[@type="application/ld+json"]' );
$json = trim( $script->item(0)->nodeValue );

$data = json_decode( $json );

的 phpFiddle demo

使用此xPath模式，您可以搜索属性 type 为“application / ld + json”的所有<script>个节点：

// Following path no matter where they are in the document script Elements <script> [@type="application/ld+json"] with attribute “tipe” as “application/ld+json”

然后检索获得第一个返回的->nodeValue节点的<script>的JSON字符串。

如果您事先不知道节点的存在和/或位置，请使用：

$jsonScripts = $xpath->query( '//script[@type="application/ld+json"]' ); if( $jsonScript->length < 1 ) { die( "Error: No script node found" ); } else { foreach( $jsonScripts as $node ) { $json = json_decode( $node->nodeValue ); // your stuff with JSON here... } }

Answer 2

$html_content = file_get_contents('https://vine.co/v/igO3EbIXDlI');

$target_class = 'script';

$dom_object = new DOMDocument;
$dom_object->loadHTML($html_content);
$xpath_object = new DOMXpath($dom_object);

$elements = $xpath_object->query("//*[contains(concat(' ', normalize-space(@class), ' '), ' {$target_class} ')]");

$output = []
foreach ($elements as $element)
{
    $output[] = $dom_object->saveHTML($element);
}

# you now have a list of strings, each containing the contents of a 
# non-overlapping script tag

使用PHP获取<script type =“application / ld + json”>的内容

2 个答案: