使用implexml_Load解析XML
我正在尝试分析XML,但在尝试获取图像URL时遇到问题。
我的XML是:
<entry>
<title>The Title</title>
<id>http://example.com/post/367327.html</id>
<summary>Some extra text</summary>
<link rel="enclosure" href="http://example.com/photos/f_0px_30px/image687.jpg" type="image/jpeg" length="" />
</entry>
到目前为止,我使用以下代码来获取其他数据:
$url = "http://msdssite.com/feeds/xml/myxml.xml";
$xml = simplexml_load_file($url);
foreach($xml->entry as $PRODUCT)
{
$my_title = trim($PRODUCT->title);
$url = trim($PRODUCT->id);
$myimg = $PRODUCT->link;
}
我如何才能从中解析href
:<link rel="enclosure" href="http://example.com/photos/f_0px_30px/image687.jpg" type="image/jpeg" length="" />
解决方案
由于您的条目似乎可以包含几个link
标记,您需要检查type
属性是否具有值image/jpeg
,以确保获得指向图像的链接:
ini_set("display_errors", "On");
$feedURL = 'http://OLDpost.gr/feeds/xml/category-takhs-xatzhs.xml';
$feed = simplexml_load_file($feedURL);
$results = array();
foreach($feed->entry as $entry) {
$result = array('title' => (string)$entry->title,
'url' => (string)$entry->id);
$links = $entry->link;
foreach ($links as $link) {
$linkAttr = $link->attributes();
if (isset($linkAttr['type']) && $linkAttr['type']=='image/jpeg') {
$result['img'] = (string)$linkAttr['href'];
break;
}
}
$results[] = $result;
}
print_r($results);
注意,像这样使用implexml(Foreach循环查找好的链接标记)不是很方便。最好使用XPath查询:
foreach($feed->entry as $entry) {
$entry->registerXPathNamespace('e', 'http://www.w3.org/2005/Atom');
$results[] = array(
'title' => (string)$entry->title,
'url' => (string)$entry->id,
'img' => (string)$entry->xpath('e:link[@type="image/jpeg"]/@href')[0][0]
);
}
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