PHP使用simpleXML解析georss命名空间

问题描述 投票:4回答:1

试图解析谷歌地图rss feed中的lat / lon:

$file = "http://maps.google.com/maps/ms?ie=UTF8&hl=en&vps=1&jsv=327b&msa=0&output=georss&msid=217909142388190116501.000473ca1b7eb5750ebfe";
$xml = simplexml_load_file($file);    
$loc = $xml->channel->item;
echo $loc[0]->title;
echo $loc[0]->point;

标题显示正常,但点没有给我任何东西。每个节点如下所示:

<item> 
    <guid isPermaLink="false">0004740950fd067393eb4</guid> 
    <pubDate>Sun, 20 Sep 2009 21:47:49 +0000</pubDate> 
    <title>Big Wong King Restaurant</title> 
    <description><![CDATA[<div dir="ltr">$4.99 full meals!</div>]]></description> 
    <author>neufuture</author> 
    <georss:point> 
      40.716236 -73.998413
    </georss:point> 
    <georss:elev>0.000000</georss:elev> 
  </item>
php xml simplexml xml-namespaces georss
1个回答
2
投票
<?php 
    $file = "http://maps.google.com/maps/ms?ie=UTF8&hl=en&vps=1&jsv=327b&msa=0&output=georss&msid=217909142388190116501.000473ca1b7eb5750ebfe";
    $xml = simplexml_load_file($file);  
    $loc = $xml->channel->item;  
    foreach ($loc->children('http://www.georss.org/georss') as $geo) { 
        echo $geo;
    } 
?>
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