How can I get url from image tag in feed with fetch_feed?

Question

I used this piece of code in functions.php to add featured image in <image> tag in feed

add_action('rss2_item', 'add_my_rss_node');

function add_my_rss_node() {
    global $post;
    if(has_post_thumbnail($post->ID)):
        $thumbnail = wp_get_attachment_image_src(get_post_thumbnail_id($post->ID), 'medium');
         $url = $thumbnail[0];
        echo("<image>{$url}</image>");
    endif;
}

then the URL of featured image displayed in feed as I expected.

Now I want to pull this URL out, I tried <pre>$item->get_title()</pre> but it didn't show anything.

Now I wonder how can I get this URL out?

this' my feed: http://techdaily.vn/feed and I follow this guide to use feed_fetch http://codex.wordpress.org/Function_Reference/fetch_feed

Answer 1

I found alternative solution for my problem. Instead of using fetch_feed, I used DOMDocument in PHP. This is my code

<?php
$rss = new DOMDocument();
$rss->load('http://techdaily.vn/feed/');
$feed = array();
foreach ($rss->getElementsByTagName('item') as $node) {
    $item = array ( 
        'title' => $node->getElementsByTagName('title')->item(0)->nodeValue,
        'link' => $node->getElementsByTagName('link')->item(0)->nodeValue,
        'image' => $node->getElementsByTagName('image')->item(0)->nodeValue,
        );
    array_push($feed, $item);
}
$limit = 1;
for($x=0;$x<$limit;$x++) {
    $title = str_replace(' & ', ' &amp; ', $feed[$x]['title']);
    $link = $feed[$x]['link'];
    $image = $feed[$x]['image'];
    ?>
<a href="http://techdaily.vn" target="_blank"><img class="logo" src="http://techdaily.s3.amazonaws.com/wp-content/uploads/2014/01/techdaily-logo1.png"></a>
<ul>
    <li class="thumb">
        <a href="<?php echo $link; ?>" class="title" target="_blank">
            <img src="<?php echo $image; ?>" alt="<?php echo $title; ?>">
        </a>
    </li>
    <li><a href="<?php echo $link; ?>" class="title" target="_blank"><?php echo $title; ?></a></li>
</ul>