• Been working on this one for a bit and haven’t been able to fully complete this task, any help would be appreciated.

    Here is the feed im working with
    https://outdoors-international.com/store/product-tag/headwear/feed/

    The page that is using the data as test is here
    https://gothunts.com/feed-test/

    I need it to display images from the feed on the page like the title is displaying.
    Snippet of xml item code im trying to call in the feed test page

    <media:content xmlns:media="https://search.yahoo.com/mrss/" url="https://outdoors-international.com/store/wp-content/uploads/2013/06/Kryptek-Koldo-Rain-Jacket.png" width="1000" height="1000" medium="image" type="image/jpeg">
    <media:copyright>Kyudo Gear by Outdoors International</media:copyright>
    </media:content>

    Here is the custom page code I have been working with (feed-test page)

    <?php // Get RSS Feed(s)
    include_once(ABSPATH . WPINC . '/rss.php');
    $rss = fetch_rss('https://outdoors-international.com/store/product-tag/headwear/feed/');
    $maxitems = 25;
    $items = array_slice($rss->items, 0, $maxitems);
    ?>
    <ul>
    <?php if (empty($items)) echo '<li>No items</li>';
    else
    foreach ( $items as $item ) : ?>
    <div style="float:left;width:150px;height:225px;padding: 10px 10px 0px 0px;text-align: center;list-style-type:none;">
    <li>
    <a href='<?php echo $item['link']; ?>' title='<?php echo $item['title']; ?>'><img src="<?php echo $item['media:content']; ?>"></a>
    <a href='<?php echo $item['link']; ?>' title='<?php echo $item['title']; ?>'> <?php echo $item['title']; ?></a>
    </li></div>
    <?php endforeach; ?>
    </ul>
    </div>

    Question is how do I get the image url from the feed into the img src code on the page?

    Thank you!

  • The topic ‘RSS Feed from one blog to another with images’ is closed to new replies.