How to get first image from a tumlbr rss feed in PHP - php

Question

0Here is the relevant part of my rss feed:
<channel>
<description></description>
<title>Untitled</title>
<generator>Tumblr (3.0; #xxx)</generator>
<link>http://xxx.tumblr.com/</link>
<item>
<title>Title</title>
<description><figure><img src="https://31.media.tumblr.com/c78c7t3abd23423549d3bb0f705/tumblr_inline_nkp9z234d0uj.jpg"/></figure></description>
<link>http://xxx.tumblr.com/post/99569244093</link>
<guid>http://xxx.tumblr.com/post/99569244093</guid>
<pubDate>Thu, 09 Oct 2014 11:19:33 -0400</pubDate>
</item>
</channel>
Using the answer from other questions on here I tried this:
$content = file_get_contents("http://xxx.tumblr.com/rss");
$feed = new SimpleXmlElement($content);
$imgs = $feed->channel->item[0]->description->xpath('//img');
foreach($imgs as $image) {
echo (string)$image['src'];
};
This is returning an empty array for $imgs
Does it have something to do with the tags being < > etc?
and if so what can I do?

Answer 1

You can get it from the description, which seems to include a HTML image tag for the image, by using a simple regular expression with preg_match:
$content = file_get_contents("http://xxx.tumblr.com/rss");
$feed = new SimpleXmlElement($content);
$img = (string)$feed->channel->item[0]->description;
if (preg_match('/src="(.*?)"/', $img, $matches)) {
$src = $matches[1];
echo "src = $src", PHP_EOL;
}
Output:
src = http://40.media.tumblr.com/58d24c3009638514325b113859ba369f/tumblr_nk0mwfhKXU1sl87kjo1_500.jpg

Answer 2

Before you can use xapth() on the description, you need to create a new XML document out of it:
$url = "http://xxx.tumblr.com/rss";
$desc = simplexml_load_file($url)->xpath('//item/description[1]')[0];
$src = simplexml_load_string("<x>$desc</x>")->xpath('//img/#src')[0];
echo $src;
Output:
http://40.media.tumblr.com/58d24c3009638514325b113859ba369f/tumblr_nk0mwfhKXU1sl87kjo1_500.jpg

Answer 3

I'm not sure if you can use this approach - as already mentioned by kjhughes as comment, your input XML does not contain any img element. But it's possible to retrieve the image source using XPath substring-functions:
substring-before(substring-after(substring-after(//item/description[contains(.,'img')],
'src='),'"'),'"')
Result:
https://31.media.tumblr.com/c78c7t3abd23423549d3bb0f705/tumblr_inline_nkp9z234d0uj.jpg