Good, I have this code of LastFM API
<?php
$xml = simplexml_load_file("http://ws.audioscrobbler.com/2.0/?method=artist.getinfo&artist=ARIANA GRANDE&api_key=b25b959554ed76058ac220b7b2e0a026");
$largeImage = $xml->xpath('/lfm/artist/image[#size="mega"]')[0];
echo '<img src="'.$largeImage.'" />';
?>
it appears used the image of Ariana Grande a php page.
Now the link of my XML file is :http://radiojoven.6te.net/playlist.xml
Well, what I am trying to do is change "ARIANA GRANDE" (the artist's name) in the link of the "simplexml_load_file" (in the php code) with the info that my XML file provides. I tried to make a code like this but to no avail.
<?php
$xml = simplexml_load_file('http://radiojoven.6te.net/playlist.xml');
$artist = urlencode($xml->Event->Song->Artist['name'];);
$url = simplexml_load_file("http://ws.audioscrobbler.com/2.0/?method=artist.getinfo&artist='.$artist.'&api_key=50ac27433c63f7298064f434f4ef6d15);
$xml2 = #simplexml_load_file($url);
$largeImage = $xml2->xpath('/lfm/artist/image[#size="mega"]')[0];
echo '<img src="'.$largeImage.'" />';
?>
Can you please help me doing this?
Thank you all in advance.
I think there are a few errors in your code.
Remove the semicolon after ['name']; in this line:
$artist = urlencode($xml->Event->Song->Artist['name'];);
Missing double quotes after api_key=50ac27433c63f7298064f434f4ef6d15 and I think '.$artist.' should be $artist in this line:
$url = simplexml_load_file("http://ws.audioscrobbler.com/2.0/?method=artist.getinfo&artist='.$artist.'&api_key=50ac27433c63f7298064f434f4ef6d15);
I don't think that you need this line:
$xml2 = #simplexml_load_file($url);
Then in this line change $xml2 to $url:
$largeImage = $xml2->xpath('/lfm/artist/image[#size="mega"]')[0];
So your code would become like this:
<?php
$xml = simplexml_load_file('http://radiojoven.6te.net/playlist.xml');
$artist = urlencode($xml->Event->Song->Artist['name']);
$url = simplexml_load_file("http://ws.audioscrobbler.com/2.0/?method=artist.getinfo&artist=$artist&api_key=50ac27433c63f7298064f434f4ef6d15");
$largeImage = $url->xpath('/lfm/artist/image[#size="mega"]')[0];
echo '<img src="'.$largeImage.'" />';
?>
Related
I want to fetch image from google using PHP. so I tried to get help from net I got a script as I needed but it is showing this fatal error
Fatal error: Call to a member function find() on a non-object in C:\wamp\www\nq\qimages.php on line 7**
Here is my script:
<?php
include "simple_html_dom.php";
$search_query = "car";
$search_query = urlencode( $search_query );
$html = file_get_html( "https://www.google.com/search?q=$search_query&tbm=isch" );
$image_container = $html->find('div#rcnt', 0);
$images = $image_container->find('img');
$image_count = 10; //Enter the amount of images to be shown
$i = 0;
foreach($images as $image){
if($i == $image_count) break;
$i++;
// DO with the image whatever you want here (the image element is '$image'):
echo $image;
}
?>
I am also using Simple html dom.
Look at my example that works and gets first image from google results:
<?php
$url = "https://www.google.hr/search?q=aaaa&biw=1517&bih=714&source=lnms&tbm=isch&sa=X&ved=0CAYQ_AUoAWoVChMIyKnjyrjQyAIVylwaCh06nAIE&dpr=0.9";
$content = file_get_contents($url);
libxml_use_internal_errors(true);
$dom = new DOMDocument;
#$dom->loadHTML($content);
$images_dom = $dom->getElementsByTagName('img');
foreach ($images_dom as $img) {
if($img->hasAttribute('src')){
$image_url = $img->getAttribute('src');
}
break;
}
//this is first image on url
echo $image_url;
This error usually means that $html isn't an object.
It's odd that you say this seems to work. What happens if you output $html? I'd imagine that the url isn't available and that $html is null.
Edit: Looks like this may be an error in the parser. Someone has submitted a bug and added a check in his code as a workaround.
I am adding an RSS feed to my website. I created the RSS.xml index file and next I want to display its contents in a nicely formatted way in a webpage.
Using PHP, I can do this:
$index = file_get_contents ($path . 'RSS.xml');
echo $index;
But all that does is dump the contents as a long stream of text with the tags removed.
I know that treating RSS.xml as a link, like this:
<a href="../blogs/RSS.xml">
<img src="../blogs/feed-icon-16.gif">Blog Index
</a>
causes my browser to parse and display it in a reasonable way when the user clicks on the link. However I want to embed it directly in the web page and not make the user go through another click.
What is the proper way to do what I want?
Use the following code:
include_once('Simple/autoloader.php');
$feed = new SimplePie();
$feed->set_feed_url($url);
$feed->enable_cache(false);
$feed->set_output_encoding('utf-8');
$feed->init();
$i=0;
$items = $feed->get_items();
foreach ($items as $item) {
$i++;
/*You are getting title,description,date of your rss by the following code*/
$title = $item->get_title();
$url = $item->get_permalink();
$desc = $item->get_description();
$date = $item->get_date();
}
Download the Simple folder data from : https://github.com/jewelhuq/Online-News-Grabber/tree/master/worldnews/Simple
Hope it will work for you. There $url mean your rss feed url. If you works then response.
Turns out, it's simple by using the PHP xml parer function:
$xml = simplexml_load_file ($path . 'RSS.xml');
$channel = $xml->channel;
$channel_title = $channel->title;
$channel_description = $channel->description;
echo "<h1>$channel_title</h1>";
echo "<h2>$channel_description</h2>";
foreach ($channel->item as $item)
{
$title = $item->title;
$link = $item->link;
$descr = $item->description;
echo "<h3><a href='$link'>$title</a></h3>";
echo "<p>$descr</p>";
}
I'm trying to print a website's favicon, as an image, not as a link to it.
I have a php script in which I extract the favicon, but now I want to show it as it is.
Here is what I've tried.
//extract favicon
$url = $_POST['url'];
$doc = new DOMDocument();
$doc->strictErrorChecking = FALSE;
$doc->loadHTML(file_get_contents($url));
$xml = simplexml_import_dom($doc);
$arr = $xml->xpath('//link[#rel="shortcut icon"]');
echo "<br>";
//echo "favicon:";
if( $arr)
{
$src = $arr[0]['href'];
echo "<img src = "$src">";//as I can see, the parameter here cannot be a variable
//second thing that I've tried: echo "<img src = "$arr[0]['href']""; it doesn't work either
}
This is what my script is echoing right now. http://i.stack.imgur.com/Wkoyj.jpg
Instead of the link to the favicon, I want the actual favicon to be displayed. I hope I explained myself correctly.
Your error is with the code:
echo "<img src = "$src">";//as I can see, the parameter here cannot be a variable
It should be
echo '<img src="'.$src.'">';
Or even
echo "<img src=\"$src\">";
I am trying to take form data (via _POST) and write it to a document using SimpleXML. This is what I have tried and I can't seem to get it to work.
<?php
$title = $_POST['title'];
$link = $_POST['link'];
$description = $_POST['description'];
$rss = new SimpleXMLElement($xmlstr);
$rss->loadfile("feed.xml");
$item = $rss->channel->addChild('item');
$item->addChild('title', $title);
$item->addChild('link', $link);
$item->addChild('description', $description);
echo $rss->asXML();
header("Location: /success.html");
exit;
?>
Any help or a point in the right direction would be much appreciated.
You use the asXML() function wrong. If you want to write your XML to a file, you must pass a filename parameter to it. Check the SimpleXMLElement::asXML manual
so your code line oututing xml should be changed from
echo $rss->asXML();
to
$rss->asXML('myNewlyCreatedXML.xml');
rather than using SimpleXMLElement you can create XML directly like this
$xml = '<?xml version="1.0" encoding="utf-8"?>';
$xml .= '<item>';
$xml .= '<title>'.$title.'</title>';
$xml .= '<link>'.$title.'</link>';
$xml .= '<description>'.$title.'</description>';
$xml .= '</item>';
$xml_file = "feed.xml";
file_put_contents($xml_file,$xml);
may this will help you
There are a few problems with your code
<?php
$title = $_POST['title'];
$link = $_POST['link'];
$description = $_POST['description'];
//$rss = new SimpleXMLElement($xmlstr); // need to have $xmlstr defined before you construct the simpleXML
//$rss->loadfile("feed.xml");
//easier to just load the file when creating your XML object
$rss = new SimpleXML("feed.xml",null,true) // url/path, options, is_url
$item = $rss->channel->addChild('item');
$item->addChild('title', $title);
$item->addChild('link', $link);
$item->addChild('description', $description);
//header("Location: /success.html");
//if you want to redirect you should put a timer on it and echo afterwards but by
//this time if something went wrong there will be output already sent,
//so you can't send more headers, i.e. the next line will cause an error
header('refresh: 4; URL=/success.html');
echo $rss->asXML(); // you may want to output to a file, rather than the client
// $rss->asXML('outfputfile.xml');
exit;
?>
I just can't seem to be able to solve this. I want to get the media:thumbnail from an RSS file (http://feeds.bbci.co.uk/news/rss.xml).
I did some research and tried to incorporate insights from
https://stackoverflow.com/questions/6707315/getting-xml-attribute-from-mediathumbnail-in-bbc-rss-feed
and from other sources.
This is what I got:
$source_link = "http://feeds.bbci.co.uk/news/rss.xml";
$source_xml = simplexml_load_file($source_link);
$namespace = "http://search.yahoo.com/mrss/";
foreach ($source_xml->channel->item as $rss) {
$title = $rss->title;
$description = $rss->description;
$link = $rss->link;
$date_raw = $rss->pubDate;
$date = date("Y-m-j G:i:s", strtotime($date_raw));
$image = $rss->attributes($namespace);
print_r($image);
}
When I run the script, all I see is a white page. If I echo or print_r any of the other variables, then it works like a charm. It's just the $image one which poses problems. Why isn't this working? Thx for any help!
OK, it works now. I replaced
$image = $rss->attributes($namespace);
with
$image = $rss->children($namespace)->thumbnail[1]->attributes();
$image_link = $image['url'];
and it works like a charm now.
Base from this blog, with post title Processing media:thumbnail in RSS feeds with php.
The solution that I found works best simply loads the xml file as a string, then find and replace 'media:thumbnail' with a correctly formatted 'thumbnail' and lastly convert it back to xml with simplexml_load_string:
$xSource = 'http://feeds.bbci.co.uk/news/rss.xml';
$xsourcefile = file_get_contents( $xSource );
$xsourcefile = str_replace("media:thumbnail","thumbnail",$xsourcefile);
$xml = simplexml_load_string( $xsourcefile );
echo $row['xtitle'] . '<BR>';
foreach ($xml->channel->item as $item) {
echo ':' . $item->title . '<BR>';
echo ':' . $item->thumbnail['url'] . '<BR>';
}
$image = $rss->attributes($namespace);
This says "Give me all attributes of this <item> element which are in the media namespace". There are no attributes on the item element (much less any in the media namespace), so this returns nothing.
You want this:
$firstimage = $rss->children($namespace)->thumbnail[0];
BTW, when you use SimpleXML you need to be careful to cast your SimpleXMLElements to string when you need the text value of the element. Something like $rss->title is a SimpleXMLElement, not a string.