创建一个通过last.fm API显示图像的PHP页面

Question

I'm trying to do a PHP page to shows an album cover via Last.FM API. However, the Artist Name and the Title of the music are provided by a XML file that a software updates via FTP.

Here is the code of Last.FM api:

<?php
$img = simplexml_load_file('http://ws.audioscrobbler.com/2.0/?method=track.getInfo&api_key=<APIKEY>&artist=cher&track=believe');
echo '<img src="';
echo $img->track[0]->album[0]->image[3];
echo '">';
?>

Now the link of my XML file is: http://summerblast.pt/avaplayer/rds.xml The info I need to the Last.FM API is in 'OnAir/CurMusic'.

Well, what I am trying to do is change "Cher" and "Believe" (the artist's name and the Title's name) in the link of the "simplexml_load_file" (in the php code) with the info that my XML file provides.

Can you please help me doing this?

Thank you all in advance.

Answer 1

Sorry was confused to what you were asking from before. I have gone ahead and corrected to do what you wanted.

You can get the Artist and the Title from rds.xml and just pass it to the URL like the following listed below (note it is important to run urlencode on them since they can have spaces and other things to break the URL):

Updated to do error checking

    $xml = simplexml_load_file('http://summerblast.pt/avaplayer/rds.xml');
    $artist = urlencode($xml->OnAir->CurMusic->Artist);     
    $track =  urlencode($xml->OnAir->CurMusic->Title);
    $url = 'http://ws.audioscrobbler.com/2.0/?method=track.getInfo&api_key=XXXXXXXXXXXXXXXXXXXXXXXXXXX&artist='.$artist.'&track='.$track;
    $xml2 = @simplexml_load_file($url);
    if ($xml2 === false) 
    {
        echo("Url failed"); // do whatever you want to do
    }
    else
    {
        if($xml2->track->album->image[3])
        {
            echo '<img src="';
            echo((string) $xml2->track->album->image[3]);    
            echo '">';
        }
        else
        {
            echo("artist does not have a image"); // do whatever you want to do
        }
    }