generating an RSS Feed from remote xml data - php

Question

I am trying to make an RSS feed which feeds live data from a remote XML file provided to me for promoting a webcam network.
I want to run a feed within my official website. My problem is, I want to output the data into an RSS feed. I am not too familiar with RSS and php so there is a lot of guess work going on.
Here is the code I have so far which will not work, it incorporates php to retrieve the xml data. Also I am unsure as to what filename/type I save this as? XML or php? Both don't seem to work.
<?php
$curlData = '<?xml version="1.0" encoding="UTF-8"?>
<SMLQuery>
<Options MaxResults="20"/>
<AvailablePerformers Exact="false" PageNum="1" CountTotalResults="true">
<Include>
<Country/>
<Headline/>
<Descriptions></Descriptions>
<Rating/>
<Age/>
<Gender/>
<Media>biopic</Media>
<FreeChatSort/>
<PerformanceSort/>
</Include>
<Constraints> <StreamType>live,recorded,offline</StreamType>
<PublicProfile/>
<Name></Name><NoKeywords/>
<RelativeURLs/>
<NoKeywords/>
</Constraints>
</AvailablePerformers>
</SMLQuery>';
$url='http://affiliate.streamate.com/?????/SMLResult.xml';
$curl = curl_init();
curl_setopt ($curl, CURLOPT_URL, $url);
curl_setopt($curl, CURLOPT_RETURNTRANSFER, 1);
curl_setopt($curl,CURLOPT_TIMEOUT,120);
curl_setopt($curl,CURLOPT_HTTPHEADER,array (
'Content-Type: application/xml; charset=utf-8',
));
curl_setopt ($curl, CURLOPT_POST, 1);
curl_setopt ($curl, CURLOPT_POSTFIELDS, $curlData);
$result = curl_exec($curl);
curl_close ($curl);
$xml = simplexml_load_string ($result);
// $json = json_encode($xml);
// $array = json_decode($json,TRUE);
?>
<?xml version="1.0" encoding="utf-8"?>
<rss version="2.0">
<channel>
<title>Website Feed</title>
<description>Website Feed coded manually</description>
<link>yourdomain.com</link>;
<?php
foreach ($xml->AvailablePerformers->Performer as $perf) {
<item>
<title><php? echo ".$perf->attributes()->Name ."?></title>
<description><php? echo " .$perf->Descriptions->About ."?></description>
<link><php? echo "<a href='http://www.example.com /cam/".$perf->attributes()->Name." /?AFNO=X-0-XXXXXX-XXXXXX&UHNSMTY=XXX?></link>
</item>
</channel>
</rss>
?>

Answer 1

As you already have an XML source and want another XML result, maybe you could try XSLT to directly convert it.
http://www.w3schools.com/xsl/