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PHP, XML, XSL

denoteone

By denoteone
in PHP Coding Help

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denoteone Advanced Member

denoteone

Posted
Posted

Trying to get everything in one file.  I have an external data source that has 3 different XML files located on it. I am writing a  php script that sits on my server and based on the GET variable I want to read a specific XML file and add a specific XSL style sheet to it then echo it to the browser.  What I cant have is 3 different .XSL files siting on my server or the data source. So if there is some way I can save the XSL info as a variable and somehow apply all from my one PHP script that would be great. So far this is what I have:

 

<?PHP

$view = $_GET['view'];

$xml_data = "http://externalserver.com/" . $view .".xml"; 
$handle = fopen($xml_data, 'r'); 
$Data = fread($handle, 512); 
fclose($handle); 

$old ='<?xml version="1.0" encoding="utf-8"?>';

//Right here is my biggest challange I dont wont the style sheets on the same server or any others for that matter since the server I am running this script on can not get out to the internet since it is behind a firewall. I was thinking of creating the XSL files on the fly and then saving them next to this script. But I would like to keep everything in just 1 file if at all possible.

$new ='<?xml version="1.0" encoding="utf-8"?>
<?xml-stylesheet href="style.xsl" type="text/xsl"?>';

$final_xml = str_replace($old, $new, $Data);

echo $final_xml;

?>

 

the url would be http://www.myserver.com/getxml.php?view=weather

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https://forums.phpfreaks.com/topic/219416-php-xml-xsl/
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denoteone Advanced Member

denoteone

Posted
  • Author
Posted

Ok so I am calling the same page to set the .XSL style sheet but am getting the dreaded "Error loading stylesheet: Parsing an XSLT stylesheet failed." so the url calls the script that gets the xml file based on the GET variable and it adds the style sheet by calling the same page with a new GET variable.

 

below is my script is this possible?

<?PHP
ini_set("display_errors","2");
ERROR_REPORTING(E_ALL);
header('Content-type: text/xml');

$view = $_GET['view'];

if(isset($_GET['xsl_style']))
{
$style = '<?xml version="1.0" encoding="ISO-8859-1"?>

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">

<xsl:template match="/">
  <html>
  <body>
  <h2>My CD Collection</h2>
    <table border="1">
      <tr bgcolor="#9acd32">
        <th>Title</th>
        <th>Artist</th>
      </tr>
      <xsl:for-each select="catalog/cd">
      <tr>
        <td><xsl:value-of select="title"/></td>
        <td><xsl:value-of select="artist"/></td>
      </tr>
      </xsl:for-each>
    </table>
  </body>
  </html>
</xsl:template>
</xsl:stylesheet>';

echo $style;

}else{

$xml_data = "http://www.website.com/STRATA/" . $view .".xml";
$handle = fopen($xml_data, 'r');
$Data = fread($handle, 10000);
fclose($handle);

$old ='<?xml version="1.0" encoding="ISO-8859-1"?>';

$new ='<?xml version="1.0" encoding="ISO-8859-1"?>
<?xml-stylesheet href="http://www.website.com/STRATA/get_view.php?xsl_style='. $view .'" type="text/xsl"?>';

$final_xml = str_replace($old, $new, $Data);

echo $final_xml;

}

?>

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https://forums.phpfreaks.com/topic/219416-php-xml-xsl/#findComment-1138239
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