What is incorrect in this script?

[fi]

Hey,

 

I run a php scipt (see down) from my server and the following error is show:

 

The XML page cannot be displayed

Cannot view XML input using XSL style sheet. Please correct the error and then click the Refresh button, or try again later.

 

 

--------------------------------------------------------------------------------

 

A string literal was expected, but no opening quote character was found. Error processing resource 'http://www.fisad.net/ph...

 

<font color=ff0000>

------------^

 

<?php
header( "cache-control: private" );
header( "pragma: no-cache" );
header("Content-Type: text/xml");
$directory = $_GET['str1']; 
$subapp = $_GET['sub1']; 
echo "<?xml version=\"1.0\" encoding=\"utf-8\"?>"."\n";
echo "<galleries>"."\n";
structure($directory,$directory,$subapp);
echo "</galleries>";
$num=1;
//function
function structure($rootdirpath, $dir, $caller)
{
//unset($root);
if($dp = @opendir($rootdirpath))
{
	for($i=0;($file=readdir($dp))!==false;$i++)
	{
		if ($file !="thumbs")
		{
			if (is_dir($rootdirpath."/".$file) && $file != "." && $file != "..")
			{
				$src= $rootdirpath."/".$file;
				$num=$num+1;
				echo "<gallery number=\"".$num."\" title=\"".$file."\">"."\n";
					ExportDirectoryXML($src,$caller);
				echo "</gallery>"."\n";
				structure($rootdirpath."/".$file,$dir,$caller);
			}
		}
	}
}
closedir($dp);
}
//
function ExportDirectoryXML($base,$subcaller)
{ 
if ($dir = @opendir($base))
{ 
	while ($file = readdir($dir))
	{ 
		if ($file{0} != ".")
		{ 
		if (is_dir($base . "/" . $file))
		{ 
			//ExportDirectoryXML($base . "/" . $file); 
		}
		else
		{ 
			$ext = @substr($file, (@strrpos($file, ".") ? @strrpos($file, ".") : @strlen($file)), @strlen($file));
			$fname = basename($file,$ext);
			$imgsrc= $base . "/thumbs/" . $file;
			$src= $base . "/" . $file;
			if (exif_imagetype($src) != false) {
				$num=rand()%1000;
				if ($subcaller !="app"){
					$imgsrc= "../../".$imgsrc;
					$src= "../../".$src;
				}
				else
				{
					$imgsrc= substr($imgsrc, 3);
					$src= substr($src, 3);
				}
				echo "<image source=\"".$src."\" thumb=\"".$imgsrc."\" descr=\"description\"/>"."\n";
				}
			}
		} 
	} 
closedir($dir); 
} 
} 
?>

 

When I load this script from http://localhost....../php/script.php

 

the result is OK 100%, variables are get from Flex application

 

But when I load from my server, the error is show.

 

I have xampp install localy with php 5.2.5 and my server have support for php 4 and 5.

 

Assets error not in my script file: "<font color=ff0000>"

 

Any help.

 

Thank in advance

 

 

 

[rhodesa]

when you open the file on the server, and you see that error, do a View Source. most likely it's a PHP error being thrown and then the XML parser is trying to validate that.

 

what do you see when you do the view source?

[fi]

I see:

 

<font color=ff0000>

Notice: Undefined index:  str1 in /usr/local/pem/vhosts/126956/webspace/httpdocs/php/gallery10.php on line 5

</font><font color=ff0000>

Notice: Undefined index:  sub1 in /usr/local/pem/vhosts/126956/webspace/httpdocs/php/gallery10.php on line 6

</font><?xml version="1.0" encoding="utf-8"?>

<galleries>

<font color=ff0000>

Warning: closedir(): supplied argument is not a valid Directory resource in /usr/local/pem/vhosts/126956/webspace/httpdocs/php/gallery10.php on line 34

</font></galleries>

 

What happen?

[rhodesa]

that means these values aresn't set:

$directory = $_GET['str1']; 
$subapp = $_GET['sub1']; 

 

-Are you supplying the values in the url for str1 and sub1? aka:

http://www.hostname.com/php/gallery10.php?str1=something&sub1=foobar

 

-If someone doesn't provide those arguments, what should $directory and $subapp be set to? Or should the script fail?

[fi]

Yes, I set from code or browse:

 

http://www.fisad.net/php/gallery10.php?str1=../images&sub1=sub

 

But the same error show:

 

The XML page cannot be displayed

Cannot view XML input using XSL style sheet. Please correct the error and then click the Refresh button, or try again later.

 

 

--------------------------------------------------------------------------------

 

A string literal was expected, but no opening quote character was found. Error processing resource 'http://www.fisad.net/ph...

 

<font color=ff0000>

------------^

 

You can try with de same line for see the error!

[fi]

yeap!

 

I get the problem!!!!!

 

The problem was in Scope of var $num=1, move after line "//unset($root);" and solved!!!!

 

Thank for your help!!!!!!