passing result from include

[chrislive]

hi everyone,

I have a script that includes a page from another server and Im having problems passing the value back to the original page.

any ideas?

this is my first code which first checks the version of a local file and then includes the server checking page.

<?php

$update_server = "http://www.sourceserver.com";

// Lets pick our filename
$filename = 'ws_gigs.php';

// Lets check the version of the local copy first
if (file_exists($filename)) {
   echo "<br>$filename current version: " . date ("Y-m-d H:i:s", filemtime($filename));
   $current_version = date ("YmdHis", filemtime($filename));
}

// Lets include the filename information and get the source server version
$file_path = "$update_server/check_updates.php?filename=$filename";
include($file_path);

echo "<br><br>";

// Lets check the difference in the version
if($current_version > $latest_source){
	echo "You have an early current version, please update.";
}else{
	echo "You have the current version";
}

echo "<br><br>";

// Display the results clearly
echo "Current Version: $current_version<br>";
echo "Latest Version: $latest_source<br>";
?>

then the server page looks like this:

<?php

$update_filename = '$filename';

if (file_exists($filename)) {
   echo "<br>$filename latest version: " . date ("Y-m-d H:i:s", filemtime($filename));
   $latest_source = date ("YmdHis", filemtime($filename));
}

?>

I need to get this $latest_source value back to the first page to do my checking.

Your suggestions would be appreciated.

Thanks
Chris

[norpel]

you could use session variables to do this quite easily.

at the start of each page just insert the line

session_start();

then on your first page at the end add

$_SESSION['latest_source']=$latest_source;

on your second page you can now access this variable by adding this somewhere after the session_start() line

$latest_source = $_SESSION['latest_source']

hope this helps,
norpel

[liquorvicar]

you don't need sessions for this. any variables in the include file are in the same scope as the calling file and hence available to it.

in your server page you have:

$update_filename = '$filename'; 

which sets $update_filename to the string literal '$filename' and not the value of $filename.
try changing it to:

$update_filename = "$filename"; 

with double quotes, PHP will parse this and replace $filename with its value.