Passing Variables back from a full URL include

hatch789

Hi guys. I spent 2 full days searching the internet and PHPBuilder for answers to my question. I keep getting close, but no one seems to have hit my problem on the head.

I need to include a FULL URL call in one of my functions. I am able to send the variables to that URL without a problem. But when I try to get a variable to come back from that "include" it comes back as only a 1.

This is consistent with my research, but there has to be a way to do this. I do NOT want to name the URL file .inc or anything risky. I want to keep it .php.

I think I'm just missing something small here so I hope you can help.

Code examples below:
test.php program
<?php
echo $LocalVersion; //this works fine
$MasterVersion = $LocalVersion + 999;
echo $MasterVersion; //this works fine also
return $MasterVersion;
?>

Main Program
<?php
$MasterVersion = include "http://domain.com/test.php?LocalVersion=20";

echo $MasterVersion; // This does not work. 1 comes back
?>

There must be a way to do this without using .inc extensions.

Thanks for the help!
-Chris

jpmoriarty

hummm... not too sure.

an alternative though (if you're open to them): why not have the include file contain a function. Then when you call the include file, the function is then there so that you can call it later on. it looks to me that this is more along the lines of what you want to achieve.

hatch789

Well I tried your suggestion but as I suspected, the function is now outside of scope once I am back in the main program.

I need to include this file with a FULL URL because I have some sensitive information in the code that I do not want others to see. This is the only way I know to keep it secure, so I am going to need another option as to how I can pass this variable back from the included file?

Suggestions, help or moral support is appreciated greatly!

-Chris

diego25

Try

test.php program
<?php
echo $LocalVersion; //this works fine
$MasterVersion = $LocalVersion + 999;
print "<? \$MasterVersion = $MasterVersion; ?>";
?>

Basically you want the script on the other server to output PHP code, so that the includer script will interpret the PHP code.

Diego