How to include only variables?

[chaking]

I have a file I'm trying to: include "update.php";

 

update.php has some echo statements if it functions properly, but it also sets the values of a lot of variables.

 

when I include update.php into update2.php, all the variables are passed very nicely, but it also prints out the rest of update.php (I have a form in update.php, so everytime I run update2.php it prints the update.php form AND the update2.php form)...

 

Basically, is there a way to only include the variable values in update2.php instead of the whole file?

[drisate]

<?php

//us a var is update.php

 

if (isset($forme_should_not_be_posted))[...]

?>

 

[chaking]

right, I understand how to include a file.

 

My question is: Is there a way to only include variables from a file instead of all the code? Because when I include all of the code from that file, it shows the form I created in the included file on the new page.

 

---OOps, ok you updated your response while I was responding.

 

So the isset function merely checks if the variable is already in use, correct?  I don't have a problem with passing the variable's value, I have a problem with the rest of the code of the included file being displayed in my final result.

[phpSensei]

You can define a CONSTANT

 

$data = "this is the data";

define("data",$data,true);

 

part 2

 

include("part1.php");

define("data",true);

echo data;

[chaking]

thanks phpsensei - 

 

I thought I might have to go that route.

[duclet]

Though I don't recommend doing this, you can try something like the following:

<?php
ob_start();
include 'update.php';
ob_end_clean();

// The rest of your update2.php file
?>

This will include the code of update.php but because of the output buffer, the output of update.php won't be shown. We use ob_en_clean to terminate the output buffer without outputting anything.

[chaking]

duclet: That was it! perfect!