I'm trying to pass a variable into an include file. My host changed PHP version and now whatever solution I try doesn't work.
I think I've tried every option I could find. I'm sure it's the simplest thing!
The variable needs to be set and evaluated from the calling first file (it's actually $_SERVER['PHP_SELF']
, and needs to return the path of that file, not the included second.php
).
OPTION ONE
In the first file:
global $variable;
$variable = "apple";
include('second.php');
In the second file:
echo $variable;
OPTION TWO
In the first file:
function passvariable(){
$variable = "apple";
return $variable;
}
passvariable();
OPTION THREE
$variable = "apple";
include "myfile.php?var=$variable"; // and I tried with http: and full site address too.
$variable = $_GET["var"]
echo $variable
None of these work for me. PHP version is 5.2.16.
What am I missing?
Thanks!
You can use the extract() function
Drupal use it, in its theme() function.
Here it is a render function with a $variables
argument.
function includeWithVariables($filePath, $variables = array(), $print = true)
{
$output = NULL;
if(file_exists($filePath)){
// Extract the variables to a local namespace
extract($variables);
// Start output buffering
ob_start();
// Include the template file
include $filePath;
// End buffering and return its contents
$output = ob_get_clean();
}
if ($print) {
print $output;
}
return $output;
}
./index.php :
includeWithVariables('header.php', array('title' => 'Header Title'));
./header.php :
<h1><?php echo $title; ?></h1>