How do you pass a parameter into an include file? I tried the following but it doesn't work.
include "myfile.php?var=123";
and in myfile.php, I try to retrieve the parameter using $_GET["var"].
include "myfile.php?var=123"; will not work. PHP searches for a file with this exact name and does not parse the parameter
for that I also did this:
include "http://MyGreatSite.com/myfile.php?var=123";but it does not also work.
Any hints? Thanks.
Wrap the contents of the included file in a function (or functions). That way you can just do
include "myfile.php";
myFileFunction($var1, $var2);
quick and dirty
$var = 123;
include "myfile.php";
in myfile just use "$var"
The same but without global variables:
function use_my_file($param) {
$var = $param;
include "myfile.php";
}
Hold on, are you trying to include the result of myfile.php, not its php code? Consider the following then
$var = 123;
readfile("http://MyGreatSite.com/myfile.php?var=$var"); //requires allow_url_fopen=on in php.ini
virtual might also work
Create a function, that's what they are for:
<?php
function doFoo($param) {
// do something with $param
}
<?php
require_once 'included.php';
doFoo('some argument');
Code in included files is executed in the same scope as the including file.
This:
// main.php
$var = 'a';
include 'inc.php';
echo $var;
// inc.php
$var = 'b';
Is for most intents and purposes exactly the same as:
// main.php
$var = 'a';
$var = 'b';
echo $var;
