Does $_POST work with a global variable? - php

Question

I have this:
page1.php
Number of students<input type="number" name="number">
page2.php
I get the data from the <input> using the $_POST method and assign it to a var $a.
<? $a = $_POST['number']; ?>
page3.php
I define a fucntion in order to calculate something using a global variable.
<? function Calculate(){
global $a;
include 'page2.php'
echo (3+10)/$a;
}?>
Now, it says Warning: Division by zero in C:/.../page3.php on line 4
But if you define a var $a = 3, for example, in the page2.php and call it in page3.php
<? function Calculate(){
global $a;
include 'page2.php';
echo (3+10)/$a;
}?>
It works perfectly.
Assuming that $_POST method gather vars, why isn't working with a global allocation? Am I doing something wrong?
Any help or suggestion will be appreciate.

Answer 1

The value of $a will be set only when a POST request is made to page2.php. But then, you won't be able to call your function as it is in page3.php.
If you go directly to page3.php, then there is no POST request and the value of $a will be 0. Thus the Division by Zero error.
If you want, you can include the page3.php in your page2.php and call your function from there when the POST request is made. But when doing this, you won't be needing that declaration of $a in the function. Instead you can pass it as an argument to the function.
function calculate($a=1) {
/* Your code */
}
The $a=1 in the function header is used to supply default value when nothing is passed when the function is called.
So your page2.php will look like
<?php
$a = $_POST['number'];
include 'page3.php';
Calculate($a);
?>
and your page3.php will look like
<?php
function Calculate($a=1) {
echo (3+10)/$a;
}
?>
Good Luck!

Answer 2

The problem is not at the global variables, I doubt if you should use them in this case.
It's because of the $_POST['number']; your code fails.
When you include a page, the data of POST-request you did before (by submitting the form) does not exist anymore, resulting in no value being assigned to $a
However, when you're hard-coding $a = 3, the code will work because it's not depending on any kind of submitted data.
I would suggest you to go from page1 directly to page3 and retrieve the form values there, because I'm not getting the point of adding an extra page in-between.