PHP Variable Scoping

[jgbishop]

Suppose I have the following code:

function one() {
var $a;
two();
}

function two() {
// I want to modify $a in this function
// Unfortunately, I cannot pass $a as a function parameter
}

I want to be able to modify variable $a in function two(). If I place the GLOBAL keyword in front of the $a in function two(), can I modify it appropriately? In that case, what happens when function one() returns? Does $a get destroyed, or is it still "global"?

 

[sourceninja]

a few ways to tacle this. You could use a static variable. That is a variable that is local to a function, but does not loose its value. or you could reference it as $GLOBALS['var'] or you could redefine the variable inside the function as global.

so we have

$a = 1;
$b = 1;
$c = 0;
function something() {
$GLOBALS['c'] = $GLOBALS['a'] + $GLOBALS['b'];
}

something();
echo $c;

That would print '2'

or we could do this

$a = 1;
$b = 1;
$c = 0;
function add() {
global $a, $b, $c;
$c = $a + $b;
}
add();
echo $c;

or finally we could do this.

or we coudl use a static.

function Test(){
static $a = 0;
echo $a;
$a++;
}

everytime you run that function a will increase by 1. So the first time a =0 then a=1 and so on.

personally, i perfer to use the $GLOBALS variable. But im curious. Why can't you pass the variable?

 

[jgbishop]

I cannot pass the variable as a parameter because function two() is a predefined callback. Unfortunately, there is no field that allows user data to be passed through (an unfortunate design). I need to manipulate a variable in function two(), and was wondering how best to do it. Thanks for the suggestions.

 

[statik213]

You can pass by ref in PHP.

Passing by Reference
You can pass variable to function by reference, so that function could modify its arguments. The syntax is as follows:


<?php
function foo (&$var)
{
$var++;
}

$a=5;
foo ($a);
// $a is 6 here
?>