Access a variable outside of a function

Question

I would elaborate further on this if I could, but the main problem I'm having is I can't access a variable outside of a function. I've tried both methods which I have below.

Here is my source:

function register($rand, $code) {
    global $rand, $access_token;
    if ($code == $access_token)
        $rand = rand('100000','1000000');
}   

echo $rand;

Should work ok, if you call the function. http://phpfiddle.org/main/code/gz9-x72 — Anirudh Ramanathan, Mar 25 '13 at 06:09
@ChristianRankin Using global doesn't mean the function will run itself. It still needs to be called. — Anirudh Ramanathan, Mar 25 '13 at 06:13

score 1 · Answer 1 · answered Mar 25 '13 at 06:18

may be it's help for you

function register($rand, $code) {
    global $rand, $access_token;
    if ($code == $access_token) {
        $rand = rand('100000','1000000');
    }



return $rand;
}

echo register($rand, $code);

score 0 · Accepted Answer · answered Mar 25 '13 at 06:11

0

function register($rand, $code) {
    if ($code == $access_token)
        $rand = rand('100000','1000000');
}

then call function first

register('1','2');
 echo $rand;

answered Mar 25 '13 at 06:11

Jhonathan H.

2,674
1
16
26

Sorry, but when calling register(1,2), what do those numbers represent? Can't I just put register($rand) – Mar 25 '13 at 06:13
means you are passing the value of $rand and $code parameter from your function.. – Jhonathan H. Mar 25 '13 at 06:14
You're method was effective, but DarkCthulhu said it best above. register(0) worked. – Mar 25 '13 at 06:18
1

@ChristianRankin don`t worry both are still the same they just differ in there parameters Darkchuchu only pass 1 parameters but here you passed 2 parameters .Because your function above state that you must pass 2 certain parameters.. – Jhonathan H. Mar 25 '13 at 06:21

score 0 · Answer 3 · edited May 23 '17 at 12:04

0

Please consider using arguments instead of using $GLOBALS or using global $var, because its recommended [take a look at these questions: one & two]

function register($rand, $code, $access_token) {
    if ($code == $access_token){
      $rand = rand('100000','1000000');
    }
    return $rand;
}

echo register($rand, $code, $access_token);

edited May 23 '17 at 12:04

Community

1
1

answered Mar 25 '13 at 06:14

score 0 · Answer 4 · answered Mar 25 '13 at 06:15

I guess it should run ok, if you are calling function before accessing $rand variable.

function register($rand, $code) {
    global $rand, $access_token;
    if ($code == $access_token) {
        $rand = rand('100000','1000000');
    }
    echo "\nInside Function = ".$rand;
    echo $access_token;
}   

register(100, '');
echo "\nOutside Function = ".$rand;

score 0 · Answer 5 · answered Mar 25 '13 at 06:19

You have to call the function before $rand is set. Global $rand inside the function does not set any value before you call register().

<?php
    function register($rand, $code) {
        global $rand, $access_token;
        if ($code == $access_token)
            $rand = rand('100000','1000000');
    }   
    register('x', 'y');

    echo $rand;
?>

If PHP would set all global values before calling functions, then think of the following. What would $rand be set to? It only makes sense when one of the functions are called.

<?php
function a() {
global $rand = 1;
}
function b() {
global $rand = 2;
}

Access a variable outside of a function

5 Answers5