PHP: Global var not being picked up inside a function

Question

This is blowing my mind...

I've got a standalone PHP file, and a simple function with a global var.

<?php
    $var = 4;

    function echoVar()
    {
        echo $var; 
    }

    echoVar();
?>

When I call echoVar() nothing is returned... However if I place the $var inside the function it will return 4.

What's going on here? Shouldn't $var be global in this case?

Take a look at this: http://php.net/manual/en/language.variables.scope.php — Babblo, Nov 29 '13 at 13:13
You seem to be confusing `PHP` with how scope works in some other languages (js for example). Please [refer to the manual to see how scope works in PHP](http://php.net/manual/en/language.variables.scope.php) — Wrikken, Nov 29 '13 at 13:14
Try like this: function echoVar() { global $var; echo $var; } — kidz, Nov 29 '13 at 13:16

score 5 · Accepted Answer · edited May 23 '17 at 12:26

5

If a variable is set outside of a function, it's not visible inside that function. To access it, you must declare it global with the global keyword. This is called a scope.

<?php

$var = 4;

function echoVar() {
    global $var;

    echo $var;
}

echoVar();

Note: This is generally considered bad practice. Read this for more information.

A good alternative would be to pass in the variable as an argument:

<?php

$var = 4;

function echoVar($var) {
    echo $var;
}

echoVar($var);

edited May 23 '17 at 12:26

Community

1
1

answered Nov 29 '13 at 13:14

Terry Harvey

9,286
1
17
22

Thank you this helped me understand it much better. I come from JS world so this seemed very strange indeed! – Allen S Nov 29 '13 at 13:27

score 3 · Answer 2 · answered Nov 29 '13 at 13:14

3

Lots of options here... like

<?php
    $var = 4;

    function echoVar($var)
    {
        echo $var; 
    }

    echoVar($var);
?>

or

<?php
    $var = 4;

    function echoVar()
    {
        global $var;
        echo $var; 
    }

    echoVar();
?>

answered Nov 29 '13 at 13:14

superphonic

7,954
6
30
63

Thanks.. fixed. As for bad practice, it's just another example on how to accomplish this. The code is there to be used. – superphonic Nov 29 '13 at 13:18

score 0 · Answer 3 · answered Nov 29 '13 at 13:19

0

You can either have $var as an argument, like this:

$var = 4;

function echoVar($var)
{
    echo $var; 
}

echoVar($var);

or use global, like this:

$var = 4;

function echoVar()
{

    global $var;
    echo $var; 
}

echoVar();

answered Nov 29 '13 at 13:19

Ana Claudia

501
6
16

score 0 · Answer 4 · answered Nov 29 '13 at 13:23

0

When you call any function it's create local variable so you have to pass argument in calling function part.

    $var = 4;

    function echoVar($var)
    {
        echo $var; 
    }

    echoVar($var);

answered Nov 29 '13 at 13:23

Martin C

395
1
7
21

score 0 · Answer 5 · answered Nov 29 '13 at 13:25

Just going to clarify since everyone seems to be posting mostly rubbish.

Do not use global $var;
Do not echo out inside of a function
Output from a function does not need to be assigned to a variable before being echo'd

This is how it "should" be done.

<?php
    $var = 4;  //set initial input var this is external to the function

    function echoVar($internalvar) {  /*notice were accepting $var as $internalvar I'm doing this to clarify the different variables so you don't end up getting confused with scope  $internalvar is local to the function only and not accessible externally*/
        return $internalvar; //now we pass the function internal var back out to the main code we do this with return you should never echo out your output inside the function
    }

    echo echoVar($var);  //call function and pass $var in as an arguement
?>

PHP: Global var not being picked up inside a function

5 Answers5