PHP using variables outside of a function

Question

I was wondering if someone could help me out.

I'm writing some PHP functions where I would like to use a variable inside a function and then use the updated variable outside of the function.

eg

$subTotal += $priceToShow * $ct_qty;
adminCharge($subTotal);
function adminCharge($loggedInfoprice, $subTotal){
    if (($loggedInfoprice == 5)||($loggedInfoprice == 6)){
    $packagingCurrency = "$";
    $packagingPrice = "63";
    $packagingPricing = $packagingCurrency.$packagingPrice;
}else {
    $packagingCurrency = "£";
    $packagingPrice = "30";
    $packagingPricing = $packagingCurrency.$packagingPrice;
}
if(isset($packagingPrice)){
    $subTotal = $subTotal + $packagingPrice;
}
}

<?php echo $subTotal; ?>

The echo'd $subTotal only seems to show the value from before the function.

Is it possible to extract both $subTotal and $packagingPricing from the function?

Thanks in advance.

Updated the function

Community · Accepted Answer · 2020-06-20 09:12:55Z

3

function adminCharge(&$subTotal){
// do things
}

or

$subTotal = adminCharge($subTotal);
function adminCharge($subTotal){
 //do things
 return $subTotal;
}

In the first case, you pass $subTotal variable as a reference, so all all changes will be reflected to "outside" variable.

In the second case, you pass it and return new value, so is easy to understand why is working.

Please, pay attention

First solution - however - can lead you to side effect that will be difficult to understand and debug

Update

If you want to "extract" (so return) two variables, you can do the following

list($subTotal,$packagingPrice) = adminCharge($subTotal);
function adminCharge($subTotal){
    if(isset($packagingPrice)){
        $subTotal = $subTotal + $packagingPrice;
    }
    return array($subTotal, $packagingPrice);
}

edited Jun 20, 2020 at 9:12

CommunityBot

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answered May 15, 2013 at 9:57

DonCallisto

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7 Comments

ty812 Over a year ago

This (pass by reference) actually is discouraged by PHP and deprecated as of PHP 5.3.0

ngplayground Over a year ago

I have updated my question to ask how to extract more than one variable from the function.

ChrisH Over a year ago

Shouldn't reference the variable, it's not an object?!

billyonecan Over a year ago

@MartinHohenberg I thought only call-time pass-by-reference was deprecated, but actually passing by reference in the function definition is perfectly okay?

billyonecan Over a year ago

@DonCallisto it's perfectly fine to pass by reference when defining the function, just not when calling it.

|

ty812 · Accepted Answer · 2013-05-15 09:57:51Z

1

function adminCharge($subTotal, $packagingPrice){
    if(isset($packagingPrice)){
        $subTotal = $subTotal + $packagingPrice;
    }

    return $subTotal;
}

Use it like this:

$subTotal = adminCharge($subTotal, $packagingPrice);

answered May 15, 2013 at 9:57

ty812

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4 Comments

ChrisH Over a year ago

You're missing $packagingPrice, it should be globalised since it's not a parameter..

ty812 Over a year ago

@ChrisH Fixed that ... Globalizing stuff is a bad idea though, it makes debugging more complicated. Best practice is to never do that.

ChrisH Over a year ago

It sure is! However, it's his function and the var isn't set, so it must be used elsewhere. It should be the second parameter, but since it's not it doesn't exist in the function, globalise the var and it does exist..

ty812 Over a year ago

For more information why globalizing variables is a horrible idea read c2.com/cgi/wiki?GlobalVariablesAreBad

ChrisH · Accepted Answer · 2013-05-15 09:58:11Z

1

You don't return anything, so you could use globals but i wouldn't recommend it.. Take a look though, because that's what your asking for: http://www.php.net/global

In my opinion, change your function:

function adminCharge($subTotal){
    global $packagingPrice
    if(isset($packagingPrice)){
         $subTotal = $subTotal + $packagingPrice;
    }
    return $subTotal; 
}

$subTotal += $priceToShow * $ct_qty;
$subTotal = adminCharge($subTotal);

Also keep in mind you didn't had $packagingPrice in your function, so i added it with global. Otherwise it wouldn't work :)

answered May 15, 2013 at 9:58

ChrisH

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Comments

jancha · Accepted Answer · 2013-05-15 10:02:09Z

0

You have to understand the variable scope read here.

In short, there are 3 options.

Use global keyword:

$a = 0;

function foo(){ global $a; $a += 1; }
Use pass by reference:

$a = 0;

function foo(&$b){ $b += 1; }

foo($a);
Collect return value of the function:

$a = 0;

function foo($b){ return $b + 1; }

$a = foo($a);

answered May 15, 2013 at 10:02

jancha

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Comments

alwaysLearn · Accepted Answer · 2013-05-15 10:20:20Z

0

you need to use pass the variable by reference or global

option 1

function adminCharge(&$subTotal){
if(isset($packagingPrice)){
    $subTotal = $subTotal + $packagingPrice;
}

return array($subTotal, $packagingPrice);

}

option 2

  function adminCharge()
  {
    global $subTotal;
    if(isset($packagingPrice))
    {
       $subTotal = $subTotal + $packagingPrice;
    }

    return  array($subTotal, $packagingPrice);

 }

edited May 15, 2013 at 10:20

answered May 15, 2013 at 10:01

alwaysLearn

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2 Comments

alwaysLearn Over a year ago

from where does $packagingPrice comes from .. ?

alwaysLearn Over a year ago

If you want to xtract both the values you have to return them as array .. see my updated answer for return

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