1

Someone please explain to me why this doesn't work, and what I am doing wrong. For some reason, when I run the function validateUsername, the $error variable remains completely unchanged, instead of evaluating to true. How is this possible?

Yet, if I remove the code within the function and run it straight without a function call, it works. The example below is so simple it is practically pseudo code, and yet it doesn't work. Is this behavior unique to PHP? I don't want to run into this again in some other language.

<?php

$username = 'danielcarvalho';
$error = false;

function validateUsername()
{
    if (strlen($username) > 10)
    {
        $error = true;
    }
}

validateUsername();

if ($error == false)
{
    echo 'Success.';
}
else
{
    echo 'Failure.';
}

?>
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3
  • 3
    Read up on variable scope: php.net/manual/en/language.variables.scope.php Commented Apr 16, 2011 at 13:37
  • 1
    In addition to variable scope, also read up on returning values from functions Commented Apr 16, 2011 at 13:40
  • 1
    It's just in a function on the same page, surely this can't be a variable scope issue? Guess I'm wrong. I'll read up. Commented Apr 16, 2011 at 13:42

2 Answers 2

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7

This isn't working because $username isn't available within the scope of your validateUsername function. (Neither is the $error variable.) See the variable scope section of the PHP manual for more information.

You could fix this by adding global $username, $error; within your function, although this isn't a particularly elegant approach, as global variables are shunned for reasons too detailed to go into here. As such, it would be better to accept $username as an argument to your function as follows:

<?php
    function validateUsername($username) {
        if (strlen($username) > 10) {
            return false;
        }

        return true;
    }

    if (validateUsername('danielcarvalho')) {
        echo 'Success.';
    }
    else {
        echo 'Failure.';
    } 
?>
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3 Comments

Thanks chap, great answer. I'm still just struggling to cope with the fact that it is a scoping issue. Seems illogical / impractical to me.
@Daniel It might seem that way initially - but just think about it as trying to keep the "locality" of variables to an absolute minimum. (In other words, as a means of obtaining the highest level of separation/isolation, which is always a good thing as you'll discover when you encounter things like object-oriented programming.) Happy travels! :-)
To be fair though, I'm very experienced in ActionScript and have done a moderate amount of C#, and both of them never had this strict scoping issue where they can't see a variable declared just outside of a function.
1

$error has local scope in function validateUsername. To access global variables, use global keyword.

Read about scopes here. Change your function to:

function validateUsername($username)
{
    global $error;
    if (strlen($username) > 10)
    {
        $error = true;
    }
}

validateUsername($username);

Better implementation using function parameter:

function validateUsername($username, &$error)
{
    if (strlen($username) > 10)
    {
        $error = true;
    }
}
validateUsername($username, $error);

Another implementation:

function validateUsername($username)
{
    if (strlen($username) > 10)
    {
        return true;
    }
    return false;
}
$error = validateUsername($username);
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6 Comments

global variables aren't the answer to all the world's problems: they're likely to cause more problems than they solve, and aren't to be recommended when function arguments and return values are a more correct approach
@Ashwini Both of your solutions are a bit awry. In the first one, you need to use &$error in the function argument (otherwise, $error won't be updated in the "calling" scope) and I'd really avoid the needless use of a global as far as the second one is concerned.
Your second example will not work as expected - variables are passed by value by default, your global $error will not change whatever you do inside the function.
I voted you up for the detailed answer as well. Feels wrong not to award you with something. Hmmm, interesting comments though, will keep these concerns in mind.
@middaparka and Maelyn. Don't know how did I write it in first place! Have edited it.
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