How can I set a variable variable from within a function in PHP?

Question

I am making a form, some of which is optional. To show the output of this form, I want to be able to check whether a POST variable has contents. If it does, the script should make a new normal PHP variable with the same value and name as the POST variable, and if not it should make a new normal PHP variable with the same name but the value "Not defined". This is what I have so far:

function setdefined($var)
{
    if (!$_POST[$var] == "")
    {
        $$var = $_POST[$var]; // This seems to be the point at which the script fails
    }
    else
    {
        $$var = "Not defined";
    }
}
setdefined("email");
echo("Email: " . $email); // Provides an example output - in real life the output goes into an email.

This script doesn't throw any errors, rather just returns "Email: ", with no value specified. I think this is a problem with the way I am using variable variables within a function; the below code works as intended but is less practical:

function setdefined(&$var)
{
    if (!$_POST[$var] == "")
    {
        $var = $_POST[$var];
    }
    else
    {
        $var = "Not defined";
    }
}
$email = "email"; // As the var parameter is passed by reference, the $email variable must be passed as the function argument
setdefined($email);

Read those three ^ and I'm pretty sure we you will find your answer. — PeeHaa
– PeeHaa, Commented Jan 4, 2013 at 18:35
Read links suggested by @PeeHaa - especially the first one, which deals with scope. Basically, your problem comes from the fact that the new variable is only available inside the setdefined function. — Wojciech Zylinski
– Wojciech Zylinski, Commented Jan 4, 2013 at 18:37

PeeHaa · Accepted Answer · 2013-01-04 18:50:01Z

2

Why don't you do it this way:

function setdefined($var)
{
    if (isset($_POST[$var]) && !empty($_POST[$var]))
    {
        return $_POST[$var];
    }
    else
    {
        return "Not defined";
    }
}
$email = setdefined('email');
echo("Email: " . $email);

The variable you create in first example is only available inside the function

edited Jan 4, 2013 at 18:50

PeeHaa

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answered Jan 4, 2013 at 18:40

Wojciech Zylinski

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3 Comments

Wojciech Zylinski Over a year ago

I wouldn't use empty tho, because it will return TRUE if $_POST[$var] == 0

microbug Over a year ago

That's a better way of doing it, I didn't know about the empty function but thanks for the introduction!

PeeHaa Over a year ago

Hmmm. At least add some check to verify the index exists AND it has a value to make sure you don't get warnings on e.g. checkboxes. Good and valid point though. In any case feel free to edit my edit :)

Sasha · Accepted Answer · 2013-01-04 18:44:07Z

0

You could make that unknown variable submitted by possibly malicious user, global one. To do so, add this to the start of your function:

global $$var ;

This is not only ugly, but maybe unsafe too.

answered Jan 4, 2013 at 18:44

Sasha

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