Undefined Variable + function not returning variable

Question

Okay so here's my code:

function payCalc($hours, $wage, $overtime, $paycheck) {
if ($hours <= 40) {
    $paycheck = $hours * $wage;
}
else {
    $paycheck = $hours * $overtime;
}
return $paycheck;

}

$name = $_POST['name'];
$dept = $_POST['dept'];
$hours = $_POST['hours'];
$wage = $_POST['wage'];

$paycheck = payCalc($hours, $wage, $overtime, $paycheck);

$overTime = $wage * 1.5;
print "Your paycheck for this period is:"; 
print $paycheck; 
print ".";

The problem is I keep getting undefined variable errors on $paycheck. As far as I know, I have the if/else set up properly, but it's not returning $paycheck. Which means that on the "print $paycheck;" line, it's not able to print. What am I missing?

EDIT: There IS more to this, but it's just echo lines to spit out the $name and $dept variables.

$hours, $wage and $overtime don't exist within the scope of your function - pass them as arguments; nor are you calling the payCalc() function anywhere; and in that function, you're not doing anything to update the value of $paycheck — Mark Baker
– Mark Baker, Commented Feb 18, 2013 at 9:40
you don't seem to be calling the payCalc() function anywhere? — Pedro del Sol
– Pedro del Sol, Commented Feb 18, 2013 at 9:41
Read PHP: Variable scope - functions don't close over variables like that. — cmbuckley
– cmbuckley, Commented Feb 18, 2013 at 9:41

fedorqui · Accepted Answer · 2013-02-18 09:58:12Z

3

You need to declare

$paycheck = payCalc();

so that you get the result coming from the function.

Also take in consideration:

if ($hours <= 40) {
    $paycheck = $hours * $wage;
}
else {
    $paycheck = $hours * $overtime;

that is, populate $paycheck.

And finally, take into account what has been advised by other users: if you want to use external variables, you need to send them to the function:

function payCall($hours, $wage, $overtime) {
   ...
}

and then use them when calling the function:

$paycheck = payCalc($hours, $wage, $overtime);

edited Feb 18, 2013 at 9:58

answered Feb 18, 2013 at 9:42

fedorqui

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8 Comments

Pachi Over a year ago

I did all that and now I'm getting 'undefined index' and 'undefined variable' errors for all of my variables.

fedorqui Over a year ago

Please update your code in the question so we can review what can be wrong now.

fedorqui Over a year ago

OK @user2082686, what you have to do is to define the function previously to calling it. That is, all the block starting on function payCalc($hours, $wage, $overtime, $paycheck) { and finishing in return $paycheck; } has to be moved to the top of the file.

Pachi Over a year ago

Did that. Updated code at the top. Same errors. Undefined index error on $name, $dept, $hours, $wage and undefined variable for $overtime and $paycheck.

Pachi Over a year ago

Also, $paycheck is returning now at the bottom (with the print line) but as 0. @fedorqui

|

Tikkes · Accepted Answer · 2013-02-18 09:41:53Z

0

In a function, you cannot call variables from outside it.

There's no way your function knows $paycheck.

do it like this:

function payCall($hours, $wage, $overtime, $paycheck) {
 //DO STUFF
return $paycheck;
}

and call it like this:

echo payCall($hours, $wage, $overtime, $paycheck);

answered Feb 18, 2013 at 9:41

Tikkes

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Comments

Nikolaj Jepsen · Accepted Answer · 2013-02-18 09:46:41Z

Well, you haven't really set the variable "paycheck" in your function.

Set the variable $paycheck before your $hours * $overtime and $hours * $wage.

However, before being able to access those, you have to set them as parameters.

The code should then be something like this:

$name = $_POST['name'];
$dept = $_POST['dept'];
$hours = $_POST['hours'];
$wage = $_POST['wage'];

$overTime = $wage * 1.5;

function payCalc($hours, $wage, $overtime) {
    if ($hours <= 40) {
        $paycheck = $hours * $wage;
    }
    else {
        $paycheck = $hours * $overtime;
    }
    return $paycheck;
}

print "Your paycheck for this period is:"; 
print payCalc($hours, $wage, $overTime); 
print ".";

The function now returns the variable "paycheck", which is set within the if-statement and the else-statement.

Does this make any sense to you?

Best,

Nikolaj Jepsen

Antonio Papa · Accepted Answer · 2013-02-18 09:49:25Z

0

Try this:

 if( isset($_POST['name']) && isset($_POST['dept']) && isset($_POST['hours']) && isset($_POST['wage']){
    $paycheck = '';
    $name = $_POST['name'];
    $dept = $_POST['dept'];
    $hours = $_POST['hours'];
    $wage = $_POST['wage'];

    $overTime = $wage * 1.5;

    function payCalc() {
        if ($hours <= 40) {
        $paycheck =$hours * $wage;
    }
    else {
        $paycheck = $hours * $overtime;
    }
    return $paycheck;
}

print "Your paycheck for this period is:"; 
print payCalc(); 
print ".";
}

edited Feb 18, 2013 at 9:49

answered Feb 18, 2013 at 9:43

Antonio Papa

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Comments

Justin L. · Accepted Answer · 2013-02-18 10:02:59Z

0

/// COPY AND PASTE THE BELOW CODE

$paycheck = '';

$name = $_POST['name'];

$dept = $_POST['dept'];

$hours = $_POST['hours'];

$wage = $_POST['wage'];

$overTime = $wage * 1.5;

function payCalc($paycheck,$hours,$wage,$overTime) {

    if ($hours <= 40) {

       $paycheck = $hours * $wage;

    }

    else {

       $paycheck =  $hours * $overTime;

    }

    return $paycheck;

}


print "Your paycheck for this period is:"; 

print payCalc($paycheck,$hours,$wage,$overTime); 

print ".";

edited Feb 18, 2013 at 10:02

Justin L.

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answered Feb 18, 2013 at 9:49

ajay

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Comments

DevMetal91 · Accepted Answer · 2013-02-18 09:42:56Z

-1

Try using global keyword.

$paycheck = '';
$name = $_POST['name'];
$dept = $_POST['dept'];
$hours = $_POST['hours'];
$wage = $_POST['wage'];

$overTime = $wage * 1.5;

function payCalc() {
    global $hours,$wage,$overtime,$paycheck;
    if ($hours <= 40) {
        $hours * $wage;
    }
    else {
        $hours * $overtime;
    }
    return $paycheck;
}

answered Feb 18, 2013 at 9:42

DevMetal91

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Comments

Omar MEBARKI · Accepted Answer · 2013-02-18 09:46:03Z

-1

You need to use global keyword to refer to global variables :

$paycheck = '';
$name = $_POST['name'];
$dept = $_POST['dept'];
$hours = $_POST['hours'];
$wage = $_POST['wage'];

$overTime = $wage * 1.5;

function payCalc() {
    global $hours, $wage, $overtime;
    $paycheck = 0;
    if ($hours <= 40) {
        $paycheck = $hours * $wage;
    }
    else {
        $paycheck = $hours * $overtime;
    }
    return $paycheck;
}


$paycheck = payCalc();
print "Your paycheck for this period is:"; 
print $paycheck; 
print ".";

answered Feb 18, 2013 at 9:46

Omar MEBARKI

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2 Comments

cmbuckley Over a year ago

This is clearly a new user of PHP - let's not advise such bad practise from the start.

Pachi Over a year ago

@cbuckley I am indeed new to this. I wouldn't know how to handle global variables at this point and I'd rather learn what's wrong (which I am through reading all this) rather than just put that (global) in without knowing what it does.

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