PHP Function Argument Default Value

Question

Hey all. I have a processForm function and a displayForm function. If there are missing form fields the processForm function returns an array of missing fields. This is all fine and dandy until I try to include this array into the displayForm function. Here's the problem:

If I don't do this:

displayForm($missingFields=array());

then my validateField function throws a warning that it is expecting the parameter to be an array. However, this overwrites the array returned by the processForm function.

I hope I'm clear. Thanks for any help.

Full Code:

if(isset($_POST['action']) && $_POST['action'] = "login")
{
    $messages = processForm();
}

processForm()

if($errorMessages)
{
     return array("errors" => $errorMessages, "missing" => $missingFields);
}
else 
{
    $_SESSION['user'] = $user;
    header("Location: index.php");
}

form.php

(!isLoggedIn())? displayForm($messages['errors']=array(),$messages['missing']=array()) : null;

These are the sections of the code I'm having trouble with.

Thanks again.

Answer 1

You don't set default argument values in the call, you set them in the signature, for example

function displayForm($arg1 = array()) {
    ...
}

---

When you write

displayForm($messages['errors']=array())

this is actually doing something like this

$messages['error'] = array(); // set $messages['error'] to an empty array
displayForm($messages['error']); // pass an empty array to displayForm

This is because in PHP, the return value from an assignment is the value assigned.

Answer 2

Why are you using this:

displayForm($messages['errors']=array(),$messages['missing']=array())

When you writing "$messages['errors']=array()" , this is setting $messeges to blank array. So the parameter is blank. You can just write:

displayForm($messages['errors'],$messages['missing'])