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.

Please show us how you're using that code, the problem is not very clear at the moment. — deceze
– deceze ♦, Commented Jan 28, 2011 at 5:06
I could be jumping the gun here by suggesting posting on codereview.stackexchange.com. It sounds like your trouble may be buried too deep in your code for a regular SO question. Otherwise, you'll have to be way more specific in showing the problem lines. — bob-the-destroyer
– bob-the-destroyer, Commented Jan 28, 2011 at 5:29
@Tablekin: also, just a suggestion for posting code examples on SO: you can write out all the code you want in a text editer such as Notepad++, select all text, hit tab once to insert proper indenting, then copy/paste into here so everything is automatically properly formatted. — bob-the-destroyer
– bob-the-destroyer, Commented Jan 28, 2011 at 5:36

Phil · Accepted Answer · 2011-01-28 05:36:15Z

3

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.

edited Jan 28, 2011 at 5:36

answered Jan 28, 2011 at 5:30

Phil

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1 Comment

Tableking Over a year ago

Thanks much. That was the problem. Cheers.

Imrul.H · Accepted Answer · 2011-01-28 05:49:45Z

0

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'])

answered Jan 28, 2011 at 5:49

Imrul.H

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