php passing and looping an array into a class method - php

Question

i'm some what new to php classes, so apologies if this is an obvious one.
i'm trying pass an array to a class method, but i get a warning when i try to foreach loop it
Warning: Invalid argument supplied for foreach() in....
the foreach does actually loop the array and gives me the desired result, so i'm confused about the warning, and i'm not finding any simple explanation on how to fix it/ do it the correct way. i also tried passing the array as a json_encode then decoding back to an array once inside the method, and the same warning occurs. what also confuses me, i get nothing if i print_r($this->payload) inside the method.
here's my class...
// class
// --------------------------
class Form {
public $form_action;
public $form_method;
public $form_class;
public $form_role;
public $payload;
function form() {
$output = '';
foreach ($this->payload as $key => $value) {
$output .= '<input type="'.$key.'" name="'.$key.'" ......>';
}
return $output;
}
}
here's the array...
// array
// --------------------------
$my_array = (object)array (
"form1" => (object)array (
"foo1" => "bar1",
"foo2" => "bar2"
)
);
and here's where i pass the array to the class...
// set class
// --------------------------
$set = new Form;
$set->form_action = '';
$set->form_method = 'post';
$set->form_class = 'form-signin';
$set->form_role = 'form';
$set->payload = $my_array->form1;
echo $set->form();

Answer 1

There is no option1 in your array (what is actually object). Anyway, why do you call it array, when its an object? So that will be $my_array->form1
And you are using the same name of your function to show the form, as the name of the class, so that will be the constructor. And there will be no value for $this->payload at the first run.
So your final code will be this:
class Form {
public $form_action;
public $form_method;
public $form_class;
public $form_role;
public $payload;
function showForm() {
$output = '';
foreach ($this->payload as $key => $value) {
$output .= '<input type="' . $key . '" name="' . $key . '" ......>';
}
return $output;
}
}
$my_array = (object) array(
"form1" => (object) array(
"foo1" => "bar1",
"foo2" => "bar2"
)
);
$set = new Form;
$set->form_action = '';
$set->form_method = 'post';
$set->form_class = 'form-signin';
$set->form_role = 'form';
$set->payload = $my_array->form1;
echo $set->showForm();

Answer 2

I can't tell from your warning message. But I think that if you were to look at the stack trace that the warning is being generated via the line $set = new Form;
From the php manual about constructors
For backwards compatibility, if PHP 5 cannot find a __construct()
function for a given class, and the class did not inherit one from a
parent class, it will search for the old-style constructor function,
by the name of the class. Effectively, it means that the only case
that would have compatibility issues is if the class had a method
named __construct() which was used for different semantics.
Because in your example class, your method shares the same name as the class and does not have a __construct method. The form() method is being called when you create the object at which point $this->payload is null.
You can fix this by either creating a __construct method or in your class definition initializing $this->payload to an empty array.