Why function parameters doesn't work with class - php

Question

I created a class. The code is below
class Another {
public $error = array();
public function set_error( $key, $value )
{
if ( isset( $key ) ) {
$sanitizedKey = sanitize_key( $key );
$this->error[ $sanitizedKey ] = wp_json_encode( $value );
return $this->error;
}
}
public function get_error( $id )
{
if ( ! is_null( $id ) ) {
return $this->error[ $id ];
}
}
public function print_error()
{
if ( $this->error ) {
foreach ($this->error as $key => $value) {
$decodeJson = json_decode( $value );
?>
<div class="ud-error">
<p class="ud-error-<?php echo $key; ?>">
<?php echo __( $decodeJson, 'ud-for-edd' ); ?>
</p>
</div>
<?php
}
}
}
}
If I invoke it in the following way it works. It echos the content as expected.
$error = new Another();
$error->set_error('gg', 'hhhh');
$error->print_error();
But if I use it with function then it doesn't work as expected. Do I have to pass parameters by reference or any other? The following way it doesn't work
function create_error($id, $val) {
$errr = new Another();
return $errr->set_error($id, $val);
}
create_error('raa', 'raashid');
$error = new Another();
$error->print_error();
I am confused about why this doesn't work. Any clue. Thanks in advance.
Steps I want the code to perform:
Create a class with 3 methods, set_error, get_error, and print_error
Then invoke the class inside the function. The function will accept two parameters because inside the class the set_error() method accepts two parameters.
In order to print the error, I will instantiate the class and call the print_error() method.
In case, if I have to print the new error. I will just call the create_error() function to do this for me. Because the function needs 2 parameters. The arguments supplied to the function must be supplied as arguments to the set_error() method.
I hope the list helps.
Update:
If I use a global variable then it works. Below is working.
$customError = new Another();
function create_error($id, $val) {
global $customError;
$customError->set_error($id, $val);
}
create_error('raa', 'rashid');
$customError->print_error();
Thanks, #roggsFolly and #El_Vanja. By understanding your tips I was able to solve the error. If there is anything wrong with the code I just said worked. Please point out.

Answer 1

The object you instantiate inside the function is not the same one you try and print the error message from.
First the object you instantiate inside the function scope is not visible outside the function.
function create_error($id, $val) {
$errr = new Another();
return $errr->set_error($id, $val);
}
create_error('raa', 'raashid');
// this instantiates a seperate Another object from the one
// you created in the function
$error = new Another();
// this tries to print from the new object taht has no message stored in it yet
$error->print_error();
To instantiate the object inside a function scope and then use that object outside the function scope you must pass that object back to the caller of the function
function create_error($id, $val) {
$errr = new Another();
$errr->set_error($id, $val);
return $errr; // this returns the object
}
$error = create_error('raa', 'raashid');
// now you can use its methods to print the stored message
$error->print_error();
Update as per your additional Information
A couple of things I think you may be getting confused about.
Each time you do $var = new ObjectName; you are creating a brand new instance of that class. This new instance has no knowledge about any other instances of that class that may or may not have been created before or may be created after that point. And more specifically to your problems, it does not have access to the properties of another version of that object.
You are I believe missing the concept of variable scope. The Object you create inside that function, will only actually exist while the function is running. Once the function completes anything created/instantiated wholly within that function is DESTROYED ( well in truth it is just no longer accessible ) but to all intent and purpose it is destroyed. you therefore cannot expect to be able to address it outside the scope of the function.
If you want the Object you instantiate within the function to be usable outside the function, you must pass a reference to that object out of the function to the calling code. This passes that reference into the scope of the calling code and keeps the object alive, global scope in your case, but that might be another function or even another object. That allows you access to that instantiation and any properties that were set within it.