How to make a fetched result fit into a model? - php

Question

Let's say I do a
$response = Http::get('http://example.com');
$i_want_to_be_a_model = $response->json();
And I have a \App\Models\Example model.
How can I make the $i_want_to_be_a_model to be an Example model?
I want to do this, because I want to add a method statusText() to the model, that is not part of the result.
class Example extends Model {
// ..
public string $statusText;
public int $status;
public function statusText() {
switch ($this->status) {
case 100:
$this->statusText = "foo";
break;
//..
default:
$this->statusText = "bar";
}
}
}
If there is a more elegant way if doing this, please let me know.

Answer 1

You can define a helper function or a Factory class to create objects of Example class.
For eg:
<?php
namespace App\Factories;
use App\Models\Example;
use Illuminate\Support\Facades\Schema;
class ExampleFactory
{
public function __construct(array $attributes)
{
$example = new Example;
$fields = Schema::getColumnListing($example->getTable());
foreach($attributes as $field => $value) {
if(in_array($field, $fields) {
$example->{$field} = $value;
}
}
return $example;
}
public static function makeFromArray(array $attributes)
{
return new static(... $attributes);
}
}
Then you can use the Factory as
// use App\Factories\ExampleFactory;
$response = Http::get('http://example.com');
$example = ExampleFactory::makeFromArray(json_decode($response->json(), true));
//Now you can do whatever you want with the instance, even persist in database
$example->save();