I'm trying to find a way to implement an Interface only when this Interface is available.
The Interface in question is
PrestaShop\PrestaShop\Core\Module\WidgetInterface
From Prestashop. It's used in a module.
The thing is, in order to be compatible with multiple version of Prestashop, the code must handle the case where WidgetInterface does not exists.
I was thinking in testing the existence of the interface and import it after, like this:
if (interface_exists('PrestaShop\PrestaShop\Core\Module\WidgetInterface')) {
use PrestaShop\PrestaShop\Core\Module\WidgetInterface
} else {
interface WidgetInterface {}
}
But of course, it's not possible to use use inside a if statement.
I then tried to do some try/catch, but that's the same issue (too bad it's not Python).
How can I do to implements WidgetInterface only when available?
You can't implement an interface dynamically, like you say, but you can write your own interface and only require it if the other does not exist.
Ie: your interface would be something like widget_interface.php, or whatever you want to call it, as long as it's not PSR-0/4 compliant, or autoloaded in whatever way you normally do.
<?php
namespace PrestaShop\PrestaShop\Core\Module;
/**
* This is the replacement interface, using the same namespace as the Prestashop one
*/
interface WidgetInterface
{
}
Then, in your class, you can do the following:
<?php
namespace App;
if (!interface_exists('\PrestaShop\PrestaShop\Core\Module\WidgetInterface')) {
require __DIR__ . '/path/to/widget_interface.php';
}
class WhateverClass implements \PrestaShop\PrestaShop\Core\Module\WidgetInterface
{
}
Your replacement interface will only be loaded if the Prestashop one doesn't exist.