John Davidson

Message:

I'm working on a Laravel project which is going to be a full API project. I've coded myself a way to do API versioning. I've structured my projects by adding version numbers in the architecture of Laravel.

For example, I'm working on the V0 of my API, I'm going to this architecture with my models : App\Models\V0\MyModel

I was wondering if my any chance, it's possible depending on the current version of the API (which is set in a middleware) to load models depending on that current variable in a use statement ? For example :

use App\Models\V0\User;

Is it possible somehow to stock "V0" into a variable and use it in the "use" to load dynamically the good version of the model ?

---

My Answer:
include() or require() functions in PHP is not possible. These functions expect a string literal as the file path parameter, so you cannot use a variable to dynamically load a class.

However, you can use the autoloading feature in PHP to dynamically load classes based on their namespace or file structure. This can be achieved by registering an autoloader function using the spl_autoload_register() function.

Here's an example of how you can use autoloading to dynamically load classes in PHP:

php
// Define the autoloader function
function myAutoloader($className) {
$file = 'path/to/classes/' . $className . '.php';
if (file_exists($file)) {
require_once $file;
}
}

// Register the autoloader function
spl_autoload_register('myAutoloader');

// Now you can create an instance of a class without explicitly including the file
$obj = new MyClass();


In this example, the autoloader function `myAutoloader()` is defined to load classes from a specific directory based on their class name. The `spl_autoload_register()` function is then used to register this autoloader function.

Now, whenever you try to create an instance of a class that has not been included in the script, the autoloader function will be called to dynamically load the class file.