I'm trying to use Repositories which implement my RepositoryInterface.
As a C# .NET developer I would create a Generic interface like this:
public interface IRepository<TModelType>
{
IEnumerable<TModelType> FindAll();
void Save(TModelType model);
}
I would implement it like this:
public class UserRepository : IRepository<User>
{
public IEnumerable<User> FindAll()
{
// Method body
}
public void Save(User model)
{
// Method body
}
}
Now my question: Is there something like this in PHP?
EDIT:
The reason I want to do this is type safety and code completion assistance and I don't want to cast the model parameter all the time. ;-)
I don't know much about C# because I don't really have any experience with it but I am guessing you want to do something like this:
interface RepositoryInterface
{
public function findAll();
public function save();
}
class UserRepository implements RepositoryInterface
{
protected $user;
public __construct(User $user)
{
$this->user = $user;
}
public function findAll()
{
// $this->user->modelMethod();
// Method body
}
public function save()
{
// $this->user->modelMethod();
// Method body
}
}
There are no return types for methods in PHP 5.6 (PHP 7 will bring return types and scalar typehints). Currently, you can type hint method arguments only for objects and arrays, int string bool (scalar data types) are not supported.
PHP is a loosely typed language so you don't really have to typehint stuff, but it supports some kind of typehinting if you want to do that.
function save(Model $model); // Method in Interface
function save(Model $model); // Method in UserRepository implementing the RepositoryInterface
When you call the method somewhere in the controller or a service class, you should be able to pass a User object and be sure that it's a User object (if that's the model object you want to pass as an argument). Probably a check which class that object belongs to is a good idea here.
@Mythos33, unfortunately PHP does not have native generics, so you'll need an interface for each type of repository if you want to be able to type hint a specific type of Model. The only other option is to be less strict and allow anything that is a Model, as @Ruffles suggested.
