Consider the following code:
class CategoryController
{
private $model;
public function __construct()
{
$this->model = new CategoryModel();
}
}
You will see that Controller depends on the Model. I've heard that doing so is not desirable and Model should be injected instead.
I question why. In my case, I build CategoryModel
specifically for CategoryController
and I don't see a problem leaving it like this inside the class. I mean, I can't inject SomeOtherModel
that's not compatible in there anyway... or can I?
Using Dependency Injection to instead inject it into the Controller seems like waste of code.
Hence, is there any reason to use DI here?
Actually in that example, the big question is what is that for?! No methods just an object holder without any way to get it!?
Yea I know, it's just an example, but thats the problem, in that example you don't need DI, actually you not even need the class at all!
Has @Mark Baker said, without the DI/IoC you can't easily test, since it's tightly cupelled. If you take sometime to read about testing and for this case also Mockery
Using Dependency Injection to instead inject it into the Controller seems like waste of code.
When in cases where you don't have something that does the DI for you, it's easy to allow the objects to pass from constructor or make the default ones, in your example would be something like: use CategoryModelInterface;
class CategoryController
{
private $model;
public function __construct(CategoryModelInterface $categoty = null)
{
$this->model = $category
? $category
: new CategoryModel();
}
}
This way you don't lose much time, and when/if/maybe you'll do some testing, or change the model completely for another, it's actually possible to do it.
I don't think you'll need DI here. However Mark Baker's comment about testing is valid. But you can get around that by using a getter method for the model. That method can then be mocked in tests:
class CategoryController
{
private $model;
public function __construct()
{
...
}
public function getModel() {
if(!$this->model) {
$this->model = new CategoryModel();
}
return $this->model;
}
}