<?php
/**
* My codebase is littered with the same conditionals over and over
* again. I'm trying to refactor using inheritance and the Factory
* pattern and I've had some success but I'm now stuck.
*
* I'm stuck because I want to derive a new class from the one
* returned by the Factory. But I can't do that, so I'm obviously
* doing something wrong somewhere else.
*/
/**
* The old implementation was as follows. There's if statements
* everywhere throughout both LayoutView and ItemIndexView and
* SomeOtherView.
*/
class LayoutView { }
class IndexView extends LayoutView { }
class SomeOtherView extends LayoutView { }
/**
* Below is the new implementation. So far I've managed to tidy
* up LayoutView (I think I have anyway). But now I'm stuck because
* the way I've tidied it up has left me not knowing how to extend
* it.
*
* For example's sake, let's say the conditions were relating to a
* type of fish: salmon or tuna.
*/
abstract class LayoutView {
protected function prepareHeader() { echo __METHOD__, "\n"; }
protected function prepareLeftHandSide() { echo __METHOD__, "\n"; }
protected function prepareFooter() { echo __METHOD__, "\n"; }
public function prepare() {
$this->prepareHeader();
$this->prepareLeftHandSide();
$this->prepareFooter();
}
}
class SalmonLayoutView extends LayoutView
{
protected function prepareLeftHandSide() { echo __METHOD__, "\n"; }
}
class TunaLayoutView extends LayoutView
{
protected function prepareLeftHandSide() { echo __METHOD__, "\n"; }
protected function prepareFooter() { echo __METHOD__, "\n"; }
}
class ViewFactory {
public static function getLayoutView($fishType) {
switch($this->$fishType) {
case 'salmon':
return new SalmonLayoutView();
break;
case 'tuna':
return new TunaLayoutView();
break;
}
}
}
/**
* Now LayoutView has been cleaned up and the condition that was once
* scattered through every LayoutView method is now in one place.
*/
$view = ViewFactory::getLayoutView( Config::getOption('fishtype') );
$view->prepare();
/**
* Now here's where I'm stuck. I want to effectively extend whatever
* class $view is an instance of.
*
* The reason being, I wish to derive a view to show an index of
* articles within the appropriate *LayoutView. The IndexView code is
* the same for Salmon and Tuna.
*
* I can do something like this:
*/
class SalmonIndexView extends SalmonLayoutView { }
class TunaIndexView extends TunaLayoutView { }
/**
* But then I'll be writing the same IndexView code twice. What I'd
* like to do is something like this:
*/
$view = ViewFactory::getLayoutView( Config::getOption('fishtype') );
class IndexView extends get_class($view) { }
/**
* But I'm pretty certain that's not possible, and even if it was
* it seems very wrong.
*
* Can someone more experienced in refactoring and OO please let
* me know where I've gone wrong and suggest a clean way to solve
* this?
*/
maceo
質問する
420 次
2 に答える
1
IndexView コードが本当に同じである場合、継承は必要ありませんが、構成が必要です。基本の LayoutView クラスに、IndexView のインスタンスを追加します。これにより、各 *LayoutView から呼び出すことができます。
継承は、オブジェクト間の関係が is-a の場合にのみ発生します。IndexView は LayoutView ではなく、LayoutView には IndexView があると推測します。
これをチェックしてください、私はそれが言うすべてに同意しませんが、それでも: http://phpimpact.wordpress.com/2008/09/04/favour-object-composition-over-class-inheritance/
于 2008-10-29T15:09:31.890 に答える
0
作成するサブビューにテンプレートをパラメーターとして渡すだけです。この場合、それは悪いことではないと思います。それが標準的なフレームワークである場合は、フォーラムで質問する方がよいかもしれませんが、このケースでは認識されていない機能を備えている可能性があるためです (通常は発生します)。
あなたは次のようなものを持つことができます
class LayoutView {
protected View $subview; //Potentially an array of views
public function prepare() {
// empty, to be filled out by derived classes
}
public function setSubView($view) { $this->subview = $view; }
public function display() {
$this->prepare();
$this->subview->prepare($this->template);
$this->template->render();
}
}
class IndexView {
protected View $subview; //Potentially an array of views
public function prepare() {
// empty, to be filled out by derived classes
}
public function prepare($template) {
//operate on it, maybe even assigning it to $this->template
}
public function setSubView($view) { $this->subview = $view; }
public function display() {
$this->prepare();
$this->subview->prepare($this->template);
$this->template->render();
}
}
于 2008-10-30T08:47:18.760 に答える