Question

我太累了，但我一直在和这个人摔跤。容器可以自己具有布局，也可以在具有布局的层次结构中的某个位置具有父级。我需要找到那个布局，无论它有多远。

对象$ container有一个属性$ parent，如果不为null，则引用另一个$ container对象（相同类型）。

我认为这段代码在while条件下有问题吗？

private function findInheritedLayout($container) {

    do {
        if ( $container->getLayoutTemplate() ) {
            return $container->getLayoutTemplate();
        }
        else {
            if ( $container->getParent() ) {
                $container = $container->getParent();
            }
            else {
                return null;
            }   
        }
    } while ( $container->getParent() ); //flawed? looks for a parent one step too far?
}

Answer 1

测试：

$test = new View('A');
$test->assign('A', new View('B'));
$test->assign('B', $test2 = new View('C'));
$test2->assign('D', $test3 = new View('E'));
$test3->assign('F', $searchTarget = new View('G'));

$root = RootViewIterator::findRootOf($searchTarget);

var_dump($root->getName());

结果：

string(1) "A"

迭代者：

class RootViewIterator
    implements Iterator
{

    protected $view;
    protected $startView;
    protected $key = 0;

    public function __construct(View $view)
    {
        $this->view = $view;
        $this->startView = $view;
    }

    public function next()
    {
        $this->view = $this->view->getParent();
        $this->key++;
    }

    public function key()
    {
        return $this->key;
    }

    public function current()
    {
        return $this->view;
    }

    public function rewind()
    {
        $this->view = $this->startView;
        $this->key = 0;
    }

    public function valid()
    {
        return null !== $this->view;
    }

    public static function findRootOf(View $view)
    {
        $result = iterator_to_array(new static($view));

        end($result);

        return current($result);
    }

}

观点：

class View
{
    protected $name;
    protected $items;
    protected $parent;

    public function __construct($name)
    {
        $this->name = $name;
    }

    public function getName()
    {
        return $this->name;
    }

    public function assign($key, $content)
    {
        if ( $content instanceof View ) {
            $content->setParent($this);
        }

        $this->items[$key] = $content;
    }

    public function setParent(View $view)
    {
        $this->parent = $view;
    }

    public function getParent()
    {
        return $this->parent;
    }

    public function hasParent()
    {
        return !empty($this->parent);
    }
}

Answer 2

你的假设是正确的。 while条件检查当前$container是否有父项，而不检查布局。

尝试以下方法：

while (!$container->getLayoutTemplate() && $container->getParent()) {
    $container = $container->getParent();
}

$template = $container->getLayoutTemplate();
return $template ?: null;

无限地寻找父母的对象

2 个答案: