繁体   English   中英

RecursiveIteratorIterator和嵌套的RecursiveArrayIterators

[英]RecursiveIteratorIterator and nested RecursiveArrayIterators

我一直在使用RecursiveArrayIterators将嵌套对象作为树来处理。 下面的代码困扰我,因为结果仅返回我期望的某些值。 主要地,根节点似乎永远不会被迭代。 我觉得我只是盯着它看了太久,但是想确保自己在正确的轨道上。

class Container extends \RecursiveArrayIterator
{
    protected $_alias;

    public function __construct( $alias = null )
    {
        if( is_null( $alias ) )
        {
            $alias = uniqid( 'block_' );
        }

        $this->_alias = $alias;
    }

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

try
{
    $root = new Container( 'root_level' );
    $block = new Container( 'first_level' );
    $child = new Container( 'second_level' );
    $child_of_child_a = new Container( 'third_level_a' );
    $child_of_child_b = new Container( 'third_level_b' );

    $child->append( $child_of_child_a );
    $child->append( $child_of_child_b );
    $child_of_child_a->append( new Container );
    $child_of_child_a->append( new Container );
    $block->append( $child );
    $root->append( $block );

    $storage = new \RecursiveIteratorIterator( $root, RecursiveIteratorIterator::SELF_FIRST );

    foreach( $storage as $key => $value )
    {
        print_r( str_repeat( ' ', $storage->getDepth() * 4 ) . $value->getAlias() . PHP_EOL );
    }
}
catch( \Exception $e )
{
    var_dump( $e->getMessage() );
}

结果是...

first_level
second_level
    third_level_a
        block_51f98b779c107
        block_51f98b779c124
    third_level_b

根节点在哪里?

回答

Sven的回答使我过度劳累的大脑无法正确处理。 这是我成功代码的最后一点,以防其他人尝试类似的事情。

class OuterContainer extends \ArrayIterator
{

}

class Container extends \ArrayIterator
{
    protected $_alias;

    public function __construct( $alias = null )
    {
        if( is_null( $alias ) )
        {
            $alias = uniqid( 'container_' );
        }

        $this->_alias = $alias;
    }

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

try
{
    $c1 = new Container( 'Base' );

    $c1_c1 = new Container( 'Base > 1st Child' );

    $c1_c2 = new Container( 'Base > 2nd Child' );

    $c1_c1_c1 = new Container( 'Base > 1st Child > 1st Child' );

    $c1_c1->append( $c1_c1_c1 );

    $c1->append( $c1_c1 );

    $c1->append( $c1_c2 );

    $outer_container = new OuterContainer;

    $outer_container->append( $c1 );

    $storage = new \RecursiveIteratorIterator( new \RecursiveArrayIterator( $outer_container ), RecursiveIteratorIterator::SELF_FIRST );

    foreach( $storage as $key => $value )
    {
        print_r( $value->getAlias() . PHP_EOL );
    }
}
catch( \Exception $e )
{
    var_dump( $e->getMessage() );
}

根节点不存在,因为您没有正确使用RecursiveArrayIterator。

它的预期用途是使一个数组具有任意级别和结构的多个子数组,并将该数组放入RecursiveArrayIterator的一个实例中,并遍历所有内容,并将其放入RecursiveIteratorIterator中。

这样,整个数组(包括顶层)都将被迭代。

因为您滥用了RecursiveArrayIterator作为信息的载体,所以不会迭代顶层对象。

建议的简单示例有些讨厌,但演示了原理:

$root = array( 'root_level' );
$block = array( 'first_level' );
$child = array( 'second_level' );
$child_of_child_a = array( 'third_level_a' );
$child_of_child_b = array( 'third_level_b' );

$child[] = $child_of_child_a;
$child[] = $child_of_child_b ;
$child_of_child_a[] = array('');
$child_of_child_a[] = array('');
$block[] = $child;
$root[] = $block ;

$storage = new \RecursiveIteratorIterator( new \RecursiveArrayIterator($root), RecursiveIteratorIterator::SELF_FIRST );

var_dump($root);
foreach ($storage as $key=>$value) {
    echo $key.": ".$value."\n";
}

结果输出:

array(2) {
    [0] =>
  string(10) "root_level"
  [1] =>
  array(2) {
        [0] =>
    string(11) "first_level"
    [1] =>
    array(3) {
            [0] =>
      string(12) "second_level"
      [1] =>
      array(1) {
                ...
            }
      [2] =>
      array(1) {
                ...
            }
    }
  }
}
0: root_level
1: Array
    0: first_level
1: Array
    0: second_level
1: Array
    0: third_level_a
2: Array
    0: third_level_b

它的输出并不完全相同,但是我的观点是,您可以遍历结构,而不必让每个节点都成为RecursiveArrayIterator的实例。 您可以添加任何内容,既可以是数组,也可以添加充当数组的对象,或者可以进行有效迭代的对象。

暂无
暂无

声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM