要解决您的问题,您需要正确地理解PHP中变量引用/别名的工作方式。
看下面的示例代码,它看起来与您的代码没有太大不同,但是利用引用来访问任何父对象,即使它已经“移动”了:
# transform $flat into a tree:foreach($flat as $id => &$value){ # check if there is a parent if ($parentId = $value[’parent’]) {$flat[$parentId][0][$id] =& $value; # add child to parentunset($flat[$id]); # remove reference from topmost level }}unset($value); # remove iterator referenceprint_r($flat); # your tree
$flat现在包含$flat- 中的所有值,但已重新排序。
解决方法我有那个数组:
$a = array( '7' => array('id' => 7,'parent' => 6 ),'6' => array('id' => 6,'parent' => 5 ),'5' => array('id' => 5,'parent' => 4 ),'4' => array('id' => 4,'parent' => 0 ),'3' => array('id' => 7,'parent' => 2 ),'2' => array('id' => 7,'parent' => 1 ),'1' => array('id' => 7,'parent' => 0 ));
我想要的结果是:
$a = array( '4' => array('id' => 4,'parent' => 0,array( '5' => array('id' => 5,'parent' => 4,array( '6' => array('id' => 6,'parent' => 5,array( '7' => array('id' => 7,'parent' => 6 )) )) )) ),'parent' => 1,array( '3' => array('id' => 7,'parent' => 2 )) ),'parent' => 0 ));
我使用的代码是这样的:
foreach($a as $v){ if(isset($a[$v[’PARENT’]])) {$a[$v[’PARENT’]][$v[’ID’]] = $v;unset($a[$v[’ID’]]); }}
我的问题是我得到了那个结果:
$a = array( '4' => array('id' => 4,'parent' => 4 )) ),'parent' => 0 ));
而不是需要它的结果。