让所有孩子获得深度多维数组

Question

我有这样的数组:

array(
    array(
        'id' => 1,
        'children' => array(
            array(
                'id' => 2,
                'parent_id' => 1
            ),
            array(
                'id' => 3,
                'parent_id' => 1,
                'children' => array(
                    array(
                        'id' => 4,
                        'parent_id' => 3
                    )
                )
            )
        )
    )
);

如果有必要,阵列会更深入.我需要为任何给定的id获取孩子.

谢谢.

Answer 1

function getChildrenOf($ary, $id)
{
  foreach ($ary as $el)
  {
    if ($el['id'] == $id)
      return $el;
  }
  return FALSE; // use false to flag no result.
}

$children = getChildrenOf($myArray, 1); // $myArray is the array you provided.

除非我遗漏了某些东西,否则迭代数组寻找与id你正在寻找的密钥和id 匹配的东西(然后返回它作为结果).您也可以迭代搜索(并给我一秒钟发布代码,这将检查parentId密钥)...

-

递归版,包括子元素:

function getChildrenFor($ary, $id)
{
  $results = array();

  foreach ($ary as $el)
  {
    if ($el['parent_id'] == $id)
    {
      $results[] = $el;
    }
    if (count($el['children']) > 0 && ($children = getChildrenFor($el['children'], $id)) !== FALSE)
    {
      $results = array_merge($results, $children);
    }
  }

  return count($results) > 0 ? $results : FALSE;
}

递归版,不包括子元素

function getChildrenFor($ary, $id)
{
  $results = array();

  foreach ($ary as $el)
  {
    if ($el['parent_id'] == $id)
    {
      $copy = $el;
      unset($copy['children']); // remove child elements
      $results[] = $copy;
    }
    if (count($el['children']) > 0 && ($children = getChildrenFor($el['children'], $id)) !== FALSE)
    {
      $results = array_merge($results, $children);
    }
  }

  return count($results) > 0 ? $results : FALSE;
}