PHP - 从阵列

Question

我试图创建一个“插件”之类的脚本添加到“菜单阵” ......我会去直接在导航/ subnav结构..

可以说我发起导航项目如下：

$sections->add_section('dashboard'); 
$DashBoardSection = $sections->load_section('dashboard'); 
$DashBoardSection->set_role(NULL); 
$DashBoardSection->set_nav_item(array(
    'identifier' => 'dashboard', 
    'text' => 'Dashboard', 
    'url' => NULL, 
    'position' => NULL 
)); 

首先创建一个新的部分并获取实例。

然后，我们设置一个“角色”，我们将在其中测试以查看用户是否已通过身份验证以查看。

设置导航项只存储数组。标识符是对项目的引用（当我们想要添加子项目时），除了“位置”之外的所有标准，它说明它位于nav中的位置，即NULL是顶层，数组（'topnav'，'subnav '）将是topnav-> subnav->仪表板。

的测试，这可以存储如下：

Array 
(
    [0] => Array 
     (
      [identifier] => dashboard 
      [text] => Dashboard 
      [url] => 
      [position] => 
     ) 
[1] => Array 
    (
     [identifier] => dashboard2 
     [text] => Dashboard2 
     [url] => 
     [position] => Array 
      (
       [0] => dashboard 
      ) 

    ) 

) 

我的问题是我怎么把它转换成以下结构：

Array 
(
    [0] => Array 
    (
     [identifier] => dashboard 
     [text] => Dashboard 
     [url] => 
     [position] => 
     [children] => Array 
      (
       [0] => Array 
        (
         [identifier] => dashboard2 
         [text] => Dashboard2 
         [url] => 
        ) 
      ) 
    ) 
) 

拉我的头发在这一项，任帮助将非常感激。

问候

我现在有

public function build_navigation($role){ 
    $role = (int)$role; 
    $nav = array(); 
    foreach($this->sections as $section) { 
     if($section->get_role() === NULL || $section->get_role() === $role) { 
      $nav_array = $section->get_nav(); 
      foreach($nav_array as $key => $nav_item) { 
       if($nav_item['position'] === NULL) { 
        $nav[$nav_item['identifier']] = $nav_item; 
       }elseif(is_array($nav_item['position'])){ 
        #...# 
       } 
      } 
     } 
    } 
    return $nav; 
} 

编辑

想象这是给定的数组（它可以在任何顺序）

Array 
(
    [0] => Array 
     (
      [identifier] => dashboard_child2 
      [text] => Dashboard Child 2 
      [url] => 
      [position] => Array 
       (
        [0] => dashboard 
       ) 

     ) 

    [1] => Array 
     (
      [identifier] => dashboard_child_child_1 
      [text] => Dashboard Child Child 1 
      [url] => 
      [position] => Array 
       (
        [0] => dashboard 
        [1] => dashboard_child1 
       ) 

     ) 



[2] => Array 
    (
     [identifier] => dashboard_child1 
     [text] => Dashboard Child 1 
     [url] => 
     [position] => Array 
      (
       [0] => dashboard 
      ) 

    ) 

[3] => Array 
    (
     [identifier] => dashboard 
     [text] => Dashboard 
     [url] => 
     [position] => 
    ) 

[4] => Array 
    (
     [identifier] => dashboard2 
     [text] => Dashboard2 
     [url] => 
     [position] => Array 
      (
       [0] => dashboard 
      ) 

    ) 

) 

一个需要格式为：

Array 
(
    [dashboard] => Array 
     (
      [text] => Dashboard 
      [url] => 
      [children] => Array 
       (
        [dashboard_child2] => Array 
         (
          [text] => Dashboard Child 2 
          [url] => 
         ) 
        [dashboard_child1] => Array 
         (
          [text] => Dashboard Child 1 
          [url] => 
          [children] => Array 
           (
            [dashboard_child_child_1] => Array 
             (
              [text] => Dashboard Child Child 1 
              [url] => 
             ) 
           ) 
         ) 
        [dashboard2] => Array 
         (
          [text] => Dashboard2 
          [url] => 
         ) 
       ) 
     ) 
) 

Answer 1

这是我对问题的看法，用递归解决。

你可以使用多个职位（我猜这就是为什么它是一个数组），如果至少有一个职位被发现，它会忽略缺少的职位，但是如果每个职位都没有，就会抱怨。

function translate($in) { 

    $out = array(); 

    // first pass, move root nodes to output 
    foreach ($in as $k => $row) { 
     if (!$row['position']) { 
      $out[$row['identifier']] = $row; 
      unset($in[$k]); 
     } 
    } 

    // while we have input 
    do { 
     $elements_placed = 0; 
     // step trough input 
     foreach ($in as $row_index => $row) { 
      foreach ($row['position'] as $pos) { 
       // build context for the node placing 
       $data = array(
        'row' => $row, 
        'in' => &$in, 
        'row_index' => $row_index, 
        'elements_placed' => &$elements_placed, 
        'pos' => $pos, 
       ); 
       // kick of recursion 
       walker($out, $data); 
      } 
     } 
    } while ($elements_placed != 0); 
    if (count($in)) { 
     trigger_error("Error in user data, can't place every item"); 
    } 
    return $out; 
} 

function walker(&$out, $data) { 
    foreach ($out as &$row) { 
     // it looks like a node array 
     if (is_array($row) && isset($row['identifier'])) { 
      // if it has children recurse in there too 
      if (isset($row['children'])) { 
       walker($row['children'], $data); 
      } 

      // it looks like a node array that we are looking for place the row 
      if ($row['identifier'] == $data['pos']) { 
       if (!isset($row['children'])) { 
        $row['children'] = array($data['row']['identifier'] => $data['row']); 
       } else { 
        $row['children'][$data['row']['identifier']] = $data['row']; 
       } 
       // report back to the solver that we found a place 
       ++$data['elements_placed']; 
       // remove the row from the $in array 
       unset($data['in'][$data['row_index']]); 
      } 
     } 
    } 
} 

$in = array (
    array (
     'identifier' => 'secondlevelchild2', 
     'text' => 'secondlevelchild2', 
     'url' => '', 
     'position' => array (
      'dashboard2', 
     ), 
    ), 
    array (
     'identifier' => 'secondlevelchild', 
     'text' => 'secondlevelchild', 
     'url' => '', 
     'position' => array (
      'dashboard2', 
     ), 
    ), 
    array (
     'identifier' => 'dashboard', 
     'text' => 'Dashboard', 
     'url' => '', 
     'position' => '', 
    ), 
    array (
     'identifier' => 'dashboard2', 
     'text' => 'Dashboard2', 
     'url' => '', 
     'position' => array (
      'dashboard', 'home', 
     ), 
    ), 
    array (
     'identifier' => 'thirdlevelchild', 
     'text' => 'thirdlevelchild', 
     'url' => '', 
     'position' => array (
      'secondlevelchild2', 
     ), 
    ), 
); 

$out = translate($in); 
var_export($out); 

在它的当前形式不一旦被放置在距离阵列节点移除identifier或position密钥。

Answer 2

您需要一个临时数组，将标识符映射到它的子数组，以便您可以将它添加到那里。

如果我看到正确的，你有类似的东西已经在这里当你添加父：

$nav[$nav_item['identifier']] = $nav_item; 

---

编辑：添加父需要如指出了一些注意事项评论：

$node = &$nav[$nav_item['identifier']]; 
$children = isset($node['children']) ? $node['children'] : array(); 
$node = $nav_item; 
$node['children'] = $children; 
unset($node); 

---

只需加给孩子，然后：

foreach($nav_item['position'] as $identifier) 
{ 
    $nav[$identifier]['children'][] = $nav_item; 
} 

你也可以使用对象，而不是数组，这样你就不会重复数据那么多（或者你可以在以后改变它），但是，只是在想，这解决您的问题一定不是必要的，所以稍后可能会有所帮助。