如何在php中显示数组父项下的子项

Question

我有一个这样的数组：

Array
(
    [0] => stdClass Object
        (
            [ID] => 1
            [menu_item_parent] => 0
            [title] => Home
            [url] => http://www.example.com/
        )

    [1] => stdClass Object
        (
            [ID] => 2
            [menu_item_parent] => 0
            [title] => Menu 2
            [url] => http://www.example.com/menu-2/
        )

    [2] => stdClass Object
        (
            [ID] => 3
            [menu_item_parent] => 2
            [title] => Sub Menu 1
            [url] => http://www.example.com/menu-2/sub-menu-1
            [target] => 
        )

    [3] => stdClass Object
        (
            [ID] => 4
            [menu_item_parent] => 0
            [title] => Menu 4
            [url] => http://www.example.com/menu-4/
            [target] => 

        )
)

现在你可以看到数组的第三个项是第二个数组项的子项(请参阅列menu_item_parent).Now我的问题是如何使用这个array.Please帮助来显示这个父项及其子项。

Answer 1

最后，在@Matt.C的帮助下解决我的问题，给@Matt.C提供link.Thanks。解决方案如下：

首先以平面数组的形式获取菜单项：

<?php
$menu_name = 'main_nav';
$locations = get_nav_menu_locations();
$menu = wp_get_nav_menu_object( $locations[ $menu_name ] );
$menuitems = wp_get_nav_menu_items( $menu->term_id, array( 'order' => 'DESC' ) );
?>

然后迭代菜单项的数组：

<nav>
<ul class="main-nav">
    <?php
    $count = 0;
    $submenu = false;

    foreach( $menuitems as $item ):
        // get page id from using menu item object id
        $id = get_post_meta( $item->ID, '_menu_item_object_id', true );
        // set up a page object to retrieve page data
        $page = get_page( $id );
        $link = get_page_link( $id );

        // item does not have a parent so menu_item_parent equals 0 (false)
        if ( !$item->menu_item_parent ):

        // save this id for later comparison with sub-menu items
        $parent_id = $item->ID;
    ?>

编写第一个父项<li>:

 <li class="item">
        <a href="<?php echo $link; ?>" class="title">
            <?php echo $page->post_title; ?>
        </a>
        <a href="<?php echo $link; ?>" class="desc">
            <?php echo $page->post_excerpt; ?>
        </a>
    <?php endif; ?>

检查此项目的父id是否与存储的父id匹配：

     <?php if ( $parent_id == $item->menu_item_parent ): ?>
Start sub-menu <ul> and set $submenu flag to true for later referance:

            <?php if ( !$submenu ): $submenu = true; ?>
            <ul class="sub-menu">
            <?php endif; ?>
Write the sub-menu item:

                <li class="item">
                    <a href="<?php echo $link; ?>" class="title"><?php echo $page->post_title; ?></a>
                    <a href="<?php echo $link; ?>" class="desc"><?php echo $page->post_excerpt; ?></a>

​

如果下一项没有相同的父id，并且我们声明了一个子菜单，那么关闭子菜单<ul>

<?php if ( $menuitems[ $count + 1 ]->menu_item_parent != $parent_id && $submenu ): ?>
        </ul>
        <?php $submenu = false; endif; ?>

<?php endif; ?>

同样，如果数组中的下一项不具有相同的父id，请关闭<li>

  <?php if ( $menuitems[ $count + 1 ]->menu_item_parent != $parent_id ): ?>
    </li>                           
    <?php $submenu = false; endif; ?>

<?php $count++; endforeach; ?>
  </ul>
</nav> 

Answer 2

试试这个:我将输入添加为数组，根据您的问题更改为对象。

$array  = Array( array("ID" => 1,"menu_item_parent" => 0,"title" => "Home","url" => "http://www.example.com/"),
                 array("ID" => 2,"menu_item_parent" => 0,"title" => "Menu 2","url" => "http://www.example.com/menu-2/"),
                 array("ID" => 3,"menu_item_parent" => 2,"title" => "Sub Menu 1","url" => "http://www.example.com/menu-2/sub-menu-1","target" =>"" ),
                 array("ID" => 4,"menu_item_parent" => 0,"title" => "Menu 4","url" => "http://www.example.com/menu-4/","target" => "")
          );

$res   = array();         
foreach($array as $val){
   if($val['menu_item_parent'] != 0){
       $res[$val['menu_item_parent']]['child'][] = $val;
   }
   else{
       $res[$val['ID']] = $val;
   }
}

echo "<pre>";
print_r($res);

输出：

Array
(
    [1] => Array
        (
            [ID] => 1
            [menu_item_parent] => 0
            [title] => Home
            [url] => http://www.example.com/
        )

    [2] => Array
        (
            [ID] => 2
            [menu_item_parent] => 0
            [title] => Menu 2
            [url] => http://www.example.com/menu-2/
            [child] => Array
                (
                    [0] => Array
                        (
                            [ID] => 3
                            [menu_item_parent] => 2
                            [title] => Sub Menu 1
                            [url] => http://www.example.com/menu-2/sub-menu-1
                            [target] => 
                        )

                )

        )

    [4] => Array
        (
            [ID] => 4
            [menu_item_parent] => 0
            [title] => Menu 4
            [url] => http://www.example.com/menu-4/
            [target] => 
        )

)

Answer 3

下面是一个非常简单的类，用于解决Wordpress特有的问题，其中包含一个返回所有子菜单项的get_submenu函数：

class NestedMenu
{
    private $flat_menu;
    public $items;

    function __construct($name)
    {
        $this->flat_menu = wp_get_nav_menu_items($name);
        $this->items = array();
        foreach ($this->flat_menu as $item) {
            if (!$item->menu_item_parent) {
                array_push($this->items, $item);
            }
        }
    }

    public function get_submenu($item)
    {
        $submenu = array();
        foreach ($this->flat_menu as $subitem) {
            if ($subitem->menu_item_parent == $item->ID) {
                array_push($submenu, $subitem);
            }
        }
        return $submenu;
    }
}

用法，构造实例：

$menu = new NestedMenu('menu_name');

迭代：

foreach ($menu->items as $item) { ...

并获取循环内的子菜单：

$submenu = $menu->get_submenu($item);

在显示子菜单之前，您可以检查它是否存在：

if ($submenu): ...