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] =>
)
)
现在您可以看到数组的第3项是第二个数组项的子项(请参阅列menu_item_parent
)。现在我的问题是如何使用此数组显示此父项与其子项。请帮忙。
答案 0 :(得分:5)
最后在@ Matt.C给出链接的帮助下解决了我的问题。谢谢@Matt.C。这是解决方案:
首先将菜单项设为平面阵列:
<?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;
?>
写下第一个父项<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>
如果下一个项目没有相同的父ID,我们宣布了一个子菜单,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,如果数组中的下一个项目没有相同的父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>
答案 1 :(得分: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] =>
)
)
答案 2 :(得分:1)
在php中使用foreach函数检查。 像
$array = array("apple" => 1, "orange" => 2);
$sep = "";
foreach($array as $key => $value) {
if($sep) {
$sep .= "<br/>key:".$key." / value:".$value;
} else {
$sep = "key:".$key." / value:".$value;
}
}
<强>输出:强>
key:apple / value:1 key:orange / value:2
答案 3 :(得分:1)
您可以迭代数组,如果对象有父对象,则将其添加到该父对象的children
数组中。例如:
$array = array(
1 => (object) array('menu_item_parent' => 0),
2 => (object) array('menu_item_parent' => 1),
3 => (object) array('menu_item_parent' => 0),
);
foreach ($array as $key => $object)
{
if (0 != $object->menu_item_parent && isset($array[$object->menu_item_parent]))
{
if (!property_exists($array[$object->menu_item_parent], 'children'))
{
$array[$object->menu_item_parent]->children = array();
}
$array[$object->menu_item_parent]->children[] = $object;
unset($array[$key]);
}
}
echo '<pre>' . print_r($array, TRUE) . '</pre>';
将转换:
Array
(
[1] => stdClass Object
(
[menu_item_parent] => 0
)
[2] => stdClass Object
(
[menu_item_parent] => 1
)
[3] => stdClass Object
(
[menu_item_parent] => 0
)
)
要:
Array
(
[1] => stdClass Object
(
[menu_item_parent] => 0
[children] => Array
(
[0] => stdClass Object
(
[menu_item_parent] => 1
)
)
)
[3] => stdClass Object
(
[menu_item_parent] => 0
)
)
然后,您可以迭代每个对象并在需要时显示其子项:
foreach ($array as $object)
{
echo 'Parent: ' . $object->title . '<br>';
if (property_exists($object, 'children') && !empty($object->children))
{
echo ' Children: ';
foreach ($object->children as $child)
{
echo $child->title . '<br>';
}
}
}
答案 4 :(得分:1)
这是一个非常简单的类,针对您的Wordpress特定问题,使用get_submenu
函数返回所有子菜单项:
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。构建实例:
$menu = new NestedMenu('menu_name');
迭代:
foreach ($menu->items as $item) { ...
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): ...
答案 5 :(得分:0)
这将是我的解决方案。将父对象下的子项移动到children
,并在名为has_child
的父项下创建一个布尔值,其值为1
。最后,取消设置并从主变量中删除子项。
$elements = wp_get_nav_menu_items("theme-location");
foreach($elements as $index => $item)
{
if($item->menu_item_parent != 0)
{
foreach($elements as $index2 => $item2)
{
if($item2->ID == $item->menu_item_parent)
{
$elements[$index2]->has_child = true;
if(!isset($elements[$index2]->children))
{
$elements[$index2]->children = array();
}
$elements[$index2]->children[] = $item;
}
}
unset($elements[$index]);
}
}