如果我正确理解您想要什么，您可以使用CSS来实现这一点。您将正常调用wp\\u nav\\u menu并让它生成所有链接，但随后您将隐藏除当前页面的子菜单之外的所有链接。

你的CSS看起来像这样，

#sidebar ul.menu li
{
    display: none;
}

    #sidebar ul.menu li.current-page-parent,
    #sidebar ul.menu li.current-page-parent ul,
    #sidebar ul.menu li.current-page-parent ul.li
    {
        display: block;
    }

ul.sub-menu
{
    display: none;
}

    #menu-main-navigation > li.current-menu-item ul.sub-menu,
    #menu-main-navigation > li.current-menu-ancestor ul.sub-menu
    {
        display: block;
    }

我为此创建了一个插件-请确保将ID作为字符串而不是整数传入。请为我的插件打分-这是我的第一个插件！：）

Download Here

我做这件事的另一种方法是直接抓取帖子，而不是使用wp\\u nav\\u菜单。这是基于实际的页面结构，而不是菜单。

功能。php：

function __construct()
{
    $this->currentPageID    = $this->getCurrentPageID();
    $this->sectionChildren    = $this->getSectionChildren();
}

function getCurrentPageID()
{
    $currentPage = $_SERVER[\'REQUEST_URI\'];

    if($currentPage == \'/\')
        $currentPage = \'/home\';
    $currentPage = get_page_by_path($currentPage);

    if($currentPage)
        return $currentPage->ID;
    else
        return -1;
}

function getSectionID()
{
    global $wpdb;

    $currentSectionID = $wpdb->get_var("
        SELECT post_parent
        FROM ". $wpdb->posts ."
        WHERE ID = ". $this->currentPageID
    );

    if($currentSectionID == 0)
        return $this->currentPageID;
    else
        return $currentSectionID;
}

function getSectionChildren()
{
    global $wpdb;

    $children = $wpdb->get_results("
        SELECT ID, post_title
        FROM ". $wpdb->posts ."
        WHERE
            post_parent = ". $this->getSectionID() ." AND
            post_type = \'page\' AND
            post_status = \'publish\'
    ", ARRAY_A);

    return $children;
}

<ul id="sub-navigation">
    <?php foreach($dc->sectionChildren as $c) : ?>
        <li <?php if($dc->currentPageID == $c[\'ID\']) echo \'class="active"\'; ?>><a href="<?php echo get_permalink($c[\'ID\']); ?>"><?php echo $c[\'post_title\']; ?></a></li>
    <?php endforeach; ?>
</ul>