Question

在我的wordpress网站中，我正在自定义菜单。因为我使用此代码获取所有菜单项

`<?php$menu_name = 'hamburger-menu'; 
if ( ( $locations = get_nav_menu_locations() ) && isset( $locations[ $menu_name ] ) ) {
$menu = wp_get_nav_menu_object( $locations[ $menu_name ] );

$menu_items = wp_get_nav_menu_items($menu->term_id);$output = '<ul>';
foreach ( (array) $menu_items as $key => $menu_item ) {
    $title = $menu_item->title;
    $output.='<li>'.$title.'</li>';

    print_r($menu_item); 
}
$output.='</select>';
}

echo $output;
                ?>`

此代码逐个打印所有菜单项和子菜单项

`<ul>
<li>Menu Item</li>
<li>Menu Item</li>
<li>Drinks</li>
<li>Sub menu item</li>
<li>Menu Item</li>
<li>Sub menu item</li>
<li>Sub menu item</li>
<li>Menu Item</li>
</ul>`

但我想要这样的输出

`<ul>
<li>Menu Item</li>
<li>Menu Item</li>
<li>Menu Item
<ul><li>Sub menu item</li></ul>
</li>
<li>Menu Item
<ul>
<li>Sub menu item</li>
<li>Sub menu item</li>
</ul>
</li>
<li>Menu Item</li>
</ul>`

可以帮助任何人

Answer 1

示例1：

$resultMainMenu = mysql_query("SELECT * FROM tblmenu WHERE parentid=0 ORDER BY menuname ASC") or die(mysql_error());
while($row = mysql_fetch_array($resultMainMenu)){
    echo $row['menutitle'] . '<br />'; // echo main menu
    $resultSubmenu = mysql_query("SELECT * FROM tblmenu WHERE parentid=" . $row['id'] . " ORDER BY menuname ASC") or die(mysql_error());
    if(mysql_num_rows($resultSubmenu) >= 1){
        while($rowSub = mysql_fetch_array($resultSubmenu)){
            echo ' -- ' . $rowSub['menutitle'] . '<br />'; // echo sub menu
        }
    }
}

示例2：

function display_children($parent, $level) {
    $result = mysql_query("SELECT a.id, a.label, a.link, Deriv1.Count FROM `menu` a  LEFT OUTER JOIN (SELECT parent, COUNT(*) AS Count FROM `menu` GROUP BY parent) Deriv1 ON a.id = Deriv1.parent WHERE a.parent=" . $parent);
    echo "<ul>";
    while ($row = mysql_fetch_assoc($result)) {
        if ($row['Count'] > 0) {
            echo "<li><a href='" . $row['link'] . "'>" . $row['label'] . "</a>";
            display_children($row['id'], $level + 1);
            echo "</li>";
        } elseif ($row['Count']==0) {
            echo "<li><a href='" . $row['link'] . "'>" . $row['label'] . "</a></li>";
        } else;
    }
    echo "</ul>";
}

你可以按照你的要求放置条件

如何单独获取菜单项和子菜单项

1 个答案: