大家好,希望大家都过得好吗?
关于如何让子菜单显示在父名称下的下拉列表中,我有点困惑。
到目前为止,我的 php mysql 代码正在做它应该做的事情,但我被困在 HTML 导航部分:(
你们中的一位可爱的人可以帮我吗?
这是我的完整菜单代码以及 php ans mysql
<div class="rd-navbar-top-panel-inner">
</div>
</div>
<div class="rd-navbar-inner">
<!-- RD Navbar Panel-->
<div class="rd-navbar-panel">
<!-- RD Navbar Toggle-->
<button class="rd-navbar-toggle" data-rd-navbar-toggle=".rd-navbar-nav-wrap"><span></span></button>
<!-- RD Navbar Brand-->
<div class="rd-navbar-brand"><a class="brand-name" href="<?php echo DIR; ?>"><img class="logo-default" src="<?php echo DIR; ?>images/logo.png" alt="" width="208" height="46"/><img class="logo-inverse" src="<?php echo DIR; ?>images/logo.png" alt="" width="208" height="46"/></a></div>
</div>
<div class="rd-navbar-aside-center">
<div class="rd-navbar-nav-wrap">
<!-- RD Navbar Nav-->
<ul class="rd-navbar-nav .margin-right">
<?php if($page_name =='index'){?>
<li class="active"><a href="<?php echo DIR; ?>index">Home</a></li>
<?php }else{ ?>
<li><a href="<?php echo DIR; ?>index">Home</a></li>
<?php } ?>
<?php
////PARENT MENU
$stmt = $db->prepare('select * from pages where parentid is null');
$stmt->execute();
?>
<?php while($menu = $stmt->fetch(PDO::FETCH_OBJ)) { ?>
<li>
<?php if($page_name ==''.$menu->slug.''){?>
<li class="active"><a href="<?php echo DIR; ?><?php echo $menu->slug; ?>"><?php echo $menu->page_name; ?></a></li>
<?php }else{ ?>
<li><a href="<?php echo DIR; ?><?php echo $menu->slug; ?>"><?php echo $menu->page_name; ?></a></li>
<?php
}
?>
<?php
//CHILD MENU
$stmt1 = $db->prepare('select * from pages where parentid = :pageid');
$stmt1->bindValue('pageid', $menu->pageid);
$stmt1->execute();
?>
<?php if($stmt1->rowCount() > 0) { ?>
<?php while($sub_menu = $stmt1->fetch(PDO::FETCH_OBJ)) { ?>
<?php if($page_name ==''.$menu->slug.''){?>
<li><a href="<?php echo DIR; ?><?php echo $sub_menu->slug; ?>"><?php echo $sub_menu->page_name; ?></a></li>
<?php }else{ ?>
<li class="active"><a href="<?php echo DIR; ?><?php echo $sub_menu->slug; ?>"><?php echo $sub_menu->page_name; ?></a></li>
<?php }} ?>
<?php } ?>
<?php } ?>
<!-- Contact -->
</ul>
</div>
</div>
<div class="rd-navbar-aside-right"><a class="button button-sm button-secondary button-nina" href="<?php echo $site['panel_book']; ?>">Book</a></div>
</div>
</nav>
</div>
非常感谢您的帮助,提前致谢