-1
我一直在試圖建立一個下拉菜單,但我沒有得到我想要的結果。這裏是我的代碼:動態下拉菜單使用PHP和MySQL
<?php require_once 'core/init.php'?>
<?php
$sql = 'SELECT * FROM categories WHERE parent = 0';
$pquery = mysqli_query($db,$sql);
?>
<?php while($parent = mysqli_fetch_assoc($pquery)):?>
<?php
$parent_id = $parent['id'];
$sql2 = 'SELECT * FROM categories WHERE parent = "parent_id"';
$cquery = mysqli_query($db,$sql2);
?>
<li class='dropdown'>
<a href='#' class='dropdown-toggle' data-toggle='dropdown'>
<?php echo $parent['id'];?><span class='caret'</span</a>
<ul class='dropdown-menu' role='menu'>
<?php while($child = mysqli_fetch_assoc($cquery)):>
<li><a href='#'><?php echo $child['parent'];?></a>
</li>
<?php endwhile; ?>
</ul>
</li>
<?php endwhile;?>
你想要什麼?通過ajax或不? – user8455694