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我有一个嵌套菜单或多级菜单的代码。但我没有得到我想要的结果。我有一个关于PHP编程语言的基础知识,请耐心等待。PHP:多级菜单或嵌套菜单
<?php
include("mycon.php");
$query1 = mysql_query("SELECT * FROM menu");
$row1 = mysql_fetch_array($query1);
$a = $row1['parent'];
?>
<nav class="navbar navbar-inverse navbar-static-top" role="navigation">
<div class="container">
<a class="navbar-brand" href="#">Title</a>
<ul class="nav navbar-nav">
<?php
$query2 = mysql_query("SELECT * from menu");
while ($fetcharray2 = mysql_fetch_array($query2)) {
$query5 = mysql_query("SELECT * from menu where menu.parent = '0'");
while ($fetcharray4 = mysql_fetch_array($query5)) {
?>
<li class="dropdown">
<?php
?>
<a href="#" class="dropdown-toggle" data-toggle="dropdown"><?php echo $fetcharray4['label']?><b class="caret"></b></a>
<ul class="dropdown-menu">
<?php
$query3 = mysql_query("SELECT * from menu where menu.id and menu.parent like '$a'");
while ($fetcharray3 = mysql_fetch_array($query2)) {
?>
<li><a href="<?php echo $fetcharray3['link'];?>"><?php echo $fetcharray3['label'];?></a></li>
<?php }?>
</ul>
</li>
<?php }}
?>
</ul>
</div>
</nav>
请解释你想达到什么? –
难道你不想'$ fetcharray3 = mysql_fetch_array($ query3)'?您已将它设置为'$ query2' – MatthewS
向我们展示您的表格结构以及您要显示的内容。 –