PHP生成UL LI,UL LI [英] PHP Generate UL LI , UL LI
本文介绍了PHP生成UL LI,UL LI的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
无法确定如何使用while循环生成此菜单.
Can't figure out how-to generate this menu using a while-loop.
这是我的代码的示例:
<ul id="nav">
<li><a href="#">Hoofdmenu 1</a>
<ul class="sub">
<li><a href="#">Submenu 1.1</a></li>
<li><a href="#">Submenu 1.2</a></li>
<li><a href="#">Submenu 1.3</a></li>
<li><a href="#">Submenu 1.4</a></li>
</ul>
</li>
<li><a href="#">Hoofdmenu 2</a>
<ul class="sub">
<li><a href="#">Submenu 2.1</a></li>
<li><a href="#">Submenu 2.2</a></li>
<li><a href="#">Submenu 2.3</a></li>
<li><a href="#">Submenu 2.4</a></li>
</ul>
</li>
</ul>
我的dbtable看起来像:
My dbtable looks like:
paginas:
id
title
content
type
当从父级输入== id时,它应该是子菜单. 在我的示例中,此方法有效,现在我必须使其动态化. 大脑不能正常工作.
When type == id from the parent it should be the submenu. In my example this works, now I've got to make it dynamic. Brains ain't working atm.
感谢您的帮助!
使用代码从db获取数据:
Used code to get data from db:
<ul id="nav">
<?php
include_once("ond/inc/php/connect.php");
$query = "SELECT * FROM paginas WHERE type = '0'";
$result = mysql_query($query);
while($row = mysql_fetch_object($result)){
echo '<li><a href="?ond='.$row->titel.'">'.$row->titel.'</a>';}
echo '<ul class="sub">';
$query2 = "SELECT * FROM paginas WHERE type = '".$row->id."'";
$result2 = mysql_query($query2);
while($row2 = mysql_fetch_object($result2))
{
echo '<li><a href="?ond='.$row2->titel.'">'.$row2->titel.'</a></li>';
}
echo '</ul>';
echo '</li>';
?>
</ul>
推荐答案
下一行解决了该问题:
<ul id="nav">
<?php
include_once("ond/inc/php/connect.php");
$query = "SELECT * FROM paginas WHERE type = '0'";
$result = mysql_query($query);
while($row = mysql_fetch_object($result)){
echo '<li><a href="?ond='.$row->titel.'">'.$row->titel.'</a>';
$query2 = "SELECT * FROM paginas WHERE type = '".$row->id."'";
$result2 = mysql_query($query2);
echo '<ul class="sub">';
while($row2 = mysql_fetch_object($result2))
{
echo '<li><a href="?ond='.$row2->titel.'">'.$row2->titel.'</a></li>';
}
echo '</ul>';
echo '</li>';}
?>
</ul>
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