PHP生成UL LI，UL LI

Question

不知道如何使用while循环生成此菜单。

这是我的代码示例：

<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看起来像这样：

paginas:
    id
    title
    content
    type

当从父菜单输入== id时，它应该是子菜单。在我的示例中，这是可行的，现在我必须将其动态化。大脑不能在自动取款机上工作。

谢谢你的帮忙!

使用代码从db获取数据：

<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>

Answer 1

接下来的几行代码解决了这个问题：

<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>

Answer 2

我会这样做：

首先，将数据作为数组提取出来，并对每个条目进行循环。然后运行如下代码：

$menuArray = array();

if (empty($type)) // If the entry has no "type", then it's a parent
{
    $menuArray[$type]['title'] = $title;
}
else // else, it's a child, so append it to the parent
{
    $menuArray[$type]['subitems'][] = $title;
}

然后，使用$menuArray循环遍历它以创建菜单：

?><ul id="nav"><?php
foreach ($menuArray as $item)
{
    ?><li><a href="#">$item['title']</a><?php
    ?><ul class="sub"><?php

    foreach ($item['subitems'] as $subItem)
    {
        <li><a href="#">$subItem</a></li>
    }

    ?></ul><?php
    ?></li><?php
}
?></ul><?php

Answer 3

function load()
{
    global $conn;


    $query = "SELECT *  FROM sub_category WHERE main_category_id='1'";
    $result = mysqli_query($conn,$query);
    while($row = mysqli_fetch_assoc($result)){
        $cat_id=$row['sub_category_id'];

    echo '<li><a href="?id='.$row['sub_category_id'].'">'.$row['sub_category_name'].'</a>';

    $query2 = "SELECT *  FROM categories WHERE sub_category_id='$cat_id'";
    $result2 = mysqli_query($conn,$query2);    
    echo '<ul class="sub">';
    while($row2 = mysqli_fetch_assoc($result2))
    {   
        echo '<li><a href="?id='.$row2['category_id'].'">'.$row2['category_name'].'</a></li>';


    }
        echo '</ul>';
    echo '</li>';}


}