当访问者单击子菜单并在新页面中打开链接时,我遇到问题,因此我想在该页面上保持该子菜单处于活动状态。
我有 css 类活动,并且有 javascript 来打开它,我需要的是用 php 让它处于活动状态。这是UL等级:
这是我的代码。可以用php或javascript来完成吗?
<ul>
<?php
$qKategori = ("SELECT * FROM kategori WHERE kprind = 0");
$rKategori = mysqli_query($dbc, $qKategori);
if ($rKategori) {
while ($exKat = mysqli_fetch_array($rKategori, MYSQLI_ASSOC)){
$emrikategorise = $exKat['kemri'];
$idkategori = $exKat['kid'];
$idprind = $exKat['kprind'];
?>
<li><a href="#"><?=$emrikategorise;?></a>
<ul>
<?php
$qPrind = ("SELECT * FROM kategori WHERE kprind = '".$idkategori."'");
$rPrind = mysqli_query($dbc,$qPrind);
while($prind = mysqli_fetch_array($rPrind)) {
?>
<li><a href="kategori.php?kid=<?=$prind['kid']?>"><?=$prind['kemri']?></a> </li>
<?php
}
mysqli_free_result($rPrind);
?>
</ul>
</li>
<?php }
mysqli_free_result($rKategori);
}
?>
</ul>
您可以在左侧的菜单中看到网站已 www.sitimobil.mk
您可能需要在输出之前构建数组,以便能够确定哪些菜单应该处于活动状态。您还可以将其与查询的优化相结合,而不必为每个类别执行 1 个查询。
像这样:
$active = isset($_GET['kid'] ? $_GET['kid'] : -1;
$tree = array();
$list = array();
$qKategori = ("SELECT * FROM kategori ORDER BY kprind");
$rKategori = mysqli_query($dbc, $qKategori);
if ($rKategori) {
while ($exKat = mysqli_fetch_array($rKategori, MYSQLI_ASSOC)){
$id = $exKat['kid'];
//To prevent numerical array with unused space
$name = 'kategori'.$exKat['kid'];
$list[$name] = $exKat;
//Calculate depth to see if the menu is a sub..sub..sub menu etc.
$parent = $list[$name]['kprind'];
if($parent == 0) {
$list[$name]['depth'] = 0;
$list[$name]['childCount'] = 0;
}
else {
$list['kategori'.$parent]['childCount']++;
$list[$name]['depth'] = $list['kategori'.$parent]['depth']+1; //Increment
}
if($id == $active) {
$list[$name]['active'] = true;
while($parent != 0) {
$parentName = 'kategori'.$parent;
$list[$parentName]['active'] = true;
$parent = $list[$parentName]['kprind'];
}
}
else
$list[$name]['active'] = false;
}
mysqli_free_result($rPrind);
//Once we have that we can output the results...
function output_menu($list, $parent = 0, $active = false)
$activeClass = $active ? ' class="active"' : '';
echo '<ul'.$activeClass.'>';
foreach($list as $row){
if($row['kprind'] != $parent) continue;
$link = $row['kprind'] == 0 ? '#' : 'kategori.php?kid='.$row['kid'];
echo '<li><a href="'.$link.'">'.$row['kemri'].'</a>';
if($row['childCount'] > 0)
output_menu($list, $row['kprind'], $row['active']);
echo '</li>';
}
echo '</ul>';
}
output_menu($list);
}
这仍然有点粗糙,但应该可以解决问题。 它可能可以优化,这样我们就不必多次浏览列表,但好处是不必请求对数据库的太多调用。 这应该会减轻数据库的工作负载和更快的输出。