Javascript 当用户单击子菜单并在新页面中打开时,子菜单将保持活动状态
当访问者单击子菜单,链接在新页面中打开时,我会遇到问题,因此我希望该子菜单在该页面上保持活动状态 我有css类活动和javascript打开它,我需要的是使它与php的活动。 这是UL认证的等级: 这是我的密码。它可以用php或javascript来完成Javascript 当用户单击子菜单并在新页面中打开时,子菜单将保持活动状态,javascript,php,css,menu,Javascript,Php,Css,Menu,当访问者单击子菜单,链接在新页面中打开时,我会遇到问题,因此我希望该子菜单在该页面上保持活动状态 我有css类活动和javascript打开它,我需要的是使它与php的活动。 这是UL认证的等级: 这是我的密码。它可以用php或javascript来完成 <ul> <?php $qKategori = ("SELECT * FROM kategori WHERE kprind = 0"); $rKategori = mysqli_query($dbc, $qKategori)
<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>
您可以看到网站左侧的菜单是www.sitimobil.mk,您可能需要在输出前构建阵列,以确定哪些菜单应处于活动状态。您还可以将其与查询优化相结合,以使每个类别不必执行1个查询 比如:
$active = isset($_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);
}
这仍然是一个有点粗糙,但应该做的把戏。它可能会被优化,这样我们就不必过多地查看列表,但它的好处是不必向数据库请求太多的调用。这将减轻数据库的工作量,提高输出速度。这不起作用,打开网站的速度非常慢。我认为最好是使用GET和put,如果还有其他选项,请使用?