php - 如果下拉菜单中的子页面已打开，我如何提供菜单项 ID ="active"？ PHP

Question

如果一个菜单项有子页面，主页本身是不可点击的——只有它的子页面可以。我的问题是，如果子页面打开，我如何将 id="active"添加到菜单项？

恐怕解决方案无法通过 URL 实现，因为子页面的 URL 是:subpage.php?id=5 而在数据库中，pages和 subpages 表具有相同的 page_id。

所以使用 (basename($_SERVER['REQUEST_URI']) == 'subpage.php?id='.$page['page_id']) 给出了错误的结果。它使项目处于事件状态，例如 $subpage['subpage_id'] = $page['page_id'] 但我需要这样的结果:$subpage['page_id'] = $page[ 'page_id']..

如何在不需要 URL 的情况下实现这一目标？

表格:pages = page_id (AI, primary), page_name;
subpages = subpage_id (AI, primary), page_id, subpage_name;

我的 subpages.php 文件:

<?php
include_once('classes.php');

$page = new Page;
$subpage = new Subpage;

$pages = $page->fetch_all();
$subpages = $subpage->fetch_all();
?>

<div class="sidenav">

<?php foreach ($pages as $page) { ?>

    <button class="dropdown-btn"
    <?php foreach ($subpages as $subpage) {
//DOWN HERE IS THE PROBLEM!
        if ($subpage['subpage_id'].$page['page_id']) {
            echo('id="active" ');
        } }?> >
        <?php echo $page['page_name']; ?>
        <i class="fa fa-caret-down"></i>
    </button>    

//DOWN HERE I MAKE SUB-PAGES ACTIVE
<div class="dropdown-container">
    <?php foreach ($childpages as $childpage): ?>
        <a 
        <?php 
        if (basename($_SERVER['REQUEST_URI']) == 'subpage.php?id='.$subpage['subpage_id'].'') {
            echo('id="active-sub" ');
        }
        ?>
        href="subpage.php?id=<?php echo $childpage['subpage_id']; ?>">
            <?php echo $childpage['subpage_name']; ?>
        </a>
    <?php endforeach; ?>
</div>

<?php } ?>

</div>

classes.php 文件:

class Page {
    public function fetch_all(){
        global $pdo;

        $query = $pdo->prepare("SELECT * FROM pages");
        $query->execute();

        return $query->fetchAll();
    }    
public function fetch_data($page_id) {
        global $pdo;

        $query = $pdo->prepare("SELECT * FROM pages WHERE page_id = ?");
        $query->bindValue(1, $page_id);
        $query->execute();

        return $query->fetch();
    }
}

class Subpage {
    public function fetch_all(){
        global $pdo;

        $query = $pdo->prepare("SELECT * FROM subpages");
        $query->execute();

        return $query->fetchAll();
    }
public function fetch_data($subpage_id) {
        global $pdo;

        $query = $pdo->prepare("SELECT * FROM subpages WHERE subpage_id = ?");
        $query->bindValue(1, $subpage_id);
        $query->execute();

        return $query->fetch();
    }
}

现在，当任何子页面打开时，所有页面都处于“事件”状态。

总结:

我需要这样的结果:$subpage['page_id'] = $page['page_id']。我如何在不需要 URL 的情况下实现这一点？

Answer 1

可能是这样，我把当前子页面的id作为变量加入::

<?php
include_once('classes.php');

$page = new Page;
$subpage = new Subpage;

$pages = $page->fetch_all();
$subpages = $subpage->fetch_all();

$currentSubpageId = intval($_GET['id']),
?>

<div class="sidenav">

<?php foreach ($pages as $page) { ?>

    <button class="dropdown-btn"
    <?php foreach ($subpages as $subpage) {
        if($currentSubpageId == $subpage['subpage_id'] && $subpage['page_id'] == $page['page_id']) 
        {
            echo('id="active" ');
        } 
    }?> >
        <?php echo $page['page_name']; ?>
        <i class="fa fa-caret-down"></i>
    </button>    

//DOWN HERE I MAKE SUB-PAGES ACTIVE
<div class="dropdown-container">
    <?php foreach ($childpages as $childpage): ?>
        <a 
        <?php 
        if (basename($_SERVER['REQUEST_URI']) == 'subpage.php?id='.$subpage['subpage_id'].'') {
            echo('id="active-sub" ');
        }
        ?>
        href="subpage.php?id=<?php echo $childpage['subpage_id']; ?>">
            <?php echo $childpage['subpage_name']; ?>
        </a>
    <?php endforeach; ?>
</div>

<?php } ?>

</div>