PHP条件，如果当前页面，则链接突出显示

Question

我有一个带有侧栏的网页，它使用一些 css 类。如果类=“active”，那么链接将突出显示。我想这样做，以便在任何页面上，脚本都会检查它所在的链接，然后决定突出显示哪个链接。下面是我所在的位置，但我真的不知道如何继续，请为我指出正确且更好的方向!如果我有 20 个链接，这个脚本将太长......谢谢!

---

<?php






$arr = array(

'1' => '/about.php', 
'2'=> '/abt-history2.php',
'3' => '/abt-shop.php',
);

$host =  $_SERVER['REQUEST_URI'];


if($host == $arr['1']) 
{
    $class1 = "active";
}

else if ($host == $arr['2']) 
{
    $class2 = "active";
}

else if ($host == $arr['3']) 
{
    $class3 = "active";
}


 ?>






   <ul id="navigationSide">
      <div style="padding-left:20px; padding-bottom:10px; font-size:16px; font-weight:800; color:#777;">TITLE</div>
      <li><a href="<?php print $arr['1'];  ?>" class="<?php echo"$class1" ?>">AA</a></li>
      <li><a href="<?php print $arr['2'];  ?>" class="<?php echo"$class2" ?>">BB</a></li>
    </ul>
    <div style="padding:5px;"></div>
    <ul id="navigationSide">
      <div style="padding-left:20px; padding-bottom:10px;  font-size:16px; font-weight:800; color:#777;">TITLE2</div>
      <li><a href="<?php print $arr['3'];  ?>" class="<?php echo"$class3" ?>" >CC</a></li>

</ul>

Answer 1

Try
  <?php
    $arr = array(

      '1' => 'about.php', 
      '2'=> 'abt-history.php',
      '3' => 'abt-shop.php',
    );


    foreach($arr as $key => $page){

      $class = "";

      $currentPage = $_SERVER['REQUEST_URI']; // check its value and get page name

      if($currentPage ==  $page){

       $class = 'class ="active"';
      }
    ?>
    <li>
        <a href ="yourdomainname/<?php echo $page; ?>" <?php echo $class; ?>>
           link<?= $key; ?>
        </a>
    </li>
    <php
    }
    ?>