如何在以下代码中设置条件以在分页中标记当前页面？

Question

I am trying to set a condition for php- mysql pagination so that it can change the current page "li" to "li class="active" " to mark the current selected page. I am new to php and searching for such pagination tutorial but no luck.

I have done so far what is working but not able to mark selected page. Here $id is for detecting the current page id. How can I set if condition ( or other) so that I can mark the current page item in the pagination? Thousands thanks for helping me.

   <ul class="pagination">
     <?php if($id > 1) {?> <li><a href="?id=<?php echo ($id-1) ?>">Previous</a></li><?php }?>
     <?php
     for($i=1;$i <= $page;$i++){
     ?>

     <?php
     if ($id>1)
     { ?>
         <li class="active"><a href="?id=<?php echo $i ?>"><?php echo $i;?></a></li>
    <?php }
     ?>
    <!--     <li><a  href="?id=<?php echo $i ?>"><?php echo $i;?></a></li>  -->
      <?php
     }
      ?>
    <?php if($id!=$page)
    //3!=4
    {?> 
      <li><a href="?id=<?php echo ($id+1); ?>">Next</a></li>
    <?php }?>
 </ul>

Answer 1

You could change your for loop from

 <?php
     for($i=1;$i <= $page;$i++){
     ?>

     <?php
     if ($id>1)
     { ?>
         <li class="active"><a href="?id=<?php echo $i ?>"><?php echo $i;?></a></li>
    <?php }
     ?>
    <!--     <li><a  href="?id=<?php echo $i ?>"><?php echo $i;?></a></li>  -->
      <?php
     }
      ?>

to:

<?php
for($i=1;$i <= $page;$i++){
  $class=($i==$id)? ' class="active"' : '';
  echo '<li'.$class.'><a href="?id='.$i.'">'.$i.'</a></li>';
}
?>

If I've understood your code properly, $page represents total pages and $id represents the current page, this will set the current page number as the active class and leave the other pages without the class