Pagination active link - php

Question

I am doing pagination with ajax for table which has some
records for which i have to show active link for each
click.Please do help me with this.
$(document).ready(function()
{
ready_data();
function ready_data(page)
{
$.ajax({
url:"connection.php",
method:"POST",
data:{page:page},
success:function(data){
$('#paginate_record').html(data);
// $('.page_links').removeClass('active');
// $(this).addClass('active');
}
}); }
$(document).on('click', '.page_links ', function(){
var page = $(this).attr("id");
ready_data(page);
});
});
Here is My pagination page
$records_per_page = 2;
$page = '';
if(isset($_POST["page"]))
{
$page = $_POST["page"];
}
else
{
$page = 1;
}
$start_from = ($page - 1)*$records_per_page;
$sql = "SELECT * FROM students LIMIT $start_from,
$records_per_page";
$result = mysqli_query($con,$sql);
$data .= '</table><br /><div align="center">';
$page_sql = "SELECT * FROM students ";
$page_result = mysqli_query($con, $page_sql);
$total_records = mysqli_num_rows($page_result);
$total_pages = ceil($total_records/$records_per_page);
for($i=1; $i<=$total_pages; $i++)
{
$data.= "<span class='page_links active'
style='cursor:pointer;margin:10px;padding:1px;border:1px
solid
;'id='".$i."'>".$i."</span>";
}
echo $data;
Now in which i want to show active link for each click.i tried by adding class on ajax succes function also add the active class on span tag but it is applying to all links.so please do help me.

Answer 1

In the last for() loop, you are applying the active class to all the output, you want to apply this only to the current page...
for($i=1; $i<=$total_pages; $i++)
{
$data.= "<span class='page_links";
if ( $i == $page ) {
$data .=" active";
}
$data.= "' style='cursor:pointer;margin:10px;padding:1px;border:1px solid;'id='"
.$i."'>".$i."</span>";
}