php pagination for search result failing to display - php

Question

i've just been learning pagination. i'm having trouble getting it to work for search results. it displays the first page correctly with the right number of links but clicking on any of the links (even on the page 1) goes to a blank page.
can somebody please tell me what i'm doing wrong:
The following code is for when the search button is clicked.
<?php
include('includes/connect-db.php');
include('includes/functions.php');
if (isset($_GET['searchbtn'])){
$product=$_GET['products'];
$status=$_GET['ticket_status'];
$order_by=$_GET['order_by'];
$ticket_type=$_GET['ticket_type'];
#check if product has been selected
if ($_GET['products'] == 'select'){
echo '<font class="error">&nbsp Please select a product to search.</font>';
}else{
if ($status == 'All' AND $order_by == 'All' AND $ticket_type == 'All' ){
$page_query="SELECT * FROM tickets WHERE product='$product' ORDER BY created DESC";
}elseif ($ticket_type == 'BOX'){
$page_query="SELECT * FROM tickets WHERE product='$product' AND status='$status' AND pbi <> '-' ORDER BY '$order_by' ";
}elseif ($ticket_type == 'PVR'){
$page_query="SELECT * FROM tickets WHERE product='$product' AND status='$status' AND inc <> '-' ORDER BY '$order_by' ";
}else{
$page_query="SELECT * FROM tickets WHERE product='$product' AND status='$status' ORDER BY created DESC";
}
}#end of else
$results_per_page=3;
$result_set=pagination($results_per_page,$page_query);
$query=$result_set['query'];
$pages=$result_set['pages'];
#SELECT, SHOW, DESCRIBE, EXPLAIN and other statements returning resultset. A resource on success or False on Error
if (!empty($query)){
$result = mysqli_query($db,$query) or die( "My query ($query) generated an error: ".mysql_error());
$num_results = mysqli_num_rows($result);
if ($num_results > 0){
displayTicket($result);
if ($pages > 1){
echo '</br>';
for($i=1;$i<=$pages;$i++){
echo ' '.$i.' ';
}
}
}else{
echo "&nbsp No Records found.";
}
}#query string is not empty
}
?>
i have put the pagination code in a separate function:
function pagination($per_page, $pages_query){
include('includes/connect-db.php');
$pagination[]=array();
#total result count
$total_result=mysqli_query($db,$pages_query);
#ceil takes a decimal number and gives the nearest whole number
$total=mysqli_num_rows($total_result);
$pages = ceil($total / $per_page);
$page = (isset($_GET['page'])) ? (int)$_GET['page'] : 1;
$start = ($page - 1) * $per_page;
$query = " LIMIT $start,$per_page";
$query = $pages_query.$query;
$pagination['query'] = $query;
$pagination['pages'] = $pages;
return $pagination;
}
Note: the pagination linking to other pages only fails when i attempt it with the search feature. i have tried it on a page that lists information in a table and the links work fine.

Answer 1

It looks to me like the problem is the information in the link: localhost/test/searchProductForm.php?page=1 tells searchProductForm which page to display, but it doesn't tell it what the search information was. Without that, your search page has no idea what to display. Try changing the pagination link to include all the information that was in the original search link (i.e., your products, ticket_status, order_by, and ticket_type parameters).

Answer 2

I know this is an old thread, but for anyone not able to solve a similar problem with the code used above, it could be to do with the lack of session_start() at the beginning of the document.
I say this because although the code is trying to obtain the session variables passed by the ?page= &products= etc to $_GET statements but they are not being retrieved into the variables from the URL but you need to place the session_start() at the begging of your PHP document to pass the information from the URL to the strings.