Pagination script doesn't work with datatables - php

Question

I've been dealing with tihs problem for more than two days but still couldn't find what's the problem. I am trying to build a pagination system for my search results. Code works perfectly fine when I run it in new php file but when it comes to display those result in a table I constantly keep getting that Error! message inside of the last else section. Plus, wherever I place for loop for page numbers it always showing before the table. Think I was so busy with dealing with this problem I am focusing to the same point. Help, please!
edit I've just deleted all conditional statements and got undefined index for all my variables that I get by POST method. That's the problem but still don't know what might be the solution for this.
<?php
if (isset($_POST['search_btn'])) {
include_once "db_connect.php";
$from = $_POST['from'];
$where = $_POST['where'];
$date = $_POST['date'];
$type = $_POST['type'];
$proc_id = $_POST['proc_id'];
if(empty($from) || empty($where) || empty($date) || empty($type)){
header("Location: index.php?search=empty");
exit();
}else{
//define how many results you want per page
$results_per_page = 10;
//number of results stored in database
"SELECT * FROM proc WHERE p_from = '".$from."' and p_where = '".$where."' and type= '".$type."' ";
$result = mysqli_query($conn, $sql);
$number_of_results = mysqli_num_rows($result);
//determine number of total pages available
$number_of_pages = ceil($number_of_results/$results_per_page);
//determine which page number visitor is currently on
if (!isset($_GET['page'])) {
$page = 1;
} else {
$page = $_GET['page'];
}
//determine the SQL LIMIT starting number for the result on the displaying page
$this_page_first_result = ($page-1)*$results_per_page;
//retrive selected results from database and display them on page $sql='SELECT * FROM proc LIMIT ' . $this_page_first_result . ',' . $results_per_page;
$result = mysqli_query($conn, $sql);
while($rows = mysqli_fetch_array($result, MYSQLI_ASSOC)) {
echo '<tr onclick="content(\''. $rows['proc_id'] .'\')">';
echo "<td>" .$rows['p_name'] . " " . $rows['p_surname']. " </td>";
echo "<td>" .$rows['p_from']. "</td>";
echo "<td>" .$rows['p_where']. "</td>";
echo "<td>" .$rows['p_date']. "</td>";
echo "<td>" .$rows['price']. "</td>";
echo "<td>" .$rows['type']. "</td>";
echo "</tr>";
}
}
//display the links to the pages
for ($page=1;$page<=$number_of_pages;$page++) {
echo '' . $page . ' ';
}
}else{
echo "Error!";
}
?>

Answer 1

I'm just going to assume that the initial display works, but that pagination is giving you problems?
You're using POST values to pass the data from the search form to your code, however, when you then click the pagination links you are transferred to the search page, and lose those values.
You could change the next page URL to be a form and you pass each necessary value as a hidden input field. However, this has the drawback that when you hit the back button in your browser that it'll complain and ask you to resubmit the form data.
Another solution would be to store these post params in a session, cookie, whatever really. But in my opinion, that's not too great of a solution for this issue either.
I'd suggest you use GET parameters and then pass those in the next page button as well. This has the added benefit of being able to bookmark your searches.
Good luck!
As a side note, instead of building your query using concatenation you should use prepared statements. See the docs for the functions I used in the converted code below: https://www.php.net/manual/en/class.mysqli-stmt.php
if ($statement = mysqli_prepare($conn, "SELECT * FROM proc WHERE p_from = ? AND p_where = ? AND type = ?")) {
mysqli_stmt_bind_param($statement, "s", $from);
mysqli_stmt_bind_param($statement, "s", $where);
mysqli_stmt_bind_param($statement, "s", $type);
$result = mysqli_stmt_get_result($statement);
}