How do I add a filter button to this pagination?

Question

Hey, I want to add a button(link), that when clicked will filter the pagination results.

I'm new to php (and programming in general) and would like to add a button like 'Automotive' and when clicked it updates the 2 mysql queries in my pagination script, seen here:

As you can see, the category automotive is hardcoded in, I want it to be dynamic, so when a link is clicked it places whatever the id or class is in the category part of the query.

1:

$record_count = mysql_num_rows(mysql_query("SELECT * FROM explore WHERE category='automotive'"));

2:

$get = mysql_query("SELECT * FROM explore WHERE category='automotive' LIMIT $start, $per_page");

This is the entire current php pagination script that I am using:

<?php

    //connecting to the database
    $error = "Could not connect to the database";
    mysql_connect('localhost','root','root') or die($error);
    mysql_select_db('ajax_demo') or die($error);

    //max displayed per page
    $per_page = 2;

    //get start variable
    $start = $_GET['start'];

    //count records
    $record_count = mysql_num_rows(mysql_query("SELECT * FROM explore WHERE category='automotive'"));

    //count max pages
    $max_pages = $record_count / $per_page; //may come out as decimal

    if (!$start)
       $start = 0;

    //display data
    $get = mysql_query("SELECT * FROM explore WHERE category='automotive' LIMIT $start, $per_page");
    while ($row = mysql_fetch_assoc($get))
    {
     // get data
     $name = $row['id'];
     $age = $row['site_name'];

     echo $name." (".$age.")<br />";

    }

    //setup prev and next variables
    $prev = $start - $per_page;
    $next = $start + $per_page;

    //show prev button
    if (!($start<=0))
           echo "<a href='pagi_test.php?start=$prev'>Prev</a> ";

    //show page numbers

    //set variable for first page
    $i=1;

    for ($x=0;$x<$record_count;$x=$x+$per_page)
    {
     if ($start!=$x)
        echo " <a href='pagi_test.php?start=$x'>$i</a> ";
     else
        echo " <a href='pagi_test.php?start=$x'><b>$i</b></a> ";
     $i++;
    }

    //show next button
    if (!($start>=$record_count-$per_page))
           echo " <a href='pagi_test.php?start=$next'>Next</a>";

    ?>

Answer 1

Example with 2 links/categories:

<a href='script.php?category=automotive'>automotive</a> <a href='script.php?category=sports'>sports</a>

Inside script.php:

$category = mysql_real_escape_string($_GET['category']);
$record_count = mysql_num_rows(mysql_query("SELECT * FROM explore WHERE category='$category'"));
...
$get = mysql_query("SELECT * FROM explore WHERE category='$category' LIMIT $start, $per_page");

Answer 2

Edit: I only posted this answer in response to the OP stating he was new to programming. Good hygiene is learned at the beginning very easily and becomes habit, or much later and is very hard to make habit.

http://us2.php.net/manual/en/security.database.sql-injection.php

please also read this article on security for databases. It would be much improved to write your queries as:

$start = mysql_real_escape_string($_GET['start']);
settype($start, 'integer');
$category = mysql_real_escape_string($_GET['category']);
$record_count = mysql_num_rows(mysql_query("SELECT * FROM explore WHERE category='%s'", $category));
...
$get = mysql_query("SELECT * FROM explore WHERE category='%s' LIMIT %d, %d", $category, $start, $per_page);

It is always worth it to write the extra secure code, even if just for practice. Basic security rules for databases are: Never trust user input, always check generated output. Since the input is from a querystring and gets run against the database, it has to be filtered.