How can I get the pagination to work here? - php

Question

I'm fairly new to coding, so I'm open to critique as well as help. I'm trying to apply pagination to my search results. I have returned the results I need, applied the limit and managed to get the pagination controls to present properly. However, when I select "next" or "previous" the pages have no results on. I'm sure there is something fundamentally wrong, but I just can't spot it.
?php
include_once("db_connex.php");
if (isset ($_POST ['search'])) {
$searchq = $_POST ['search'];
$count_query = mysql_query("SELECT * FROM activejobs WHERE jobtitle LIKE '%$searchq%' OR region LIKE '%$searchq%' AND status= '1'");
$count = mysql_num_rows($count_query);
// pagination starts here
if (isset($_GET['page'])) {
$page = pre_replace("#[^0-9]#","",$_GET['page']);
}
else {
$page = 1;
}
$perPage = 3;
$lastPage = ceil($count / $perPage);
if ($page < 1) {
$page = 1;
}
else if ($page > $lastPage) {
$page = $lastPage;
}
$limit = "LIMIT " . ($page - 1) * $perPage. ", $perPage";
$query = mysql_query("SELECT * FROM activejobs WHERE jobtitle LIKE '%$searchq%' OR region LIKE '%$searchq%' AND status= '1' ORDER BY jobid DESC $limit");
}
if ($lastPage != 1) {
if ($page != $lastPage) {
$next = $page + 1;
$pagination .= 'Next';
}
if ($page != 1) {
$prev = $page - 1;
$pagination .= 'Previous';
}
}
while ($row = mysql_fetch_array($query)) {
$jobtitle = $row['jobtitle'];
$region = $row['region'];
$salary = $row['salary'];
$jobdescription = $row ['jobdescription'];
$joburl = $row ['joburl'];
$output .= '<div id= "searchresults"><div id= "applybutton">Details</div><font id= "resultstitle">'.$jobtitle.'&nbsp-&nbsp'.$region.'&nbsp-&nbsp'.$salary.'</font><br>'.$jobdescription.'</div>';
}
?>

Answer 1

This kinds of if-else hell:
// pagination starts here
if (isset($_GET['page'])) {
$page = pre_replace("#[^0-9]#","",$_GET['page']);
}
else {
$page = 1;
}
can be solved like this (default it at start):
// pagination starts here
$page = 1;
if (isset($_GET['page'])) {
$page = pre_replace("#[^0-9]#","",$_GET['page']);
}
or even this (if you feel adventurous):
$page = (isset($_GET['page']) ? pre_replace("#[^0-9]#","",$_GET['page']) : 1);
Making a lot of If-else is easy at start and hard later, so keep it simple by reducing when you have nothing to do. Making your code smaller in a step closer to any solution.
Also this is common:
...
while ($row = mysql_fetch_array($query)) {
$jobtitle = $row['jobtitle'];
$region = $row['region'];
...
Why assigning $jobtitle to $row['jobtitle']; ?
It doesnt make your code easier, it just adds more code and making you read harder.
Give $row['X'] directly.
Also, as #ojovirtual stated you need to pass "$search" parameter everytime, otherwise your entire code block will be ignored ("$search" is not set)
Finally, when working with MySQL you need to check the values you feed your queries with,
in this example the $searchq. A malicious coder could make the $searchq look like a part of the query.
There is a simple fix for that:
Instead of plain:
$count_query = mysql_query("SELECT * FROM activejobs WHERE jobtitle LIKE '%$searchq%' OR region LIKE '%$searchq%' AND status= '1'");
make it a habit doing this:
$searchq = mysql_real_escape_string($searchq);
$count_query = mysql_query("SELECT * FROM activejobs WHERE jobtitle LIKE '%$searchq%' OR region LIKE '%$searchq%' AND status= '1'");
Not a universal solution but a starter before dive into new technologies as a starter.
This is a must for fields like username, password etc.
Finally, change from:
if (isset ($_POST ['search'])) {
$searchq = $_POST ['search'];
to:
if (isset ($_GET['search'])) {
$searchq = $_GET['search'];

Answer 2

You need to pass search again in your next and previuos buttons. A quick fix would be:
Change $_POST to $_REQUEST:
if (isset ($_REQUEST ['search'])) ...
Add the search to your next and previousbuttons:
$pagination.= 'Next';
Same with the previous button.
As someone stated, you should do some input sanitation before any database query.