PHP Pagination issue mysql and php showing same data on all pages - php

Question

I am having issues with some pagination with data i get from a MySQL database in PHP.
My code is below. Basically what happens is it creates the right amount of pages however each page shows the same data and doesn't even show only 5 rows per page.
I'm really stuck. This is my first time trying pagination. Any help would be greatly appreciated.
$pagenum = $_GET['pagenum'];
if (!(isset($pagenum)))
{
$pagenum = 1;
} else
{
$pagenum = $_GET['pagenum'];
}
//Count the number of results.
$data = mysql_query("SELECT * FROM `articles` WHERE `content` = '' AND `requestedby` != '$id'") or die(mysql_error());
$rows = mysql_num_rows($data);
//Set the number of results to be displayed per page.
$page_rows = 5;
//This tells us the page number of our last page
$last = ceil($rows/$page_rows);
//this makes sure the page number isn't below one, or more than our maximum pages
if ($pagenum < 1)
{
$pagenum = 1;
} elseif ($pagenum > $last)
{
$pagenum = $last;
}
//This sets the range to display in our query
$max = 'LIMIT ' . ($pagenum - 1) * $page_rows .',' .$page_rows;
//This is the query again, the same one... the only difference is we add $max into it.
$data_p = mysql_query("SELECT * FROM `articles` WHERE `content` = '' AND `requestedby` != '$id' '$max'") or die(mysql_error());
//Work out writers earnings based on prices.
//100 Words - $1.25
//300 Words - $2.50
//500 Words - $4.00
//700 Words - $5.50
//1000 Words - $8.00
$_100earnings = "0.65";
$_300earnings = "1.25";
$_500earnings = "2.50";
$_700earnings = "3.00";
$_1000earnings = "5.00";
?>
<!-- main -->
<div id="main">
<center><h2>Write Articles</h2></center>
<br />Available Projects:
<table border="1">
<tr>
<td>Title:</td>
<td>Length:</td>
<td>Writers Earnings:</td>
</tr>
<?php
//This is where you display your query results
while($info = mysql_fetch_array($data_p))
{
echo "<tr>";
echo "<td>" . $info['keywords'] . "</td>";
echo "<td>" . $info['length'] . "</td>";
switch ($info['length'])
{
case 100:
$writersearnings = $_100earnings;
break;
case 300:
$writersearnings = $_300earnings;
break;
case 500:
$writersearnings = $_500earnings;
break;
case 700:
$writersearnings = $_700earnings;
break;
case 1000:
$writersearnings = $_1000earnings;
break;
}
echo "<td>$" . $writersearnings . "</td>";
//echo $info['Name'];
echo "</tr>";
}
?>
</table>
<br /><br />
<?php
// This shows the user what page they are on, and the total number of pages
echo " --Page $pagenum of $last-- <p>";
// First we check if we are on page one. If we are then we don't need a link to the previous page or the first page so we do nothing. If we aren't then we generate links to the first page, and to the previous page.
if ($pagenum == 1)
{
} else
{
echo " <a href='{$_SERVER['PHP_SELF']}?pagenum=1'> <<-First</a> ";
echo " ";
$previous = $pagenum - 1;
echo " <a href='{$_SERVER['PHP_SELF']}?pagenum=$previous'> <-Previous</a> ";
}
//just a spacer
echo " ---- ";
//This does the same as above, only checking if we are on the last page, and then generating the Next and Last links
if ($pagenum == $last)
{
} else
{
$next = $pagenum + 1;
echo " <a href='{$_SERVER['PHP_SELF']}?pagenum=$next'>Next -></a> ";
echo " ";
echo " <a href='{$_SERVER['PHP_SELF']}?pagenum=$last'>Last ->></a> ";
}

Answer 1

use this code, remove ' around $max in query, when you add ' around $max the query becomes select.... from.... where..... 'LIMIT.....' and it fails the query.
$data_p = mysql_query("SELECT * FROM `articles` WHERE `content` = '' AND `requestedby` != '$id' $max") or die(mysql_error());

Answer 2

You need to specify LIMIT in your query LIMIT offset,count
$data_p = mysql_query("SELECT * FROMarticlesWHEREcontent= '' ANDrequestedby!= '$id' '$max'") or die(mysql_error());
Remove the ' quotes from '$max'
$data_p = mysql_query("SELECT * FROM articles WHERE content= '' AND requestedby!= '".$id."' ".$max) or die(mysql_error());