PHP Paging Not Updating Results - php

Question

I am creating a page that lists celebrities on Twitter. The page displays 10 different results as it should for the first loop but it doesn't seem to update the $result to pull the next 10 values from the database (for the next set of pages etc).
<?php
// connect to the database
$con = mysql_connect("localhost","someuid","somepwd");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("celebrity_twitter", $con);
$result = mysql_query("SELECT * FROM celebrities");
// number of results to show per page
$per_page = 10;
$total_results = mysql_num_rows($result);
$total_pages = ceil($total_results / $per_page);
// check if the 'page' variable is set in the URL (ex: view-paginated.php?page=1)
if (isset($_GET['page']) && is_numeric($_GET['page']))
{
$show_page = $_GET['page'];
// make sure the $show_page value is valid
if ($show_page > 0 && $show_page <= $total_pages)
{
$start = ($show_page -1) * $per_page;
$end = $start + $per_page;
}
else
{
// error - show first set of results
$start = 0;
$end = $per_page;
}
}
else
{
// if page isn't set, show first set of results
$start = 0;
$end = $per_page;
}
// display data in table
echo "<table class='table table-striped'>";
echo "<tr> <th>Avatar</th> <th>Celebrity Name</th> </tr>";
// loop through results of database query, displaying them in the table
for($i = $start; $i < $end; $i++ && $row = mysql_fetch_array($result) )
{
// make sure that PHP doesn't try to show results that don't exist
if ($i == $total_results) { break; }
// echo out the contents of each row into a table
echo "<tr>";
echo "<td><img height='73' width='73' src=" . "http://api.twitter.com/1/users/profile_image?screen_name=" . $row['avatar'] . "&size=bigger></td>";
echo "<td><a href=" . $row['url'] . " target='_blank'>" . $row['uid'] . "</td></a>";
echo "</tr>";
}
// close table>
echo "</table>";
// display pagination
echo "<strong>Page: </strong>";
for ($i = 1; $i <= $total_pages; $i++)
{
echo "<a href='index.php?page=$i'>$i</a> ";
}
?>

Answer 1

I'm using this at the moment and it's working perfectly fine for me, I hope it works for you too : PHP Pagination

Answer 2

You are not positioning yourself in the database, you always read the first elements:
// loop through results of database query, displaying them in the table
for($i = $start; $i < $end; $i++ && $row = mysql_fetch_array($result) )
{
// make sure that PHP doesn't try to show results that don't exist
if ($i == $total_results) { break; }
The proper way of doing this would be using the LIMIT clause of MySQL, with a special clause for recovering the true number of records
SELECT SQL_CALC_FOUND_ROWS * FROM celebrities LIMIT $start, $resultsperpage;
SELECT FOUND_ROWS(); // This is number of celebrities
This means that you need to calculate $start "before" running the query, i.e., "before" you know whether $start is a valid start at all. So you have to prepare for the case in which no rows are returned.
A quick alternative, which is pretty wasteful, is to retrieve everything and display only wanted rows:
// loop through results of database query, displaying them in the table
for ($i = 0; $i < $total_results; $i++)
{
$row = mysql_fetch_array($result);
if ($i < $start) continue;
if ($i == $end) break;
LIMIT implementation
mysql_select_db("celebrity_twitter", $con);
// number of results to show per page
$per_page = 10;
// We REALLY ought to move on to PDO: mysql_* will be deprecated sooner or later
$query = mysql_query('SELECT SQL_CALC_FOUND_ROWS * FROM celebrities');
// check if the 'page' variable is set in the URL (ex: view-paginated.php?page=1)
if (isset($_GET['page']) && is_numeric($_GET['page']))
{
$show_page = (int)$_GET['page'];
// make sure the $show_page value is valid
if ($show_page > 0)
$start = ($show_page -1) * $per_page;
else
$start = 0;
$query .= " LIMIT $start, $per_page;";
$paged = True;
}
else
{
$paged = False;
// The query having no LIMIT, it will retrieve everything
// if (!$paged), later we will display a link "Show Paged".
}
// Now run the query
$result = mysql_query($query);
// Also fetch REAL number of rows in table {{{
$exec = mysql_query("SELECT FOUND_ROWS() AS results;");
$tuple = mysql_fetch_assoc($exec);
$results= $tuple['results'];
mysql_free_result($exec);
unset($exec, $tuple);
// }}}
// If paging, calculate page number. Later, if ($paged), we'll display the pager.
if ($paged)
{
$maxpages = ceil($results / $per_page);
if ($page > $maxpages)
$page = $maxpages+1; // We fetch nothing.
else
$page = floor($start / $per_page)+1;
}
// display data in table. It is best to calculate $html first
// and then output it all together at end if no error.
$html = "<table class='table table-striped'>";
$html .= "<tr> <th>Avatar</th> <th>Celebrity Name</th> </tr>";
// loop through results of database query, displaying them in the table
// Here, we get all and only good results, so we need not check anything
while($row = mysql_fetch_array($result))
{
// echo out the contents of each row into a table
$html .= "<tr>";
}