Basicly, I planned on doing a "next/previous" button on my website, but they wont work, I think the problem lies in the server being in safe mode, I was hoping someone could show me how to get around it (keeping safe mode on) cheers heres my code
if(!isset($_GET['start'])) $start = 0;
// connect to the server
@mysql_connect($host, $username, $password) or die ("SERVER CONNECTION ERROR");
@mysql_select_db($database) or die ("DATABASE CONNECTION ERROR");
// retrieve all the rows from the database//
$results = mysql_query("SELECT * FROM news ORDER BY id DESC LIMIT " . $start . ", 2") or die ("TROUBLE GETTING INFORMATION FROM THE DATABASE");
$num = mysql_num_rows($results);
$query = "SELECT count(*) as count FROM news";
$result= mysql_query($query);
$row = mysql_fetch_array($result);
$numrows= $row['count'];
// print out the results
?>
<div align="center">
all messages are displayed in Decending order (newest first)<br>
<table>
<?
if ($num == 0) {
echo "no contacts to display";
}
else {
while ($contact = mysql_fetch_row($results))
{
// print out the info
$id = $contact[0];
$title = $contact[1];
$message = $contact[2];
echo"<tr><td></td><td><b>";
echo" $title</b></td></tr><tr><td></td><td>";
echo"$message<hr></td></tr>";
}
}
echo "<br>";
if($start > 0)
echo "<a href=\"" . $PHP_SELF . "?start=" . ($start - 4) .
"\">Previous </a>\n";
echo "||";
if($numrows > ($start + 4))
echo "<a href=\"" . $PHP_SELF . "?start=" . ($start + 4) .
"\">Next</a>\n";
?>