What do i need to do to get the most recent to be displayed on the top

popcop

how can i make new details which have been sent through a form to appear on the top of the page rather than the bottom

have a look at this... http://www.popcop.co.uk/ca/form.htm

heres the code im using

<?php

$host = "localhost";
$user = "X";
$password = "X";
$dbname = "X";
$tablename = "X";
$currentTime = time();

$link = mysql_connect($host, $user, $password) or die ("could not connect to database");
mysql_select_db ($dbname, $link) or die ("error");

/* Build SQL query to insert our information into table */
$query = "INSERT INTO $tablename values ('$name', '$comments', '$currentTime')";

/* Execute query */
$db_result = mysql_query($query);

/* If there was an error executing query... */
if (!$db_result) {

/* Return error information to Flash and quit! */
print "database error";
exit;
} else {


$query = "SELECT name, comments FROM $tablename";


$result = mysql_query($query);
$numItems = mysql_num_rows($result);
?>


<html>
<head>
<title>news</title>
</head>
<body bgcolor="#0099CC" leftmargin="0" topmargin="0" marginwidth="0" marginheight="0">
  <table width="26" height="115" border="0" cellpadding="0" cellspacing="0">
    <tr> 
      <td align="center" valign="middle"><img src="top.gif" width="300" height="115"></td>
    </tr>
  </table>

<?

print"<table width='300' height='49' border='0' cellpadding='0' cellspacing='0'>";
while(list($name, $comments) = mysql_fetch_row($result)) {
  print"<tr> 
    <td colspan='3'><img src='spacer.gif' width='100' height='5'></td>
  </tr>
  <tr bgcolor='#1DB7EB'> 
    <td width='10' rowspan='2'></td>
    <td width='274' height='20' valign='middle' class='unnamed1'><img src='name.gif' width='40' height='8'>$name</td>
    <td width='16' rowspan='2'></td>
  </tr>
  <tr> 
    <td height='20' valign='middle' bgcolor='#1DB7EB' class='unnamed1'><img src='said.gif' width='65' height='8'>$comments</td>
  </tr>";
}
}
print"</table>";
?>

LordShryku

$query = "SELECT name, comments 
          FROM ".$tablename."
          ORDER BY time_submitted DESC";

popcop

see when i replace

$query = "SELECT name, comments FROM $tablename";

with what you've just given me...

$query = "SELECT name, comments 
          FROM ".$tablename." 
          ORDER BY time_submitted DESC";

i seem to get these errors....

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /home/popcop/public_html/ca/senddata.php on line 34

AND

Warning: mysql_fetch_row(): supplied argument is not a valid MySQL result resource in /home/popcop/public_html/ca/senddata.php on line 52

Line 34 is..

$numItems = mysql_num_rows($result);

and line 52 is..

while(list($name, $comments) = mysql_fetch_row($result)) {

LordShryku

time_submitted was a guess at the name of the column. You need to replace that with the real name.