I am these 2 errors come up in my database and I cant figure out the problem with my code....can someone help please? here is the code.

<body>
<?php

include("../misc.inc");
mysql_connect($host,$user,$pass);
mysql_select_db($database);
if($HTTP_GET_VARS['id'])
{
$query = "SELECT * FROM guestlist WHERE eventID = '".$HTTP_GET_VARS['id']."'";
$result = mysql_query($query);
$nrows = mysql_num_rows($result);
if($nrows > 0){
$i = 0;
$myGuests = array();
while($row = mysql_fetch_array($result))
{
extract($row);
$totalGuests += $guests;
$myGuests[$i] = "<tr><td>".($i+1)."</td><td>$guestName</td><td>$guests</td><td>$email</td></tr>";
$i++;
}
echo "<p>$nrows on guestlist, ".($nrows+$totalGuests)." expected for event.</p>";
echo "<table width =\"600\" border=\"1\">";
echo "<tr><th>&nbsp;</th><th>Name</th><th width=\"50\">Guests</th><th>Email</th></tr>";
for($i=0;$i<sizeof($myGuests);$i++)
{
echo $myGuests[$i];
}
echo "</table>";
}
else
{
echo "<p>No guests.</p>";
}
}
else
{
$currentDate = date("Y-m-d",time());
$query = "SELECT * FROM events_entries WHERE eventDate >= '$currentDate' ORDER by eventDate";
$result = mysql_query($query);
$nrows = mysql_num_rows($result);
while($row = mysql_fetch_array($result))
{
extract($row);
echo "<p>-<a href=\"getguests.php?id=$id\">$eventDate: $eventName @ $eventLocation</a></p>";
}
}

?>
<p>-<a href="index.php?<?php echo session_id(); ?>">Back To Main</a><br />
-<a href="logout.php?<?php echo session_id(); ?>">Logout</a></p>
</body>

I am getting these 2 error messages:
Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource
Please advise....thanks a lot guys!!

hi
this is because $result doesnot contain resource.
after $result = mysql_query($query);
put following code

if (!$result) {
    die('Invalid query: ' . mysql_error());
}

it will not allow further execution if $result is null.
if it is so check you $query string by echoing it.

OK I tried that and the result is the same....should it look like this?

$query = "SELECT * FROM guestlist WHERE eventID = '".$HTTP_GET_VARS."'";
$result = mysql_query($query);
if (!$result) {
die('Invalid query: ' . mysql_error());
}
$nrows = mysql_num_rows($result);
if($nrows > 0){
$i = 0;
$myGuests = array();
while($row = mysql_fetch_array($result))
{
extract($row);


hi
this is because $result doesnot contain resource.
after $result = mysql_query($query);
put following code

if (!$result) {
    die('Invalid query: ' . mysql_error());
}

it will not allow further execution if $result is null.
if it is so check you $query string by echoing it.

hey
try this...

<?php

include("../misc.inc");
$variable = $HTTP_GET_VARS['id'];
mysql_connect($host,$user,$pass);
mysql_select_db($database);
if($HTTP_GET_VARS['id'])
{
$query = "SELECT * FROM guestlist WHERE eventID = '$variable'";
$result = mysql_query($query);
$nrows = mysql_num_rows($result);
if($nrows > 0){
$i = 0;
$myGuests = array();
while($row = mysql_fetch_array($result))
{
extract($row);
$totalGuests += $guests;
$myGuests[$i] = "<tr><td>".($i+1)."</td><td>$guestName</td><td>$guests</td><td>$email</td></tr>";
$i++;
}
echo "<p>$nrows on guestlist, ".($nrows+$totalGuests)." expected for event.</p>";
echo "<table width =\"600\" border=\"1\">";
echo "<tr><th>&nbsp;</th><th>Name</th><th width=\"50\">Guests</th><th>Email</th></tr>";
for($i=0;$i<sizeof($myGuests);$i++)
{
echo $myGuests[$i];
}
echo "</table>";
}
else
{
echo "<p>No guests.</p>";
}
}
else
{
$currentDate = date("Y-m-d",time());
$query = "SELECT * FROM events_entries WHERE eventDate >= '$currentDate' ORDER by eventDate";
$result = mysql_query($query);
$nrows = mysql_num_rows($result);
while($row = mysql_fetch_array($result))
{
extract($row);
echo "<p>-<a href=\'getguests.php?id=$id\'>$eventDate: $eventName @ $eventLocation</a></p>";
}
}

?>

it wont show me the errors in my browser?
can you copy in the full code, and the error messages please.
see if i can help :D

This is the error message


Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /home/scorpio/public_html/getguests.php on line 57

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/scorpio/public_html/getguests.php on line 58

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