Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource

[dlc2000]

Someone can help me with this?

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /home/username/public_html/include/fuseronline.php on line 18

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/username/public_html/include/fuseronline.php on line 19

<?php
if( getenv('HTTP_X_FORWARDED_FOR') != '' )
{
$ipnumb = ( !empty($SERVER['REMOTE_ADDR']) ) ? $SERVER['REMOTE_ADDR'] : ( ( !empty($ENV['REMOTE_ADDR']) ) ? $ENV['REMOTE_ADDR'] : $REMOTE_ADDR );

if ( preg_match("/^([0-9]+.[0-9]+.[0-9]+.[0-9]+)/", getenv('HTTP_X_FORWARDED_FOR'), $iplist) )
{
  $iplocal = array('/^0./', '/^127.0.0.1/', '/^192.168..*/', '/^172.16..*/', '/^10..*/', '/^224..*/', '/^240..*/');
  $ipnumb = preg_replace($iplocal, $ipnumb, $iplist[1]);
}

}
else
$ipnumb = ( !empty($SERVER['REMOTE_ADDR']) ) ? $SERVER['REMOTE_ADDR'] : ( ( !empty($ENV['REMOTE_ADDR']) ) ? $ENV['REMOTE_ADDR'] : $REMOTE_ADDR );

include("dbconnection.php");
$dataip = mysql_query("select from utnonline where ipaddress='".$ipnumb."';");
$numip = mysql_num_rows($dataip);
$donline = mysql_fetch_array($dataip);
$newtime = @time()+605;
if ($numip>0)
{
$tnow = @time();
$told = @time()-60*60;
if ($told > $donline['timecon'])
{
$SESSION['login']=0;
mysql_query("delete from utnonline where ipadress='".$ipnumb."';");
mysql_query("insert into utnonline (id, utente, ipaddress, timecon) values (NULL, NULL,'".$ipnumb."','".$newtime."');");
}
else
{
mysql_query("update utnonline set timecon='".$newtime."' where ipaddress='".$ipnumb."';");
}
mysql_query("delete from utnonline where timecon<='".$tnow."';");
}
else
{
if ($SESSION['login'] == 1 && $SESSION['user'] != "")
mysql_query("insert into utnonline (id, utente, ipaddress, timecon) values (NULL,'".$SESSION['usermio']."','".$ipnumb."','".$newtime."');");
elseif ($SESSION['login'] == 0 && $SESSION['user'] != "")
mysql_query("insert into utnonline (id, utente, ipaddress, timecon) values (NULL, NULL,'".$ipnumb."','".$newtime."');");
elseif ($SESSION['login'] == 0 && $SESSION['user'] == "")
mysql_query("insert into utnonline (id, utente, ipaddress, timecon) values (NULL, NULL,'".$ipnumb."','".$newtime."');");
}

mysql_close();
?>

[leatherback]

I am assuming this:

$dataip = mysql_query("select * from utnonline where ipaddress='".$ipnumb."';");

causes the error.

Try when you replace it for:

$dataip = mysql_query("select * from utnonline where ipaddress='$ipnumb' "); 

[laserlight]

unless the semi-colon within the query is leading to an error, which apparently it doesnt for most ppl, that shouldnt make much of a difference.

perhaps the database connection is not present?

still, try leatherback's suggestion, it may be slower in parsing (minimally) but is easier to read

you might also want to change it to:

$dataip = mysql_query("SELECT * FROM utnonline WHERE ipaddress='$ipnumb'")
	OR die(mysql_error());

to help in debugging

[jimson]

do u have records in the database ? table utnonline ?
or is the table name correctly... ?

try the console one...
show tables;

$numip = mysql_num_rows($dataip); 
$donline = mysql_fetch_array($dataip); 

why count the number of rows when u will using just one row?