Question

i created a php code to show at admin who is online on website and store the ip of users connected at website and delete from database when user is no longer on website after 10 min.its work but its show me this error and i dont know where i make wrong. somebody can help me with this thing?

the error is:

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /home/a7078942/public_html/online.php on line 17

and here is my code:

<?php

require "connect.php";
require "functions.php";

// We don't want web bots scewing our stats:
if(is_bot()) die();



$stringIp = $_SERVER['REMOTE_ADDR'];
$intIp = ip2long($stringIp);

// Checking wheter the visitor is already marked as being online:
$inDB = mysql_query("SELECT 1 FROM tz_who_is_online WHERE ip=".$stringIp);

if(!mysql_num_rows($inDB))
{
    // This user is not in the database, so we must fetch
    // the geoip data and insert it into the online table:

    if($_COOKIE['geoData'])
    {
        // A "geoData" cookie has been previously set by the script, so we will use it

        // Always escape any user input, including cookies:\
        list($city,$countryName,$countryAbbrev) = explode('|',mysql_real_escape_string(strip_tags($_COOKIE['geoData'])));
    }
    else
    {
        // Making an API call to Hostip:

        $xml = file_get_contents('http://api.hostip.info/?ip='.$stringIp);

        $city = get_tag('gml:name',$xml);
        $city = $city[1];

        $countryName = get_tag('countryName',$xml);
        $countryName = $countryName[0];

        $countryAbbrev = get_tag('countryAbbrev',$xml);
        $countryAbbrev = $countryAbbrev[0];

        // Setting a cookie with the data, which is set to expire in a month:
        setcookie('geoData',$city.'|'.$countryName.'|'.$countryAbbrev, time()+60*60*24*30,'/');
    }

    $countryName = str_replace('(Unknown Country?)','UNKNOWN',$countryName);

    // In case the Hostip API fails:

    if (!$countryName)
    {
        $countryName='UNKNOWN';
        $countryAbbrev='XX';
        $city='(Unknown City?)';
    }
    mysql_query("   INSERT INTO tz_who_is_online (ip,city,country,countrycode)
                    VALUES('".$stringIp."','".$city."','".$countryName."','".$countryAbbrev."')");
}
else
{
    // If the visitor is already online, just update the dt value of the row:
    mysql_query("UPDATE tz_who_is_online SET dt=NOW() WHERE ip='".$stringIp."'");
}

// Removing entries not updated in the last 10 minutes:
mysql_query("DELETE FROM tz_who_is_online WHERE dt<SUBTIME(NOW(),'0 0:10:0')");

// Counting all the online visitors:
list($totalOnline) = mysql_fetch_array(mysql_query("SELECT COUNT(*) FROM tz_who_is_online"));

// Outputting the number as plain text:
echo $totalOnline;

?>

Answer 1

I think the problem may solve by this change:

$inDB = mysql_query("SELECT 1 FROM tz_who_is_online WHERE ip='".$stringIp."'");

But if this error occurs again please change the line :

$inDB = mysql_query("SELECT 1 FROM tz_who_is_online WHERE ip=".$stringIp);

to :

$inDB = mysql_query("SELECT 1 FROM tz_who_is_online WHERE ip=".$stringIp) or die("Error: ". mysql_error(). " with query ". $query);

to see whats the error make this problem and put the error in comment.

notice: its better ask these question in stackoverflow instead of serverfault