What am I doing wrong here?

[N9ne]

This line:

PHP Code:

$adminonline = mysql_fetch_array(mysql_query("SELECT COUNT(userid) as total FROM session LEFT JOIN user USING(userid) WHERE user.usergroupid=6")); 


is giving this error:

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

[Xenon]

hmm, should work normally...
maybe just try this

PHP Code:

$adminsql = mysql_query("SELECT COUNT(userid) as total FROM session LEFT JOIN user USING(userid) WHERE user.usergroupid=6");
$adminonline=mysql_fetch_array($adminsql); 


normally it shouldn't make any differenze, but as said, i don't know the php mysql commandos very well..

[mr e]

do it like this, and change whatever you want to repeat inside the while

PHP Code:

$getadminonline = mysql_query("SELECT COUNT(userid) as total FROM session LEFT JOIN user USING(userid) WHERE user.usergroupid=6");

while ($adminonline=$DB_site->fetch_array($getadminonline)) {
    if ($adminonline['total'] > 1) {
        echo("There are $adminonline[total] admins online");
    } elseif ($adminonline['total'] == 1) {
        echo("There is $adminonline[total] admin online");
    } else {
        echo("There are no admins online");
    }
} 


[Xenon]

@ mr e:
as long as $DB_site is not defined this can't work