SQL problem :(

[xyuri]

Iwant to find the numer of people who have accessed my site in the last five minutes. (but i abviously having difficulties). at the moment i have been trying and experimenting with the following code:

$on_online =  mysql_query("SELECT count(*) FROM activity WHERE lactive>('$date $time'-'0000-00-00 00:00:05')"); 

but doesnt to anything for me.

how can it be done ? (preferable through SQL and not too much PHP).

[Mincer]

I'm presuming lactive is a DATETIME column with the last active time??

$sql =  "SELECT COUNT(*) FROM activity" ;
$sql .= " WHERE UNIX_TIMESTAMP( NOW() ) - UNIX_TIMESTAMP( lactive ) <= 300" 

Matt.

[xyuri]

yes, it is a datetime column. but that codxce didnt work is there maybe a mysql functions or something, maybe a tutorial that you could point me to that i might be able to learn from ? (i dont generaly use C&P'd code)

I'm using the following code now:

                    $sql =  "SELECT COUNT(*) FROM activity";
                    $sql .= " WHERE UNIX_TIMESTAMP( NOW() ) - UNIX_TIMESTAMP( lactive ) <= 300";
                    
                    $on_online =  mysql_query($sql);
                    ECHO mysql_result($no_online, 0, "count(*)");
                    //$c_user->no_of_reg_users(); 

[Mincer]

You have a couple of errors, I think your code should be:

$sql =  "SELECT COUNT(*) FROM activity";
$sql .= " WHERE UNIX_TIMESTAMP( NOW() ) - UNIX_TIMESTAMP( lactive ) <= 300";
                    
$no_online =  mysql_query($sql); // did say '$on_online'
ECHO mysql_result($no_online, 0, 0 ); // simplified, not sure if your method would work or not 

Matt.

[xyuri]

YAY ! it works. thanx matt

[Mincer]

Originally posted by xyuri
YAY ! it works. thanx matt

Of course it does!