output registered user online

[rutin]

hi there,
trying to output how many registered users is online !
this is my sintax, and i wonder if this is correct ?

registering session like this : just to show !

 
$loggedin = true;
$upwd = $password;
$msg = "Welcome $uname, you are now logged in.  <a href=index.php>Click here</a> to return to the front page.";

session_register("loggedin");
session_register("upwd");
session_register("uid");
session_register("uname");

code for registered users online:

<?

if (logincheck($uid, $upwd)) {
$sql_query = "SELECT * FROM users ORDER BY userid LIMIT 5";
$result = mysql_query($sql_query);
if(mysql_num_rows($result))
{
echo ("<hr>");
echo ("<span class=\"style1\">online now:</span></br>");
echo ("<span class=\"style5\">$uname</span><br>");

}
//if no fields exist
else
{
echo "no values in the database";
}
}
 ?>

I wonder if this will output right, and have it like those bug guys, registered users browsing this site?

Thank you

[AcousticJames]

What you need to do is loop through your result set. Like so:

echo "<hr><br>";
echo "Online now: <br>";

$sql_query = "SELECT * FROM users ORDER BY userid LIMIT 5";
$result = mysql_query($sql_query);
if ($row = mysql_fetch_array($result)) { // make sure you have something to work with
   do {
      printf($row['uname'] . ", ");
   } while ($row = mysql_fetch_array($result));
}

The printf() statement simply prints the name followed by a comma and a space, so you end up with a nice comma separated list of users. But your SQL statement will select all users from your database, regardless if they are online. You might want something more like:

$sql_query = "SELECT * FROM users WHERE logged_in = 'Y' ORDER BY userid LIMIT 5";

Where you would have a column in your table named "logged_in" that would contain a 'Y' if the user is currently logged in or a 'N' if the user is not logged in (keeping track of this status would be done elsewhere, of course). But this should give you a general idea of how to go about this.

Hope that helps!

James

[rutin]

hi there,

Thank you very much for your quick reply.
Your sugested a good thing.
I tried that, but doesnt work

I create new table item called: loged_in, and folowed your intstruction, its just simply wont update table item.

my code: for login script

<?php


  $sql = "select * from users where login = '$login' and password = '$password'";
  $result = mysql_query($sql ,$db);

  if ($myrow = mysql_fetch_array($result)) {


     do {

        $uid = $myrow["userid"];
        $uname = $myrow["login"];

     } while ($myrow = mysql_fetch_array($result));

     $loggedin = true;
     $upwd = $password;
     $msg = "Welcome $uname, you are now logged in.  <a href=index.php>Click here</a> to return to the front page.";

$sql = "update users set logged_in ='Y' where login = '$login' and password = '$password'";


  } else {
     $loggedin = false;
     $upwd = "";
     $uid = "";
     $uname = "";
     $msg = "Sorry, that login and password is not valid.  Click back on your browser to try sgain.  If you have forgotten your password <a href=forgot.php>click here</a>.  If you are a new user you will need to <a href=newuser.php>create a new login</a>";


  }

session_register("loggedin");
session_register("upwd");
session_register("uid");
session_register("uname");

?>