Already exists code

[Mike4x4s]

Hello, just trying to make my page output user is already online if there name exists in the onlineusers field ive already selected the table anyone jknow the rest of the code?:

$query = mysql_query("SELECT * FROM onlineusers")
or die("Could not insert data because ".mysql_error());
$qry = mysql_fetch_array( $query );

[guido2004]

Mike, are you serious? After all the questions you've been asking in the PHP forum, and after all these months, you still don't know how to write a simple 'if' that confronts a form value with a value extracted from the DB?

[decowski]

// $username is the username of current user

$result = mysql_query("SELECT
                          `username`
                       FROM
                          `onlineusers`
                       WHERE
                          `username` = '$username'")
    or die("Could not insert data because ".mysql_error());

if (mysql_num_rows($result) == 1) {
    // user online
} 

[Mike4x4s]

hmm you sure on that code?

[decowski]

Yes, I’m sure. Try it before asking and if it doesn’t work then give more details. No need to be so secretive.

Your code selects all online users. You don’t need to do that if you’re checking for one particular user. My query looks for that specific username—if it returns a row then it means the user is online (is in the onlineusers table, anyway), if it doesn’t then they’re not.

[guido2004]

Mike, just to state the obvious :
if the field in your onlineusers table that contains the user id isn't called 'username', then you'll have to change the code decowski posted. Idem if the variable that contains the user id in the script isn't called $username.