php - User Online Remains 1 -

everything working , communicating database regardless of how many users online stays 1

i have helper functions created num rows , query

online users

function users_online(){       $session = session_id();     $time = time();     $time_out_in_seconds = 60;     $time_out = $time = $time_out_in_seconds;      $sql = "select * users_online session = '$session'";     $result = query($sql);     $count = row_count($result);      if ($count == null) {          $sql= "insert users_online(session, time) values ('$session','$time')";         query($sql);     } else {          $sql= "update users_online set time = '$time' session = '$session'";         query($sql);     }      $sql = "select * users_online time > '$time_out'";         return $count_user = row_count($result); } 

db config helper functions

$connection = mysqli_connect('localhost', 'root', '', 'danger_db');  function row_count($result) {     return mysqli_num_rows($result); }  function escape($string) {     global $connection;     return mysqli_real_escape_string($connection, $string); }  function query($query) {     global $connection;     return mysqli_query($connection, $query); }  function confim($result) {     global $connection;     if (! $result) {         die("query failed" . mysqli_error($connection));     } }  function fetch_array($result) {     global $connection;     return mysqli_fetch_array($result); } 

database table

create table if not exists `users_online` (   `id` int(11) not null,   `session` varchar(255) not null,   `time` int(11) not null ) engine=myisam  default charset=utf8 auto_increment=31 ;  -- -- dumping data table `users_online` --  insert `users_online` (`id`, `session`, `time`) values (30, 'n3kppn4lp0n9h4gdm7a6rh84b2', 1452969896), (29, 'mqhrkr2b3f2rga1vhqh4t6pe23', 1452969689), (28, 'dpg2nmsrvsvottj9k7ot1gvcm6', 1452969893); 

    $time = time();     $time_out_in_seconds = 60;     $time_out = $time = $time_out_in_seconds; 

what trying here? setting time_out and $time equal $time_out_in_seconds. you're setting 3 values 60 static value.

then later in code saving value 60 database , calling time > '$time_out' sql you're setting user online being 60 , asking database return rows greater than 60.

solution

replace above code with:

    $time = time();     $time_out_in_seconds = 60;     $time_out = $time - $time_out_in_seconds;     //edit further down page show users      //updated within last 60 seconds. 

this mean sql query check time of session against current time updating sessions with current time well. should give better results.

you have poor coding @ end of code here:

$sql = "select * users_online time > '$time_out'";     return $count_user = row_count($result); 

the return takes value , passes calling script, value value want give number of users online, value of return? made of row_count($result). $result? last $sql row 1 need results from, row has no relation result variable.

you have missed line , need reinsert line $result = query($sql); so:

$sql = "select * users_online time > '$time_out'";     $result = query($sql); return row_count($result); } 

this give numeric value number of rows returned of online users, result of function users_online();