User Online List PHP - php

Question

I'm trying to make a online user list output all my usernames in a row like this (username, username, username)
The code I'm trying:
$result2 = $db->query("SELECT * FROM `online`");
$fetched2 = $result2->fetch_assoc();
$user_id = $fetched2['user_id'];
$result = $db->query("SELECT * FROM `users` WHERE id = $user_id");
while ($fetched = $result->fetch_assoc())
{
$users_arr[] = array("username"=> $fetched['username']);
}
But it's only returning 1 username from online table and not all in a row, how can i do this?
return json_encode(array("status" => 200, "message" => $users_arr)));
This only outputted 1 username, and also it's outputted in json format which i do not want, i want it outputted like so: User, User, User and not like [{'username': 'vanilla'}]"
Can someone please show me the correct way of using an array or a solution to my problem, thanks!
UPDATE:
Tried this and it worked, but now the dilemma is how do i make the outputted json array to look like User, User instead of ['user1', 'user2']
$users = array();
$result = $db->query('SELECT user_id, ident, u.group_id FROM '.$db->prefix.'online LEFT JOIN '.$db->prefix.'users AS u ON (ident=u.username) WHERE idle=0 ORDER BY ident', true) or error('Unable to fetch online list', __FILE__, __LINE__, $db->error());
while ($user_online = $result->fetch_assoc())
{
$users[] = $user_online["ident"];
}
I tried to use:
$trimmed = trim($users, '[]');
But this threw a warning/error:
PHP Warning: trim() expects parameter 1 to be string
Also tried:
$string = implode('[]', $users);
It does output 2 users, but its formated/outputted like:
user1[]user2

Answer 1

(Moved solution to answer space on behalf of the question author to move it from the question).
Fixed by using:
$string = implode(", ", $users);