确定用户是否在线的最简单方法是什么？ (PHP/MYSQL)

Question

Is there a way I can piggy back sessions to know if the user is online?

Ie: use logs on, I set a $_SESSION variable, user times out- cookie Garbage collector updates the database to update their status as offline.

EDIT: I want a solution that does not involve times or dates. I want something to ride on sessions or something similar. Guessing if someone is online is not good enough for what I need.

Answer 1

Don't bother with figuring out the differences between timezones. That's not necessary.

Whenever the user accesses a page, update a field in their record of the Users table last-updated-time. Then do a query for all users having a last-updated-time within the last 5 minutes. Anything more than this, and they are considered "offline."

If you use your server-time, via the NOW() function in MySQL, you'll side-step calculating differences between timezones.

This is the standard way of tracking how many users are presently online (Meaning, active within the last couple of minutes).

Constantly Updated

If you would like to know they are still active even when they're not jumping from page to page, include a bit of javascript to ping your server every 60 seconds or so to let you know they are still alive. It'll work the same way as my original suggestion, but it will update your records without requiring them to be frantically browsing your site at least once every five minutes.

var stillAlive = setInterval(function () {
    /* XHR back to server
       Example uses jQuery */
    $.get("stillAlive.php");
}, 60000);

Answer 2

What you are asking for (after the clarification) is, by definition, impossible. HTTP is a connectionless protocol, so as soon as a user has hit a page and all the content comes back from the server to the user's browser, there is no connection between the two. Someone is "online" with your website for less than a second.

One thing you could do is to have JavaScript on your web page make AJAX requests back to your server on a regular basis which includes identifying information, and a different AJAX request when the user leaves the page, using window.onbeforeunload .

Answer 3

My way may not be the best way but since my site and userbase is all in mysql DB, when a user logins into my site,

1. I update the user table to say they are online
2. Insert them into an Online table
3. Then I set a session with the current time

Then on every page load I check for the online time session, if it exist, I check to see how old it is, if it is less then 5 minutes old, I do nothing, if it is older then 5 minutes, then I update the session time again with current time and also update the online users table with the time

Then I have a cron job that runs every 10 - 15 minutes that deletes any uses from online table and marks there user table as offline if there online time has updated within X amount of minutes

Answer 4

It sounds like your "is the person online" comparison logic isn't taking into account the timezones. If you are dealing with timezones in this manner, I strongly recommend that you store all your times in GMT, and convert them to local time for your users right before you display them. This will make any comparison operations very simple.

There's a nice SO thread about timezones here .

Answer 5

One thing I would advise is to store this kind of information in memory with for example memcached or mysql heap or redis. Because otherwise the database will be hit a lot.

Answer 6

Store the time-zone in the table, and use it in a calculation to find that users local time in comparison to the local time of the user viewing the page.

Edit:

Better yet, store all times as the server time, and base all calculations relative only to the server time.

Answer 7

Depends on your situation, it may be better for you to create a separate table "whose-online" with the columns: "ip" and "last-updated-time" and query/update this table every time a user loads a page.

On page load queries may include :

1. Update/insert "whose-online" table for current user based on ip.
2. Delete "expired" rows (can also be done periodically using cronjob).
3. Count "active" users.

Benefits of using this technique :

1. If a user is not logged in, he/she is still being counted/tracked.
2. Depends on the amount of users you have, querying this table may be quicker.

Note: If you use this you should want to take into consideration that any pageview will create a row in your table so based on useragent you can disregard bots or only count the popular ones (Firefox, IE, Safari, Opera, etc).

Answer 8

The solution that I have implemented for this is to, on every page load by an authenticated user, set/reset a memcache var such as "user_{userid} isonline" => true and expire it in 5 minutes. Then check if the var is in the cache when I access the user's info. Depending on the size of your user base, if you want to get a list of everyone online, you could use a memcache getmulti call with an array of "user {userid}_isonline" keys for all of your users.

of course, this really depends on how often a user will change pages on your site... to get a more accurate representation of the users online, you could implement an ajax xmlhttprequest call on your page running at a small interval (30 seconds or so) that resets the memcache var, and have the memcache var expire in less time (1 minute to account for possible browser issues). This is not COMPLETELY accurate, but as http does not have a persistent connection to the server, you are pretty limited on what you can do.

IF you require an up to the second representation of who is online, you could maybe have a flash app loaded in your page that connects to a jabber server, then just check if that particular user is logged in on the server.

Answer 9

Here's a way on how to get the difference between two dates from a database in minutes, then check for the difference and set the online/offline status.

$query = 'SELECT * FROM Users';
$result = mysqli_query($mysqli, $query);

foreach($result as $user){
    // date from the database
    $dbLastActivity = date("d-m-Y h:i:s a", strtotime($user['lastOnline']));
    // date now
    $now = date("d-m-Y h:i:s a", $date);

    // calculate the difference
    $difference = strtotime($now) - strtotime($dbLastActivity);
    $difference_in_minutes = $difference / 60;

    // check if difference is greater than five minutes
    if($difference_in_minutes < 5){
        // set online status
        $updateStatus = 'UPDATE Users SET Status="online" WHERE lastOnline="'.$user['lastOnline'].'"';
    } else {
        // set offline status
        $updateStatus = 'UPDATE Users SET Status="offline" WHERE lastOnline="'.$user['lastOnline'].'"';
    }

    // check if mysqli query was successful
    if (!$mysqli->query($updateStatus)){
        printf("Error Message %s\n", $mysqli->error);
    }
} 

Answer 10

You can also use the onunload tag... On the body More information here ( https://www.w3schools.com/jsref/event_onunload.asp )

This way you don't have to wait 4 -5 minutes for the user to get offline.

But you should still use the ajax update method in-case the user's browser crashes...

Edit: As @Quentin pointed out in the comments you may not be able to send out requests but there is a way you can do it... Explained in more detail in this Article .
So basically they are using the sendBeacon() Here is an Example

window.addEventListener("unload", function logData() {
  navigator.sendBeacon("/log", analyticsData);
});

NOTE: Although it should be kept in mind that this should not be the only method you should depend upon, and should implement for redundancy...