Call function when PHP-page gets closed

Question

Is there a way to call a function when the php-page gets closed?
I am saving the online/offline status in my MySQL db.
With my function

Users::isOffline(Users::getUserName())

I'm writing "offline" into the db.
Now I want to call this function when someone is closing the page with the close button.
I've tried this:

<script language="JavaScript" type="text/javascript">
window.onbeforeunload = confirmExit;
function confirmExit()
{
var offline= <?php echo Users::isOffline(Users::getUserName())?>;
}
</script>

But then, the function gets called when the php-page is loaded....
Any ideas? :)

Just curious, any reason you are not using something like a last_activity time? is it a game? or...? — Clay, Jan 15 '16 at 08:20
@ChrisBanks No, its a website with an filebrowser on it. I am using a last_activity_time... but I also wanna use the online/offline status :) — npit, Jan 15 '16 at 08:23

score 4 · Answer 1 · answered Jan 15 '16 at 08:25

4

It is because PHP code doesn't run on client side - you need to use JS.

function confirmExit(){
//here you send request to your server
}

And on serverside you recieve request and mark user as offline

answered Jan 15 '16 at 08:25

Manasov Daniel

1,348
9
17

But when I send the request in this javascript function, I have to open a new db connection and stuff like that. Thats what I not wanted to to ... Thats why I wrote my Users class – npit Jan 15 '16 at 08:32
1

@npit java != javascript – online Thomas Jan 15 '16 at 08:33
@npit you can use your class method while recieving request from JS `confirmExit` – Manasov Daniel Jan 15 '16 at 08:35
@ManasovDaniel Can you explain this a little bit more ? How should this look like? – npit Jan 15 '16 at 08:38
@npit as Thomas sais Java is not JavaScript, like tea is not teapot – Manasov Daniel Jan 15 '16 at 08:38
@npit you can invoke `Users::isOffline(Users::getUserName())` when you got request. I don't know what is inside your app so I can't explain more ;) – Manasov Daniel Jan 15 '16 at 08:40
@ManasovDaniel Ah now I get what you think... It is already JS ` – npit Jan 15 '16 at 08:42
@npit yeah, that's right. But my comment was about you said Java in first comment. It's little bit offensive for JS-developers when people say Java instead JavaScript :D – Manasov Daniel Jan 15 '16 at 08:46
@ManasovDaniel yeah, that was my fault ^^ just forgot the script behind it. But I corrected it :) – npit Jan 15 '16 at 08:47

score 2 · Answer 2 · edited May 23 '17 at 12:31

What you try to achieve is not that easy using only php and javascript. Usually when you want to have a good online/offline status check you will have to use websockets.

There is a free subscription service called Pusher, i have used it many times and it's very easy to setup.

And also, here you can read how to set it up for online/offline: Show the connecting user login status using Websocket and Pusher

score 2 · Answer 3 · answered Jan 15 '16 at 08:48

You can post user's information to another php file before page close.

Button would be like this:

<button type='button' data-userid='<?php echo Users::isOffline(Users::getUserName())?>'>click1</button>

You can post userid to a php file to update the status.

$('button').click(function(){
    var userid = $(this).attr('data-userid';
    $.ajax({
      type: "POST",
      url: "gooffline.php",
      data: {userid: userid},
      success: function(data)
      {
          alert("Come back soon!");
      }
   });
});

Php file would be like this:

$userid = $_POST["userid"];
$status = userLoggingOut($userid); //your db update function
if($status){
    echo json_encode(array("status"=>"success","message"=>"user info updated"));
}

Hope this solves.

Andriy Ivaneyko · Accepted Answer · 2016-01-15T09:09:00.813

2

You cannot do it because php could be executed on backend, you can achieve desired result in next ways:

Preferable

If you need to do it in right way see @Philip Andersson answer. Also you could get an clear idea of how to use websockets in Create Live Group Tutorial.

Workaround

What you want to achieve can be done by applying next steps:

create php file which would perform Users::isOffline(Users::getUserName()) while handling request from your browser
Create javascript handler onbeforeunload, which would send AJAX request to newly created php file mentioned before.

Tricky moment: It's not trivial way to send Ajax Request from onbeforeunload handler. To do it use code provided below:

window.onbeforeunload = function() {
    var x = sendUserOfflineAjax();
    return x;
};

function sendUserOfflineAjax(){
        // Perform sending ajax request
}

edited Jan 15 '16 at 09:09

answered Jan 15 '16 at 09:01

Andriy Ivaneyko

20,639
6
60
82

I tried this : `` But it doesnt work ... – npit Jan 15 '16 at 09:13
Ok, I had another mistake with my including of AJAX. Your solution with sending an AJAX request works very well. Thanks. – npit Jan 19 '16 at 14:06

Call function when PHP-page gets closed

4 Answers4