I'm not awesome enough to write a chat application, and I'm trying to get one to work, and I've recently downloaded one from here, it's pretty good so far, as I've tested it out on XAMPP, but I have a slight problem. I'm trying to generate a list of online users to give it a more practical application-like feel, but the problem with that, is I have no clue how to do it easily.
When users login to my site, a session named g_username is created, (the chat says 'username', but I'll fix that) and from what I see so far, the easiest method would be to store their username in a database called OnlineUsers and call that data via Ajax, but, the other problem, is that it's session based, and sometimes the users can just leave, without logging out, and I intended to run a script to logout the user from both the OnlineUsers table, and by deleting the session.
If they leave without logging out, they'd be online forever! I could potentially suffix a bit of code on every page, that toggled an ajax event on page close, the event being a script that kills their OnlineUsers table record, but then again, that would load the server with useless queries as users jump between pages, as far as I'm aware.
Creating my entire site in Ajax isn't really an option, as it's a load of different sites combined in to 1 'place' with a social 'layer' (if you will) from a social service.
Does anyone see a way to do this that would make sense, and be easy to integrate, and do with Apache, without command line access?
You could so something like storing a timestamp of the users last action in a database, comparing that timestamp when outputting online users and making sure that it was done at most 1 min ago.
Run on all/vital pages:
(Deciding if the last action is outdated, you could also check if it was done for one minute ago to reduce the database-load)
if($user['lastAction'] < time()) {
//update into database, last action is outdated
}
When calculating the amount of users online and is within the loop of each timestamp
//If the users last action was within a minute, the user is most likely online
if(($row['lastAction']- time()) > 60*60)
//count user as online
you could have a cron job [if you have cpanel] running on the server once every 60secs or so, that checks when a user last sent anything via the chat if they have not in the last lets say 5mins then remove their entry from the online users list.
Related
I've created an iOS and android app that uses a database where users can login which in my case "User_active" gets changes to 1 and logged out gets changed to 0. The only problem I have is that I cannot check to see if the user has left the app, or if they have closed it, so what I want to do is create an event on phpmyadmin to change the "User_active" to 0 after 30min, but I can't seam to get things right. I'm using Awardspace as my host and I've only seen tutorials for localhost. Errors that I'm getting are that my user account does not have the write privileges to create the events and I cannot find any user tabs to change user privileges. If anyone can send me off in the right direction that would be great or even better a solution.
phpMyAdmin is a fantastic interactive tool, but it is not the tool for this.
In some ways, this is a fairly common problem. Typical web based applications don't know when a user has left, they only know when a user does something. One method is to do the following:
1 - Add a field to the user table (the same table that has the "user_active" field) to track the last activity by that user of any type. This should be a DATETIME or TIMESTAMP field and updated always with the current time.
2 - Add a cronjob to check any records in the user table to see if they are (a) still logged in (user_active=1) and the timestamp is more than 30 minutes old. If so, set user_active to 0 - essentially force a logout after 30 minutes. You may want to have an additional field to indicate it was a forced automatic logout rather than a user-initiated logout.
If your hosting service allows cronjobs, then run this periodically (perhaps every 5 minutes) and you are done. If your hosting service doesn't allow cronjobs then one option is to add a call to the "automatic user logout" function as part of every regular page. That won't work if you have thousands of active users as the overhead would be too much (and the delay on each page display). But if you have a small number of active users it will get the job done - I have done plenty of system housekeeping using that method.
I'll try to explain my question the best way I can.
I'm not asking for codes, only for the best method of doing it.
I want to create a browser game and use time for upgrading stuff, building etc.
For example, to build 1 house will take 1 hour.
So I will began with saving the timestamp+(60*60) at the moment the user did his action.
My question is, how to update it the best way?
One way I thought of was to add function that check every page view of the user if it's done.
But then if he's not logged in the update wont happen.
Second way i thought about is for every page view of any user to check for every user registered. But it's not effective and there is a problem if no user is logged in.
Any suggestions?
I had my game doing it simply, without crons.
When a player wanted something that takes time, i just updated his database information with the appropriate time of ending that job (columns are just examples)
UPDATE player SET jobend = UNIX_TIMESTAMP() + (60*60*4) # ending in 4 hours
Then, every page that had an information about the remaining time, i just used something like this:
SELECT (jobend - UNIX_TIMESTAMP()) AS jobremaining FROM player
I formatted correctly the time using strftime and i displayed that to the user.
In the case the remaining time was negative, the job was done.
There was no-need for absolute counting since user was able to do something with the job when he was connected.
When the player just changed pages or doing something else i had a function where i just checked all timely events while the user was online (so to catch any negative timer), then presented with javascript any change (i posted javascript counters for every page)
Now, if you talk about updating in real-time, cron is the way but are you sure you're going to need it for a game? I asked that question myself too and the answer was not.
EDIT
If another player sees the buildings on schedule page (an hypothetical page) i am doing the same calculations; if a time just got negative for a specific player (regardless if another player see the page), i just reward him with the building (in database i make all the changes), even if he's offline. There's no harm in this, since he can't do anything anyway. The other players will just see he has a building. The key here is that i execute the required updating PHP code regardless of player's connection to the game; as long at least ONE player is logged-in i'm executing the progress function for everything.
This isn't so slow as it sounds (updating all players by using just a connected player that visits a specific page). You just have a table of 'jobs' and check timers against the current time. More like a single query of getting the negative ones.
I'm currently trying to display all online users on my SITE'S userpage using the php session variables. To do this, whenever a user logs in or out, a column in a database gets set to "1" or "0".. However this doesn't entirely work since the database doesn't get updated when the user closes their browser (and therefor destroys the session). So is there another way of checking if a certain sessionid is set??
I also want to know how the twitter and facebook handle this ?
You almost have it. The way that the vast majority of sites deal with the issue is to have a table like you do, but they add a timestamp to it and update the timestamp when a new page is loaded.
When querying the table, you simply look for say the last five minutes of active users and they are the "live" users on the site.
Technically, you don't even need to keep the "logged in/out" value in that table. If they have been logged in within the lat five minutes, they are probably still about.
There is no guaranteed, sure-fire, totally bullet-proof way of checking if a user is there or not. You can do some tricky JS to ping on and off, you can add even more JS that will try to alert the db when the user navigates away from the page - but at the end of the day, you cannot do anything if a browser is closed unexpectedly, or if that user loses power, or network.
On top of that web browsing is by default stateless and doesn't maintain a connection to the user after the server has finished sending code. The best we can efficiently do is update a table when the user does something new and assume they will be around for a few minutes at least.
I haven't checked but Twitter and Facebook most likely have Javascript code which notifies the server when somebody closes the page, probably coupled with a periodic heartbeat and timeout.
Check the onunload event and XMLHTTPRequest to see how you can make a request to your PHP application notifying of an user leaving (a library like jQuery might help you do this much more easily).
Add a field "last_visit" to user's Table and update it every time when user visit your site
When user login to your site find "last_visit" time and current time,after that use this function
$time_period = floor(round(abs($current_time - $last_visit)/60,2));
if ($time_period <= 10)
$online_offline_status = 1;
else
$online_offline_status = 0;
}
and then print your final result
<?php if (
$online_offline_status == 0){ ?>
<span style="color:#FF0000;">Offline</span>
<?php } else if ($online_offline_status == 1) {?>
<span style="color:#669900;">Online</span>
<?php }?>
I know the title is complicated, but i was looking for some advise on this and found nothing.
Just want to ask if i'm thinking the right way.
I need to make a top facebook shared page with about 10 items or so for my website items (images, articles etc.)
And this is simple, i will just get the share count from facebook graph api and update in database, i don't want to make it in some ajax call based on fb share, it could be misused.
Every item has datetime of last update, create date and likes fields in database.
I will also need to make top shared url in 24h, 7 days and month so the idea is simple:
User views an item, every 10 minutes the shared count is obtained from fb graph api for this url and updated in database, database also stores last update time.
Every time user is viewing the item, the site checks last update datetime, if it is more than 10 minutes it makes fb api call and updates. It is every 10 minutes to lower fb api calls.
This basically works, but there is a problem - concurrency.
When the item is selected then in php i check if last update was 10 minutes ago or more, and only then i make a call to fb api and then update the share count (if bigger than current) and rest of data, because a remote call is costly and to lower fb api usage.
So, till users view items, they are updated, but the update is depending on select and i can't make it in one SQL statement because of time check and the remote call, so one user can enter and then another, both after 10 minutes and then there is a chance it will call fb api many times, and update many times, the more users, the more calls and updates and THIS IS NOT GOOD.
Any advise how to fix this? I'm doing it right? Maybe there is a better way?
You can either decouple the api check from user interaction completely and have a separate scheduled process collect the facebook data every 10 minutes, regardless of users
Or, if you'd rather pursue this event-driven model, then you need to look at using a 'mutex'. Basically, set a flag somewhere (in a file, or a database, etc) which indicates that a checking process is currently running, and not to run another one.
I have an inline chat application which I got from Ajax Chat, which is working brilliantly. The application allows a user to chat with users that are registered on the system. Ie:
Now I need to show if the user is online or offline.
So my question is how do I show online users using PHP?
Thank You
Basically what you need is a way to register users activity.
One way you can do this is doing it by sessions within PHP, and you log these. There are tons of ways to register then your activity in a log. If the activity is not updated for example in 5 minutes, the user is offline. Bassically you just need then a sessionId, and a timestamp (and i would recommend this also to hang to a userid). If offline, there is no userId assigned and when online you add a userId. If you have those, its pretty easy. Its a matter of updating them constantly when a new page is loaded and if they log out, you simply destroy the session, or update it so it wont be linked to the user.
It may not be the best system, but it works, and it might help you.
I don't know your specific needs. Pardon me, If I am wrong.
If Jabber support is there with Ajax Chat, why not try ejabberd kind of XMPP servers rather than re-inventing the wheels on your own. And you could have a look at Apache Vysper too, since it has support of extension modules too. If XMPP server is there, users presence handling and message transfer would become a cake walk.
What you need is a constantly update for a table in your database that save the last change in an user and save the date time... so if that date is more than 5 or 10 min, the user ir off..you can do it with ajax...
What i would do is have a script that the clients run to do an ajax call to update a entry in your database with a time stamp for last seen. Not too often or you will overload your server.
you can also put some if statements where it checks for keystrokes, mouse movement, and if the window is active if you really want to get technical and do a away status.
then in active chats just check the time stamp for active messages or when the user list is open. anything outside a acceptable range will show the user as off line. 5 minutes seems pretty long to me. poll for a check every 10 seconds maybe?