Thread: how we can do that when one user is log suuceefully (servlets)

hii to all

first i want to elaborate the scenario
it is like that when one user for example user A HAS USERNAME:ANURAG & PASSWORD: UITK

HE IS ALREADY LOGIN SUCCESSFULLY IN APPLICATION HE SURFING THE WEBSITE.

THEN if another person B TRYING TO LOGIN INTO THE APPLICATION BY THE USE OF THE PERSON A USERNAME AND PASSWORD

then during this i want that person B GET A MESSAGE LIKE THIS "U ARE ALREADY LOGIN " .

MEANS HOW WE ENSURE THAT SEESION MANAGEMENT VERIFY THE UNIQUE USER_ID MEANS NO OTHER CAN BE LOG ON WHEN THAT USERID IS ALREADY LOG IN.



the session because each client will gets unique session and they cannot be cross-checked. You'll either need to keep information in the application context to check against, or perhaps even in the DB.


means let us suppose we has a table current_user

which has field user_id , application_id , session _id ,

and the sessionid is generated from container when user A is login for their user_id and this information is save in this table like this
test , school, jsesso12345 ,

m eans when user is login then we can check it for user_id exist or not in that table but my doubt is that when the second time user is login then this user_id is already exist for same user so how can he will be able to login for the second time .

please suggest me

Well you would probably need another db fields as a timedate

`logged_in` TIMESTAMP NOT NULL DEFAULT CURRENT_TIMESTAMP ON UPDATE CURRENT_TIMESTAMP,

select id from table where username='blah' and
(unix_timestamp(now())) time1 ,(unix_timestamp(logged_in)+300)




String id = session.getId();
//get current id - compare this to output of your db - if not found then its a new user.

Also when they log out - before doing log out - to store a status in the same table to say logged_out=1;

But they may close the browser so by checking the session id if the same


Learn Java by Examples - How do I count number of online users?

Have a look here for some more guidance on active sessions.