How to detect if user open two tabs for same session?

Question

I\'ve done a booking application done using CakePHP which involves a few steps before the checkout page. In between these steps I store the information in the session.

Answer 1

As Benjamin pointed out, you can do this by coming up with a per-window ID which you then report to the server with each request. If you want to enable the multiple window use case, I think this is the most reasonable approach.

I don't think you need to poll the server via ajax to do this though. As long as every request to the server includes the window ID, you can disambiguate what session on the server side to associate the request with. That will mean that your server-side code has to change to include the window ID as part of the session.

However, that approach has serious drawbacks. For instance, if the user closes both Tab A and Tab B, and then opens up a new Tab C... what happens? Which set of data should they reconnect with?

The approach of restricting the user to a single tab sounds pretty reasonable if you need to prevent the intermingling of multiple tabs worth of state.

You can do this via Local Storage, btw. If you generate a random ID on the client and push it into Local Storage via a call like window.localStorage.setItem('window_id', ); then you can check for this value on page load. If the value exists when the page loads, you've got a pretty clear indicator that at least one other window is open. You'll have to make sure to remove this value on page unload, otherwise dead sessions might trigger false positives.