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Posted by Tony Marston on 10/05/06 16:33
Read http://www.tonymarston.co.uk/php-mysql/client-clones.html for a
description of how I solved this very same problem.
--
Tony Marston
http://www.tonymarston.net
http://www.radicore.org
"cendrizzi" <cendrizzi@gmail.com> wrote in message
news:1160057473.202722.19300@i3g2000cwc.googlegroups.com...
> Hi all,
> I've read some stuff on this but can't seem to come up with a solution
> that works right. I have a semi-mature (yet very large and robust)
> internal web application that currently only works with one window open
> at a time.
>
> Maybe I'm abusing the use of $_SESSION but I have data entry processes
> split up in several steps (which is required since depending on what
> was put before determines what pages will be shown after). To store
> this information between the steps I use the session, which is the only
> way I know that this would be possible (uses arrays in arrays). So if
> I were to open up a window and start the data entry it would overwrite
> the last one.
>
> So one thought was trying to create an instance id that that is
> randomly generated that is passed between the pages using a $_GET
> variable. Using this id I would create and reference custom session
> variables making sure each instance is completely unique. This seems
> to be consistent with what others have tried but I don't think it would
> work since if someone middle clicked on a page link it would open up a
> new tab (in firefox and IE7) with the same instance id with. There
> doesn't seem to me to be an easy way to ensure that each window in a
> browser has it's own, unique, instance.
>
> Surely this isn't unique to me so how in the world can I ensure that
> each window/tab gets a unique id of some sort so that my session is
> unique for each one?
>
> Lastly, if necessary, I wouldn't mind trying to suppress a new window
> but this is plagued with the same issues (how do I know it's a new
> window/tab?). I know this is not ideal but given the nature of the
> application this wouldn't be a big problem.
>
> Thanks in advance.
>
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