When a user revisits a page on our site via a menu dropdown, I'm required to display the page as it was when the user last left. (By the way, the browser back button is disabled.) The approach we use now is to rebuild the page with information stored in the cache (and differentiated with a sessionid) . This seems like needless much work for the server. I'm wondering if there is a way to use varybycustom feature of the output cache to preserve the most recent version of the page on a session level
I've had no success with these:
<%@ OutputCache Duration=“300" VaryByParam="None" VaryByCustom=“sessionid”%>
Public Overrides Function GetVaryByCustomString(
ByVal context As System.Web.HttpContext,
ByVal custom As String) As Strin
If custom = "sessionid" The
return = context.session.sessionid
End I
End Functio
In GetVaryByCustomString, the session object is nothing.
Even if it were not "nothing," I'm unsure about what version of the page would be in the output cache - the original version when first called up, or the version after editing
John Saunders - 16 Mar 2004 20:02 GMT
> When a user revisits a page on our site via a menu dropdown, I'm required to display the page as it was when the user last left. (By the way, the
browser back button is disabled.)
Interesting. Did you also disable the address bar, favorites and shortcuts?
I and many other readers of this newsgroup would like to know how you did
that.

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John Saunders
John.Saunders at SurfControl.com
jt - 17 Mar 2004 02:01 GMT
I used "disable" as a shortcut term. The developer who did the work used "window.history.forward," so the button is there, but in effect it does nothing.
Ravichandran J.V. - 23 Mar 2004 12:40 GMT
How did you disable the backward button?
with regards,
J.V.Ravichandran

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