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From cpanon <cpa...@yahoo.com>
Subject Re: ajax tomcat caching
Date Sun, 24 Aug 2008 19:41:40 GMT

Hello
Update
There seems to be another set of outputs.  If I do not run the web first and
just start it and then run my ajax app, I execute the call to the webapp,
but there is no parameter, even though I check both
XMLHttpRequest.open(a,b,c) and XMLHttpRequest.send(query).  This is a check
right before, I know it the proper url.  This sort of explains the initial
problem cited below, but not how I loose the parameter.  Ideas2? Thanks.

cpanon wrote:
> 
> Hello
> I trying to access an running webapp that itself works fine.  I can see
> the results of the app within the output of my IDE when I just run the app
> separately.  Problem is I am trying to test an ajax app and when I execute
> the XMLHttpRequest.send(query) after proper setup with a known parameter I
> see the app just re-executing the same page with same parameter that it
> had; not the one in my ajax call.  It is as if my XMLHttpRequest.send()
> was just executing a refresh.  Obviously I then dont get anything aback. 
> I have tried from a different browser window. This is not exotic, just
> self testing an ajax app against a localhost webapp.  Ideas?  tia.
> 

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