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. > -- View this message in context: http://www.nabble.com/ajax-tomcat-caching-tp19133882p19133988.html Sent from the Tomcat - User mailing list archive at Nabble.com. --------------------------------------------------------------------- To start a new topic, e-mail: users@tomcat.apache.org To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED]
