Thanks Christopher. It solved my problem. I was not setting the content-type
explicitly. However, I did workaround my problem in another way i.e. i sent
the parameter info as extra-path info in the request and didn't use
getParameter().

Thanks again,
-Sourabh

----- Original Message -----
From: Christopher K. St. John <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Wednesday, September 05, 2001 9:09 PM
Subject: Re: getQueryString() and getParameter()


> > Sourabh Kulkarni wrote:
> >
> > I am trying to send an object to servlet.
> > URL url = new
URL("http://localhost:8080/context/servletname?param=value";);
> > 2] If i use request.getParameter("param") and then go ahead with reading
> > the object, I get the following excetpion on server.
> >
> >    at javax.servlet.http.HttpUtils.parseQueryString(HttpUtils.java:151)
> >    at javax.servlet.http.HttpUtils.parsePostData(HttpUtils.java:254)
> >    at
org.apache.tomcat.util.RequestUtil.readFormData(RequestUtil.java:101)
> >    at
org.apache.tomcat.core.RequestImpl.handleParameters(RequestImpl.java:691)
> >    at
org.apache.tomcat.core.RequestImpl.getParameterValues(RequestImpl.java:259)
> >    at
org.apache.tomcat.core.RequestImpl.getParameter(RequestImpl.java:250)
> >
>
>  The servlet container is trying to read your POST'ed object
> as a form. That should only happen if your content-type is
> "application/x-www-form-urlencoded", which is (always?) the
> default if you don't set it to anything else.
>
>  Are you setting the Content-Type?
>
>
> --
> Christopher St. John [EMAIL PROTECTED]
> DistribuTopia http://www.distributopia.com
>
>
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