You are right. When you try to use the forward method, you'll give another
resourse the responsibility of manipulating the Response object. So you
can't use forward if in your code you've already accessed a PrintWriter or
ServletOutputStream object. May wanna try "include" instead?

Regards,

-Yue

-----Original Message-----
From: Michael Jager [mailto:[EMAIL PROTECTED]]
Sent: Friday, May 11, 2001 11:44 AM
To: [EMAIL PROTECTED]
Subject: AW: response already commited


To me it looks like the problem with your code is, that you try to forward
after already having committed a response to the out-object (via your
mailForm, which instantiates a PrintStream), which is not possible.

Hope this helps

Michael

-----Ursprungliche Nachricht-----
Von: A mailing list about Java Server Pages specification and reference
[mailto:[EMAIL PROTECTED]]Im Auftrag von Patrick Pierra
Gesendet: Freitag, 11. Mai 2001 14:07
An: [EMAIL PROTECTED]
Betreff: Re: response already commited


  String mailForm (HttpServletRequest req) throws JspException{
   StringBuffer message = new StringBuffer();

  Enumeration fields = fieldList.elements();

    while (fields.hasMoreElements()) {
      formField field = (formField)fields.nextElement();

      message.append(field.name)
             .append(" : ")
             .append(req.getParameter(field.name))
             .append("\n");
    }
     System.out.println(message.toString());
    //  Nouveau JavaMail here.

    try {
          SmtpClient client = new SmtpClient(mailhost);
          client.from(senderAddress);
          client.to(formRecipient);
          PrintStream msg = client.startMessage();
          msg.println();
          msg.println();
          msg.println("Sujet : Feedback Web");
          msg.println();
          msg.println(message.toString());
          msg.println("--");
          msg.println("Message from [EMAIL PROTECTED]");
          client.closeServer();
          } catch (IOException e) {
               e.printStackTrace();
          }

    return (message.toString());
  }

%>

<%
     if (isValidForm(request)) {
     User tUser = new User();  // here is the problem

     /*tUser.setMUserName(request.getParameter("UserName"));
     tUser.setMPassword(request.getParameter("Password"));
     session.setAttribute("user", tUser);*/
    mailForm(request);

    // need to create a 'thanks for your comments' page
    //response.sendRedirect("footer_help/thanks.jsp");
     %>
     <jsp:forward page="thanks.jsp"/>
 <% }
%>

> Hello,
>
> I' try to create an Object in a JSP :
>
> <%
>      User tUser = new User();
> ......
> %>
>
> When i run this JSP on Tomcat i receive this error message :
>
> Error: 500
> Location: /odata1/Registration.jsp
> Internal Servlet Error:
> java.lang.IllegalStateException: Response has already been committed
>            java.lang.Throwable(java.lang.String)
>            java.lang.Exception(java.lang.String)
>            java.lang.RuntimeException(java.lang.String)
>            java.lang.IllegalStateException(java.lang.String)
>            void
> org.apache.tomcat.core.HttpServletResponseFacade.sendError(int,
> java.lang.String)
>            void
>
org.apache.jasper.runtime.JspServlet.unknownException(javax.servlet.http.Htt

pServletResponse,
>
> java.lang.Throwable)
>            void
>
org.apache.jasper.runtime.JspServlet.service(javax.servlet.http.HttpServletR

equest,
>
> javax.servlet.http.HttpServletResponse)
>            void
> javax.servlet.http.HttpServlet.service(javax.servlet.ServletRequest,
> javax.servlet.ServletResponse)
>            void
>
org.apache.tomcat.core.ServletWrapper.handleRequest(org.apache.tomcat.core.R

equest,
>
> org.apache.tomcat.core.Response)
>            void
>
org.apache.tomcat.core.ContextManager.service(org.apache.tomcat.core.Request

,
>
> org.apache.tomcat.core.Response)
>            void
>
org.apache.tomcat.service.http.HttpConnectionHandler.processConnection(org.a

pache.tomcat.service.TcpConnection,
>
> java.lang.Object [])
>            void org.apache.tomcat.service.TcpConnectionThread.run()
>            void java.lang.Thread.run()
>
>
> any explication ?
>
> Patrick PIERRA
>
>
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