Granny's Programming Pearls
"inside of every large program is a small program struggling to get out"
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Still confused of forward()

 
Tang Yue
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"In the doGet() of FirstServlet:
PrintWriter out = response.getWriter();
out.println("<html><body>Page 1");
RequestDispatcher rd = request.getRequestDispatcher("NextServlet");
rd.forward(request, response);
out.println("<br>Page 3</body></html>");
In the doGet() of SecondServlet:
PrintWriter out = request.getWriter();
out.println("<br>Page 2");
"

The answer provided by J+web simulator is "IllegalStateException" with the explanation as follows:
"
The servlet specification mandates that when a request is "forwarded" to another resource, the buffer should be cleared of the contents generated by the forwarding resource. Therefore, only Page2 will be sent to the client. However, calling forward() means that the request processing is delegated to the callee resource permanently. So, although the control does return to the caller resource [because rd.forward() is just a regular method call], the caller resource cannot generate any output after the call to forward().
In this case, FirstServlet is trying to send output to the client after calling forward() and so an IllegalStateException will be thrown. "



Is this true? Can't the caller call print() after a forward() ?
 
Arun Kumar Gupta
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yes i think the explanation is true ...the forward did return to firstservlet but you cannot print anything.
try using include if you want to print page3.
 
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