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Servlet Query

 
Ranch Hand
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Hi All
Please help me in solving the below mentioned Question:

Q: Consider the following code snippets. What will be displayed on the browser when a GET request is sent to FirstServlet assuming that the buffer is large enough to hold all the data before sending the data to the client?
Code Snippet:

Option 1: Only Page2
Option 2: IllegalStateException at Runtime

Ans: 2

As per my understanding Option 1 should be correct as I have tested this case.

Code Snippet
===================
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");

 
pie sneak
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You make it hard to believe that you tested it when we see request.getWriter() in the code.
 
Amit Tayal
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I did not test the same code but tested the code similar to it.
Please read request.getWriter() as response.getWriter()

Now what will be the answer for the question??

Amit
 
Amit Tayal
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Hi to All
Someone please help me out of this question

Amit
 
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I think once you forward to a page with rd.forward(..) you shouldn't attempt to print like "out.println()" right after forwarding to a different page. It may generate exception...

Tazim
 
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The output of "Page 2" is correct.

After any Requestdispatch or sendRedirect or jsp:forward, all the contents of the forwarding page are lost. That's what is happening for the Page1. Also, after the RequestRedirect, the control from page1 is transferred to the called jsp, so "page3" never gets printed.
The reason why the IllegalStateexception won't occur, is that, the response is never "committed" or "sent to the browser". Hope it helps.

Rajat
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