Question

我使用doGet方法获取参数，然后对其进行处理。发送答复后，它包含状态数据。同样，当第二个客户端连接时，答案包含来自第一个客户端的数据。我要在第二个客户端连接后发送新数据。重置缓冲区并将新数据发送到第二个客户端，或者在还原页面后发送新数据。

它不起作用。客户端再次检索页面，并向我发送旧数据和新数据。我只需要将新数据发送到客户端。

class HttpTest extends HttpServlet {

    private static final long serialVersionUID = 1L;



    /**
     * @see HttpServlet#HttpServlet()
     */
    public HttpTest() {
        super();
        // TODO Auto-generated constructor stub
    }

    @Override
    protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {



        PrintWriter out = response.getWriter();

        System.out.println(request.getQueryString());

        String[] parameters = request.getQueryString().split(";");
        String out_data="";


        /*
         * algorithm for generate out data
         */

        out.println(out_data);

        out.flush();

    }


    /**
     * @see HttpServlet#doPost(HttpServletRequest request, HttpServletResponse response)
     */
    @Override
    protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
        // TODO Auto-generated method stub
        doGet(request, response);

    }

}

Answer 1

尝试一下可能会帮助

private void doGet(HttpServletRequest request, HttpServletResponse response) throws IOException {
    try (PrintWriter writer = response.getWriter()) {
        System.out.println(request.getQueryString());
        String[] parms = request.getQueryString().split(";");
        String outData = "";
        /*
         * algorithm for generate out data
         */
        writer.println(outData);
    }
}

为新的获取数据重置响应后的HttpServletresponse

1 个答案: