如何解决提交响应后无法调用sendRedirect（）的问题？

Question

我正在尝试使用以下代码将值从servlet传递到jsp页面：

public void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {

    PrintWriter out = response.getWriter();
    try {
        System.out.println("try");

        String taskid=request.getParameter("id");   
        convty = new Connectivity();
        con = convty.setConnection();
        st=con.createStatement();

        query="select * from task_table";

        rset=convty.getResultSet(query, con);
        while(rset.next()) {
            System.out.println("while");
            lst.add(rset.getString("task_id"));
            lst.add(rset.getString("date"));
            lst.add(rset.getString("project_name"));    
        }
        rset.close();
        System.out.println("after while");

    } catch(Exception e) {

        RequestDispatcher dd= request.getRequestDispatcher("error.jsp");
        dd.forward(request, response);

    } finally {
        System.out.println("finally1");
        HttpSession sessionvar = request.getSession();
        sessionvar.setAttribute("TaskData", "lst");

        response.sendRedirect("usertaskpage.jsp");

        lst.clear();
        out.close();            
    }
}

当我运行页面时，我得到：

错误：

java.lang.IllegalStateException: Cannot call sendRedirect() after the response has been committed
    org.apache.catalina.connector.ResponseFacade.sendRedirect(ResponseFacade.java:483)
    src.Taskservlet.doPost(Taskservlet.java:108)
    javax.servlet.http.HttpServlet.service(HttpServlet.java:647)
    javax.servlet.http.HttpServlet.service(HttpServlet.java:728)
    org.apache.tomcat.websocket.server.WsFilter.doFilter(WsFilter.java:51)

我尝试使用：

RequestDispatcher dd=request.getRequestDispatcher("usertaskpage.jsp");
dd.forward(request, response);

但是我遇到了同样的错误。

如何解决这个错误？

Answer 1

您已经在catch代码块中转发了响应：

RequestDispatcher dd = request.getRequestDispatcher("error.jsp");

dd.forward(request, response);

因此，您不能再次调用：

response.sendRedirect("usertaskpage.jsp");

因为它已经转发（提交）。

因此，您可以做的是：保留一个字符串以分配需要转发响应的位置。

    String page = "";
    try {

    } catch (Exception e) {
      page = "error.jsp";
    } finally {
      page = "usertaskpage.jsp";
    }

RequestDispatcher dd=request.getRequestDispatcher(page);
dd.forward(request, response);