java - 我正在开发 servlet，只有一种 deGet 和 DoPost 方法。我在那里写了一种新方法。 。我如何直接访问该新方法？

Question

我已经在doGet下调用了这个方法。请帮助我摆脱困境。这是我自己的方法，我想调用它。

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

        HttpSession session = request.getSession();
        String[] checkedQues = request.getParameterValues("ttom");
        List<String> checkedQuesList = Arrays.asList(checkedQues);
        Map<String, String> preferences = new LinkedHashMap<String, String>();
        if (session.getAttribute("username") != null) {
            List<Question> questionsList = (List<Question>) session
                    .getAttribute("restaurantQuestionList");

            List<Question> questionsListTemp1 = new ArrayList<>();
            for (int i = 2; i < 4; i++) {
                questionsListTemp1.add(questionsList.get(i));
            }
            session.setAttribute("tomtomRestaurantQuestionList1",
                    questionsListTemp1);

            for (Question question : questionsList) {
                String questionId = String.valueOf(question.getId());
                if (checkedQuesList.contains(questionId)) {
                    String answerId = request.getParameter(questionId);
                    // PreferenceDAO.storePreferences(questionId, answerId,
                    // CATEGORY);
                    preferences.put(questionId, answerId);
                    System.out.println("queid : " + questionId + "answerid : "
                            + answerId);

                }
            }

            String username = (String) session.getAttribute("username");
            PreferencesProcessor.process(preferences, username);

            RequestDispatcher requestdp = request
                    .getRequestDispatcher("WEB-INF/jsp/table.jsp");
            requestdp.forward(request, response);
        } else {
            RequestDispatcher requestdp = request
                    .getRequestDispatcher("WEB-INF/jsp/login.jsp");
            requestdp.forward(request, response);
        }

    }

    protected void doPost(HttpServletRequest request,
            HttpServletResponse response) throws ServletException, IOException {
        // TODO Auto-generated method stub
    }
}

/**
 * @see HttpServlet  doPost(HttpServletRequest request, HttpServletResponse
 *      response)
 */

Answer 1

Servlet 将 HTTP 请求 header 映射到预定义的方法，例如 doGet()、doPost() 等。

https://tomcat.apache.org/tomcat-5.5-doc/servletapi/javax/servlet/http/HttpServlet.html

由于您的方法会修改数据，因此您应该使用 POST 调用它。

最简单的方法是将 doPost() 转发到此方法:

public void doPost(HttpServletRequest request, HttpServletResponse response) {
    doYourThingHere(request, response);
}

通常会发生的情况是，您将向 doPost 添加一些路由逻辑，如下所示:

public void doPost(...) {
   String action = request.getParameter("action");

   switch (action) {
      case "doSomething":
          doSomething(request, response);
          break;
      case "somethingElse":
          doSomethingElse(request, response);
          break;
      ...
    }
}