【发布时间】:2016-03-09 11:52:08
【问题描述】:
我想在 Spring MVC 中重写 servlet 控制流,这是我的 doGet 进入 Servlet
protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
String action = request.getParameter("action");
String path = request.getParameter("path");
if (path != null && path.equals("register")) {
RequestDispatcher view = request.getRequestDispatcher("/WEB-INF/views/system/registeruser.jsp");
view.forward(request, response);
} else if (path != null && path.equals("usermang")) {
RequestDispatcher view = request.getRequestDispatcher("/WEB-INF/views/system/manageuser.jsp");
view.forward(request, response);
}
else {
PrintWriter out = response.getWriter();
out.print("Served at: " + request.getContextPath());
RequestDispatcher view = request.getRequestDispatcher("/WEB-INF/views/system/index.jsp");
view.forward(request, response);
}
}
我想使用模型和视图将上面的 doGET 转换为 Spring RequestMapping 示例
@RequestMapping(value ="/grcon" ,method = RequestMethod.GET)
public ModelAndView getGrcon()
ModelAndView modegeron = new ModelAndView("index");
if (path != null && path.equals("register")) {
view = request.getRequestDispatcher("/WEB-INF/views/system/registeruser.jsp");
return modegeron;
}
}
【问题讨论】:
-
很难理解你想要做什么,因为标题和描述都没有提出任何问题。在 Spring MVC 中有一个现成的 servlet,称为 DispatcherServlet,您应该使用它。逻辑由控制器处理。
-
请浏览Spring MVC文档。一些有用的博客链接mkyong.com/tutorials/spring-mvc-tutorials。你应该可以得到这个
标签: java spring spring-mvc model-view-controller