如何在spring mvc中处理浏览器后退按钮 [英] How handle browser back button in spring mvc

问题描述

当用户登录时，会话信息被存储。
当用户登录
时，会话信息被删除。但是当我点击浏览器的后退按钮时，会显示用户信息。由于会话已经消失，但我们无法确定
是否执行了用户登录操作。我如何解决这个
问题？

When a user is logged on session information is stored. And session information is erased when the user is logged out . But when I hit the browser 's back button user information is displayed. Since session is gone but we can not be sure the user login operation is carried out. How do I resolve this issue ?

  ----------------------------log out -------------------------------

   @RequestMapping(value="logout.htm",method = RequestMethod.GET)
   public void logOut(HttpSession session,HttpServletResponse                 
   response,HttpServletRequest request) throws IOException{
    final String refererUrl = request.getHeader("Referer");
    response.setHeader(refererUrl, "no-cache");
    response.setHeader("Cache-Control", "no-cache");
    response.setDateHeader("Expires", 0);
    session.removeAttribute("user");
    session.invalidate();
    response.sendRedirect("index.htm");
   }
    ---------------------------------- login ---------------
  @RequestMapping(value="/userLogin",method=RequestMethod.POST)

  public @ResponseBody JsonResponse
 login(@ModelAttribute(value="user") User user, BindingResult     result,HttpServletRequest request,HttpSession session,ModelMap model) throws    UnsupportedEncodingException{

    JsonResponse res = new JsonResponse();

    if(!result.hasErrors()&& userService.findUser(user, request)){
        res.setStatus("SUCCESS");
        session.setAttribute("user",
      new String(user.getUsername().getBytes("iso-  8859-1"), "UTF-8"));
      }
         else{
        res.setStatus("FAIL");
        result.rejectValue("username","1");
        res.setResult(result.getAllErrors());
       }
      return res;
   }
   --------------------------profile --------------------------------------

    @RequestMapping(value="myProfile.htm",method = RequestMethod.GET)
   public String showmyProfile(@ModelAttribute(value="addUser") User user,Model          model,HttpServletRequest request,
        HttpServletResponse response,
         HttpSession session) throws IOException{

        if(session.getAttribute("user")== null){
        response.sendRedirect("index");
    }

推荐答案

我使用此方法。
首先创建一个实现Filter并覆盖doFilter（）方法的类。 doFilter（）的
代码是：

i use this method. first create one class that implements Filter and override doFilter() method. code of doFilter() is:

 @Override
 public void doFilter(ServletRequest req, ServletResponse res, FilterChain chain) throws IOException, ServletException {
HttpServletResponse hsr = (HttpServletResponse) res;
hsr.setHeader("Cache-Control", "no-cache, no-store, must-revalidate"); // HTTP 1.1.
hsr.setHeader("Pragma", "no-cache"); // HTTP 1.0.
hsr.setDateHeader("Expires", 0); // Proxies.
chain.doFilter(req, res);
 }

在web.xml中使用过滤器后。这个过滤器就是这个。

after use filter in web.xml. this filter is this.

  <filter>
    <filter-name>noCacheFilter</filter-name>
    <filter-class>com.example.NoCacheFilter</filter-class>
 </filter>
 <filter-mapping>
  <filter-name>noCacheFilter</filter-name>
  <url-pattern>/secured/*.jsp</url-pattern>// urls that not cached 
 </filter-mapping>

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