HTTP Status 500 - Cannot create a session after the response has been committed

Question

我想通过filter获取session的值来让用户不随意登入网页， 但是当我打开不存在的页面的时候， 就会出现这个错误，
下面是我的代码filter.java public void doFilter(ServletRequest req, ServletResponse res,
FilterChain chain) throws IOException, ServletException
{
// TODO Auto-generated method stub
System.out.println("before filter");

req.setCharacterEncoding("utf-8");

chain.doFilter(req, res);
 

req=(HttpServletRequest)req;
HttpServletRequest request= (HttpServletRequest)req;

HttpSession hs=request.getSession();

Object obj_username= hs.getAttribute("username");
System.out.println("username="+obj_username);
System.out.println("requestURI"+request.getRequestURI());
String reqURI=request.getRequestURI();
if(obj_username==null&&!reqURI.equals("/ajax1.3/filter/login.jsp")&&!reqURI.equals("/ajax1.3/login.do")){
//String username=(String)obj_username;
System.out.println("非法登入");
req.getRequestDispatcher("/filter/login.jsp");
 
}
 
 

//req.getRequestDispatcher("filter/filter1.jsp").forward(req, res);
}
 

Answer 1

<filter>
   <filter-name>CharEncodingFilter</filter-name>
   <filter-class>com.filter.CharEncodingFilter</filter-class>
  </filter>
  <filter-mapping>
   <filter-name>CharEncodingFilter</filter-name>
   <url-pattern>/*</url-pattern>
  </filter-mapping>zhege?

Answer 2

public void doFilter(ServletRequest req, ServletResponse res,
        FilterChain chain) throws IOException, ServletException
{
    // TODO Auto-generated method stub
    System.out.println("before filter");
     
    req.setCharacterEncoding("utf-8");
     
    chain.doFilter(req, res);
  
 
    req=(HttpServletRequest)req;
    HttpServletRequest request= (HttpServletRequest)req;
     
    HttpSession hs=request.getSession();
     
    Object obj_username= hs.getAttribute("username");
    System.out.println("username="+obj_username);
    System.out.println("requestURI"+request.getRequestURI());
    String reqURI=request.getRequestURI();
    if(obj_username==null&&!reqURI.equals("/ajax1.3/filter/login.jsp")&&!reqURI.equals("/ajax1.3/login.do")){
        //String username=(String)obj_username;
        System.out.println("非法登入");
        req.getRequestDispatcher("/filter/login.jsp");
        return;
          
    }
  
      
     
    //req.getRequestDispatcher("filter/filter1.jsp").forward(req, res);
}

Answer 3

能不能给我讲解一下啊，
为什么这里加个return就不报错了，

Answer 4

安全，你可以看下api，经验之谈。

Answer 5

你不加return   你把chain.doFilter(req, res); 去掉试试看报错么