2

我正在尝试进行登录操作，但是当我使用过滤器检查会话是否有效时，我得到了异常：- java.lang.IllegalStateException，在响应提交后无法创建会话。

Myservlet 过滤器代码是：-

 public class SessionFilter implements javax.servlet.Filter 
    {

private ArrayList<String> urlList;

     public void destroy() {

}

public void doFilter(ServletRequest req, ServletResponse res, FilterChain chain) 
    throws IOException, ServletException {

HttpServletRequest request = (HttpServletRequest) req;
    HttpServletResponse response = (HttpServletResponse) res;
    String url = request.getServletPath();
    boolean allowedRequest = false;


    if(urlList.contains(url)) {
        allowedRequest = true;

    }

    if (!allowedRequest) {
        HttpSession session = request.getSession(false);
        if (null == session) {
            response.sendRedirect("Login.jsp");
        }
    }

  chain.doFilter(req, res);



}


public void init(FilterConfig fConfig) throws ServletException {

    String urls = fConfig.getInitParameter("avoid-urls");

    //StringTokenizer token = new StringTokenizer(urls, ",");

    StringTokenizer token = new StringTokenizer(urls);

    urlList = new ArrayList<String>();

    urlList.add(token.nextToken());

    /*while (token.hasMoreTokens()) {
        urlList.add(token.nextToken());

    }*/
}

}



  Filter mapping in web.xml is :-

<filter>
   <filter-name>SessionFilter</filter-name>
   <filter-class>Filter.SessionFilter</filter-class>

    <init-param>
      <param-name>avoid-urls</param-name>
      <param-value>Login.jsp,</param-value>
    </init-param>
</filter>

<filter-mapping>

    <filter-name>SessionFilter</filter-name>
    <url-pattern>/*</url-pattern>

</filter-mapping>

现在，当我请求时，login.jsp我得到了这个异常。请帮助我自从过去 2 天以来一直卡在我的项目上。

4

1 回答 1

3

做完后不要继续过滤sendRedirect。sendRedirect会将标头提交给客户端，之后您不应该写任何东西。

并将 login.jsp 设置为您允许的 url。

  if (!allowedRequest) {
    HttpSession session = request.getSession(false);
    if (null == session || ) {
        response.sendRedirect("Login.jsp");
        return;
    }
}

参考：

Servlet“响应已提交”的原因

于 2013-03-25T10:37:10.733 回答