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,我得到的异常为: - java.lang.IllegalStateException
,在提交响应后无法创建会话。
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login.jsp
时,我得到了这个例外。请帮助我
自从过去2天以来一直困在我的项目上。
答案 0 :(得分:3)
执行sendRedirect
后不要继续使用过滤器。 sendRedirect
会将标头提交给客户端,之后你不应该写任何东西。
将login.jsp保存到您允许的网址。
if (!allowedRequest) {
HttpSession session = request.getSession(false);
if (null == session || ) {
response.sendRedirect("Login.jsp");
return;
}
}
<强>参考文献:强>