带有forward的Spring Security SuccessHandler错误

Question

当我使用自定义SuccessHandler尝试Http POST或Http PUT时，我不断收到Http 400错误。具体地说，在我转发请求之后，会出现错误。我将请求转发如下：

public void onAuthenticationSuccess(HttpServletRequest request,
        HttpServletResponse response, Authentication authentication)
        throws IOException, ServletException {
        String context = request.getContextPath();
        String fullURL = request.getRequestURI();
        String url = fullURL.substring(fullURL.indexOf(context)
                + context.length());
        request.getRequestDispatcher(url).include(request, response);
        request.getRequestDispatcher(url).forward(request, response);
}

我的任何其他不包括RequestBody的调用都不会产生Http400错误。可能是什么问题？

Answer 1

由于请求调度器在同一服务器上为servlet工作，请尝试request.getRequestDispatcher（conetext）。。。其中"context"是您在服务器上前进的相对目标，即/webapp/login参见

• http://docs.oracle.com/javaee/7/api/javax/servlet/ServletRequest.html#getRequestDispatcher%28java.lang.String%29
• http://www.jguru.com/faq/view.jsp?EID=206736

Answer 2

显然我不理解RequestDispatcher的include和forward方法。通过移除request.getRequestDispatcher(url).forward(request, response);，我让它开始工作。有关原因的解释，请访问：http://www.theserverside.com/tip/Include-vs-Forward-of-the-Servlet-RequestDispatcher.