Spring Security benutzerdefinierte filter

Möchte ich anpassen, Spring security 3.0.5 und ändern login-URL /login anstelle von /j_spring_security_check.

Was ich tun müssen, ist, um zu ermöglichen, Anmeldung im "/" Verzeichnis und sichern "/admin/report.html" Seite.

Zunächst erstelle ich meinen eigenen filter mit tutorial und Spring Security source code:

public class MyFilter extends AbstractAuthenticationProcessingFilter {
    private static final String DEFAULT_FILTER_PROCESSES_URL = "/login";
    private static final String POST = "POST";
    public static final String SPRING_SECURITY_FORM_USERNAME_KEY = "j_username";
    public static final String SPRING_SECURITY_FORM_PASSWORD_KEY = "j_password";
    public static final String SPRING_SECURITY_LAST_USERNAME_KEY = "SPRING_SECURITY_LAST_USERNAME";

    private String usernameParameter = SPRING_SECURITY_FORM_USERNAME_KEY;
    private String passwordParameter = SPRING_SECURITY_FORM_PASSWORD_KEY;

    protected MyFilter() {
        super(DEFAULT_FILTER_PROCESSES_URL);
    }

    @Override
    public Authentication attemptAuthentication(HttpServletRequest request,
                                                HttpServletResponse response) 
                          throws AuthenticationException, IOException, ServletException {
        String username = obtainUsername(request);
        String password = obtainPassword(request);

        if (username == null) {
            username = "";
        }

        if (password == null) {
            password = "";
        }

        username = username.trim();
        UsernamePasswordAuthenticationToken authRequest = new UsernamePasswordAuthenticationToken(username, password);
        HttpSession session = request.getSession(false);
        if (session != null || getAllowSessionCreation()) {
            request.getSession().setAttribute(SPRING_SECURITY_LAST_USERNAME_KEY, TextEscapeUtils.escapeEntities(username));
        }
        setDetails(request, authRequest);

        return this.getAuthenticationManager().authenticate(authRequest);
    }

    protected void setDetails(HttpServletRequest request, UsernamePasswordAuthenticationToken authRequest) {
        authRequest.setDetails(authenticationDetailsSource.buildDetails(request));
    }

    @Override
    public void doFilter(ServletRequest req, ServletResponse res,
                         FilterChain chain) throws IOException, ServletException {
        final HttpServletRequest request = (HttpServletRequest) req;
        final HttpServletResponse response = (HttpServletResponse) res;
        if (request.getMethod().equals(POST)) {
            //If the incoming request is a POST, then we send it up
            //to the AbstractAuthenticationProcessingFilter.
            super.doFilter(request, response, chain);
        } else {
            //If it's a GET, we ignore this request and send it
            //to the next filter in the chain.  In this case, that
            //pretty much means the request will hit the /login
            //controller which will process the request to show the
            //login page.
            chain.doFilter(request, response);
        }
    }

    protected String obtainUsername(HttpServletRequest request) {
        return request.getParameter(usernameParameter);
    }

    protected String obtainPassword(HttpServletRequest request) {
        return request.getParameter(passwordParameter);
    }
}

nachdem Es ich mache die folgenden änderungen in xml -

 <security:http auto-config="true">
        <!--<session-management session-fixation-protection="none"/>-->
        <security:custom-filter ref="myFilter" before="FORM_LOGIN_FILTER"/>
        <security:intercept-url pattern="/admin/login.jsp*" filters="none"/>
        <security:intercept-url pattern="/admin/report.html" access="ROLE_ADMIN"/>
        <security:form-login login-page="/admin/login.jsp" login-processing-url="/login" always-use-default-target="true"/>
        <security:logout logout-url="/logout" logout-success-url="/login.jsp" invalidate-session="true"/>
    </security:http>   
<security:authentication-manager alias="authenticationManager">
  <security:authentication-provider>
    <security:password-encoder hash="md5" />
    <security:user-service>
    <!-- peter/opal -->
      <security:user name="peter" password="22b5c9accc6e1ba628cedc63a72d57f8" authorities="ROLE_ADMIN" />
     </security:user-service>
  </security:authentication-provider>
</security:authentication-manager>
<bean id="myFilter" class="com.vanilla.springMVC.controllers.MyFilter">
<property name="authenticationManager" ref="authenticationManager"/>
</bean>

und dann habe ich JSP mit meinem code.

<form action="../login" method="post">
    <label for="j_username">Username</label>
    <input type="text" name="j_username" id="j_username" />
    <br/>
    <label for="j_password">Password</label>
    <input type="password" name="j_password" id="j_password"/>
    <br/>
    <input type='checkbox' name='_spring_security_remember_me'/> Remember me on this computer.
    <br/>
    <input type="submit" value="Login"/>
</form>

wenn Sie versuchen zu navigieren Sie zu " /admin/report.html ich bin Weiterleitung auf die login-Seite.
aber nach dem Absenden der Anmeldeinformationen ich bin immer:

HTTP Status 404 - /SpringMVC/login/

type Status report

message /SpringMVC/login/

description The requested resource (/SpringMVC/login/) is not available.

Sieht es aus wie ich haben problem in der Konfiguration, aber ich kann nicht herausfinden was dies verursacht.
Können Sie mir helfen?