Spring Security自定义过滤器

Question

我想自定义Spring security 3.0.5并将登录URL更改为/ login而不是/ j_spring_security_check.

我需要做的是允许登录到"/"目录并保护"/admin/report.html"页面.

首先,我使用教程和Spring Security源代码创建自己的过滤器:

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);
    }
}

之后我在xml中进行了以下更改

 <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>

然后我用我的代码编写JSP.

<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>

当我尝试导航到/admin/report.html时,我被重定向到登录页面.但在提交凭证后我得到了:

HTTP Status 404 - /SpringMVC/login/

type Status report

message /SpringMVC/login/

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

看起来我的配置有问题,但我无法弄清楚导致这种情况的原因.你能帮我吗？

Answer 1

我迟到了大约12个月,但是为了自定义Spring Security表单登录的登录URL,您不需要创建自己的过滤器.form-login标签的一个属性允许您设置自定义URL.实际上,您还可以使用form-login标记的属性更改默认的j_username和j_password字段名称.这是一个例子:

<form-login login-page="/login" login-processing-url="/login.do" default-target-url="/" always-use-default-target="true" authentication-failure-url="/login?error=1" username-parameter="username" password-parameter="password"/>

Answer 2

我认为 @Ischin 对表单操作 url 的疑问是正确的。尝试输入完整路径，看看是否有效。如果确实如此，您可以从那里找出不匹配的内容。

我唯一能想到要检查的是 web.xml 中的过滤器映射。由于您正在点击登录页面，因此您已经进行了此设置，但我会检查您是否不仅拦截具有特定扩展名的网址等。

另外，仅供参考，如果您希望请求（一旦登录表单对用户进行身份验证）转到受保护的资源（在本例中为/admin/report.html），那么您应该删除表单：login always-use-默认目标=“真”。将此标志设置为 true 将导致请求始终转到默认目标 url，这通常不是您想要的。来自春季安全文档：

映射到 UsernamePasswordAuthenticationFilter 的 defaultTargetUrl 属性。如果未设置，则默认值为“/”（应用程序根目录）。如果用户在尝试访问安全资源时没有被要求登录，则登录后将被带到此 URL，此时他们将被带到最初请求的 URL。