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>

然后我有 JSP 和我的代码。

<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>

当尝试导航到/admin/report.html 时，我被重定向到登录页面。 但在提交凭据后我得到:

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，但我无法弄清楚是什么原因造成的。 你能帮忙吗？

最佳答案

我在这方面晚了大约 12 个月，但为了自定义 Spring Security 表单登录的登录 URL，您不需要创建自己的过滤器。 form-login 标记的属性之一允许您设置自定义 URL。事实上，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"/>

关于java - Spring Security 自定义过滤器，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/5971453/