java - 添加带有 spring security 的自定义登录 Controller

Question

从 spring petclinic sample app 构建的应用程序使用自定义登录表单添加了 spring 安全性。

该应用没有 this tutorial 建议的 WebMvcConfiguration.java 类.相反，它在 mvc-core-config.xml 中有以下行:

<mvc:view-controller path="/login" view-name="login" />

我在 eclipse 中执行了 Ctrl-H 并在整个工作区中对术语 /login 进行了关键字搜索，但没有 Controller 可见。我还查看了上面教程链接中提到的 messages-jc 示例项目，但在那里也找不到“/login” Controller 。

我如何添加一个 Controller ，它将使用标准用户名和密码执行 spring 身份验证，但当“/login”url 处的登录表单时，这也将允许我随后向身份验证过程添加其他代码提交了吗？

是否像将以下内容添加到 SomeOtherController.java 一样简单:

@RequestMapping(value = "/login", method = RequestMethod.GET)
public String showLoginForm(Model model) {
        //what goes here?       
    return "public/loginform";
}

@RequestMapping(value = "/login", method = RequestMethod.POST)
public String processLoginForm(HttpSession session, @ModelAttribute("user") User user,
        BindingResult result, Model model, final RedirectAttributes redirectAttributes)
{
        //what goes here?
    return "secure/main";
}

Answer 1

在spring-security-core jar中，有一个接口(interface)UserDetailsS​​ervice，它有一个方法

UserDetails loadUserByUsername(String username) throws UsernameNotFoundException;

您可以实现此接口(interface)并创建您自己的代码，例如

@Service("userDetailsService")
public class UserDetailsServiceImpl implements UserDetailsService {

@Transactional(readOnly = true)
public UserDetails loadUserByUsername(String username) {
    User user = userService.findUserByUsername(username);
    if (user != null) {
        String password = user.getPassword();
        boolean enabled = user.getActive();
        boolean accountNonExpired = user.getActive();
        boolean credentialsNonExpired = user.getActive();
        boolean accountNonLocked = user.getActive();

        Collection<GrantedAuthority> authorities = new ArrayList<GrantedAuthority>();
        for (Role r : user.getRoles()) {
            authorities.add(new SimpleGrantedAuthority(r.getAuthority()));
        }
        org.springframework.security.core.userdetails.User securedUser = new org.springframework.security.core.userdetails.User(
                username, password, enabled, accountNonExpired,
                credentialsNonExpired, accountNonLocked, authorities);
        return securedUser;
    } else {
        throw new UsernameNotFoundException(
                "Unable to find user with username provided!!");
    }
}

然后使用 DaoAuthenticationProvider 创建一个对象

<bean id="daoAuthenticationProvider"
    class="org.springframework.security.authentication.dao.DaoAuthenticationProvider">
    <property name="userDetailsService" ref="userDetailsService"></property>
</bean>

最后，将这个 DaoAuthenticationProvider 提供给 ProviderManager

<bean class="org.springframework.security.authentication.ProviderManager">
    <constructor-arg>
        <list>
            <ref bean="daoAuthenticationProvider" />
        </list>
    </constructor-arg>
</bean>

<security:authentication-manager>
    <security:authentication-provider
        user-service-ref="userDetailsService">
        <security:password-encoder hash="plaintext"></security:password-encoder>
    </security:authentication-provider>
</security:authentication-manager>

添加 web.xml 细节

<listener>
    <listener-class>
        org.springframework.web.context.ContextLoaderListener
    </listener-class>
</listener>

<context-param>
    <param-name>contextConfigLocation</param-name>
    <param-value>classpath:spring-config/spring-*.xml</param-value>
</context-param>


<filter>
    <filter-name>springSecurityFilterChain</filter-name>
    <filter-class>org.springframework.web.filter.DelegatingFilterProxy</filter-class>
</filter>

<filter-mapping>
    <filter-name>springSecurityFilterChain</filter-name>
    <url-pattern>/*</url-pattern>
 </filter-mapping>