Question

人，

人们，我需要帮助。尝试登录用户时出现错误。我使用Spring Security，但是需要手动实现该功能。显然，登录正常进行，但是当我重定向到主页时，出现错误，并且菜单未显示。如果刷新页面，则会显示菜单，但不会显示更多错误。

为什么会出现此错误？在方法中，我得到SecurityContextHolder.getContext并设置了authenticateManager认证的用户。这是不对的？我必须先做其他事情才能重定向到页面？

登录服务

@Service
public class LoginService {

    @Autowired
    protected AuthenticationManager authenticationManager;

    public void authenticateUserAndSession(String username, String password, HttpServletRequest request) {

        try {
            UsernamePasswordAuthenticationToken token = new UsernamePasswordAuthenticationToken(username, password);

            request.getSession();

            token.setDetails(new WebAuthenticationDetails(request));
            Authentication authenticationUser= authenticationManager.authenticate(token);

            SecurityContextHolder.getContext().setAuthentication(authenticationUser);

        } catch (Exception e) {
            System.err.println("Error: "+e);
        }   
    }

菜单index.xhtml

<sec:authorize ifAnyGranted="ROLE_USER, ROLE_ADMIM">
    <h:form>

        <nav class="user">
            <ul>

            </ul>
        </nav>
        <nav>
            <a href="#" id="menu-icon"></a>
            <ul>
                <li class="current-user"><p:graphicImage library="images" name="img-profile2.JPG" width="25" height="25" styleClass="profile-img-menu" />   </li>
                <li><h:commandLink value="#{userMB.currentUser.pflUser.name}" action="#{userMB.showCurrentUserData}" /></li>
                <li><h:commandLink value="Home" action="/index.xhtml?faces-redirect=true" /></li>
                <li><h:commandLink value="create" /></li>
                <li><h:commandLink value="Work" /></li>
                <li><h:commandLink value="Sair" action="#{loginMB.logout}" /></li>
            </ul>
        </nav>
    </h:form>
</sec:authorize>

SecurityWebConfig

@Configuration
@EnableWebSecurity
public class SecurityWebConfig extends WebSecurityConfigurerAdapter{

    private static final String USER_BY_MAIL = "SELECT  mail, password, enabled FROM user WHERE mail=?";

    private static final String PERMISSION_BY_USER = "SELECT u.mail, a.role FROM user_auth ua "
            + "JOIN user u ON u.id_user = ua.user_id "
            + "JOIN authority a ON a.id_auth = ua.auth_id "
            + "WHERE u.mail=?";

    @Autowired
    private DataSource dataSource;

    @Override
    public void configure (HttpSecurity http) throws Exception{
        http
        .authorizeRequests()
            .antMatchers("/resources/**/*","/css/**","/javax.faces.resource/**","/template.jsf", "/account/signup.jsf","/account/signup**").permitAll()
            .anyRequest().authenticated()
        .and()
            .formLogin()
            .loginPage("/login.jsf")
            .loginProcessingUrl("/login")
            .permitAll()
            .defaultSuccessUrl("/index.jsf",true)
            .failureUrl("/login.jsf?error=true")
        .and()
        .logout().logoutSuccessUrl("/login.jsf")
        .and()
            .csrf().disable();

    }

    @Override
    public void configure(AuthenticationManagerBuilder builder) throws Exception{
        builder
            .jdbcAuthentication()
            .dataSource(dataSource)
            .passwordEncoder(new BCryptPasswordEncoder())
            .usersByUsernameQuery(USER_BY_MAIL)
            .authoritiesByUsernameQuery(PERMISSION_BY_USER)
            .rolePrefix("ROLE_");

    }

谢谢！

错误：org.springframework.security.authentication.BadCredentialsException：凭据错误

0 个答案: