2015-07-10 83 views
0

我正在使用Spring-Security进行身份验证/授权的Java应用程序,该应用程序连接到Spring-MVC服务器。登录部分工作正常,并且我在Java应用程序中返回了JSESSIONID,但是当我向安全资源发出请求时,它失败,Spring-Security无法找到任何已登录的用户。我在这里做错了什么?Spring-Security:不确认在RestTemplate中设置的Cookie

安全的applicationContext.xml:

<security:http pattern="/resources/**" security="none"/> 

    <security:http create-session="ifRequired" use-expressions="true" auto-config="false" disable-url-rewriting="true"> 
     <security:form-login login-page="/login" login-processing-url="/j_spring_security_check" 
          default-target-url="/dashboard" always-use-default-target="false" 
          authentication-failure-url="/denied"/> 
     <security:remember-me key="_spring_security_remember_me" user-service-ref="userDetailsService" 
           token-validity-seconds="1209600" data-source-ref="dataSource"/> 
     <security:logout delete-cookies="JSESSIONID" invalidate-session="true" logout-url="/j_spring_security_logout"/> 
     <!--<security:intercept-url pattern="/**" requires-channel="https"/>--> 
     <security:port-mappings> 
      <security:port-mapping http="8080" https="8443"/> 
     </security:port-mappings> 
     <security:logout logout-url="/logout" logout-success-url="/" success-handler-ref="myLogoutHandler"/> 


     <security:session-management session-fixation-protection="migrateSession"> 
      <security:concurrency-control session-registry-ref="sessionRegistry" max-sessions="5" expired-url="/login"/> 
     </security:session-management> 

    </security:http> 

    <security:authentication-manager alias="authenticationManager"> 
     <security:authentication-provider ref="restaurantauthenticationprovider"/> 
     <security:authentication-provider ref="userauthenticationprovider"/> 
    </security:authentication-manager> 

    <beans:bean id="encoder" 
       class="org.springframework.security.crypto.bcrypt.BCryptPasswordEncoder"> 
     <beans:constructor-arg name="strength" value="11"/> 
    </beans:bean> 

    <beans:bean id="restaurantauthenticationprovider" 
       class="org.springframework.security.authentication.dao.DaoAuthenticationProvider"> 
     <beans:property name="userDetailsService" ref="LoginServiceImpl"/> 
     <beans:property name="passwordEncoder" ref="encoder"/> 
    </beans:bean> 

    <beans:bean id="userauthenticationprovider" 
       class="org.springframework.security.authentication.dao.DaoAuthenticationProvider"> 
     <beans:property name="userDetailsService" ref="UserLoginServiceImpl"/> 
     <beans:property name="passwordEncoder" ref="encoder"/> 
    </beans:bean> 

正如我有2代表从登录到检查,我有2个DAOAuthenticationProviders。

UserLoginServiceImpl:

@Transactional 
@Service("loginuserDetailsService") 
public class UserLoginServiceImpl implements UserDetailsService { 


    @Autowired 
    private PersonDAO personDAO; 
    @Autowired 
    private UserAssembler userAssembler; 

    @Override 
    public UserDetails loadUserByUsername(String username) throws UsernameNotFoundException,DataAccessException { 
     System.out.println("Username is "+username); 
     Person person = this.personDAO.findPersonByUserName(username.toLowerCase()); 
     if(person == null) { throw new UsernameNotFoundException("Wrong username or password");} 
     return userAssembler.buildUserFromUserEntity(person); 
    } 
} 

汇编:

@Service("userassembler") 
@Transactional 
public class UserAssembler { 

    @Transactional 
    User buildUserFromUserEntity(Person userEntity){ 
     System.out.println("We are in Userassembler"+userEntity.getEmail()); 
     String username = userEntity.getUsername().toLowerCase(); 
     String password = userEntity.getPassword(); 

     boolean enabled = userEntity.isEnabled(); 
     boolean accountNonExpired = userEntity.isAccountNonExpired(); 
     boolean credentialsNonExpired = userEntity.isCredentialsNonExpired(); 
     boolean accountNonLocked = userEntity.isAccountNonLocked(); 

     Collection<GrantedAuthority> authorities = new ArrayList<>(); 
     authorities.add(new SimpleGrantedAuthority("ROLE_USER")); 

     return new User(username,password,enabled,accountNonExpired,credentialsNonExpired,accountNonLocked,authorities); 
    } 
} 

上面是配置,现在我要把这是失败的其余代码:

Thread thread = new Thread(new Runnable() { 
      @Override 
      public void run() { 
       Log.d("Username is ", username); 
       String jsessionid = rest.execute("http://192.168.178.60:8080/j_spring_security_check", HttpMethod.POST, 
         new RequestCallback() { 
          @Override 
          public void doWithRequest(ClientHttpRequest request) throws IOException { 
           request.getBody().write(("j_username=" + username + "&j_password=" + password).getBytes()); 
          } 
         }, new ResponseExtractor<String>() { 
          @Override 
          public String extractData(ClientHttpResponse response) throws IOException { 
           List<String> cookies = response.getHeaders().get("Cookie"); 
           if (cookies == null) { 
            cookies = response.getHeaders().get("Set-Cookie"); 
           } 
           String cookie = cookies.get(cookies.size() - 1); 
           System.out.println("Cookie is " + cookie); 
// The method below gets me which user is logged in, and I always get null for Controller method. 
           reply = rest.getForObject(
             "http://192.168.178.60:8080/dashboard", String.class); 

           int start = cookie.indexOf('='); 
           int end = cookie.indexOf(';'); 
           return cookie.substring(start + 1, end); 
          } 
         }); 

      } 
     }); 
     thread.start(); 

更新 最后,这个代码ked:

//我从服务器获取cookie,我为每个请求手动设置cookie,cookie是一个静态易失性字符串。

HttpHeaders requestHeaders = new HttpHeaders(); 
requestHeaders.add("Cookie", "JSESSIONID=" + StaticRestTemplate.jsessionid); 
HttpEntity requestEntity = new HttpEntity(null, requestHeaders); 

ResponseEntity rssResponse = rest.exchange(
            "http://192.168.178.60:8080/dashboard", 
            HttpMethod.GET, 
            requestEntity, 
            String.class); 

String abc = (String) rssResponse.getBody(); 
+0

你的人服务是什么样的?如果将方法更改为: 字符串checkWichUserLoggedIn(Principal p){ return p == null? null:p.getName(); } –

+0

@RobWinch:我得到它为空...因为找不到用户匿名用户。我已经更新了我的主帖中的代码,顺便说一句,当我打印认证详细信息时,会话Id为空......我明白,RestTemplate在您进行身份验证时在内部设置了Cookie,但似乎并不是这样。 –

+0

你真的得到了JSESSIONID吗?什么是HTTP状态码?您使用的是哪种版本的Spring Security? –

回答

1

Spring的RestTemplate默认不会跟踪cookie。这可以确保您不会意外地从一个用户代表另一个用户传递cookie(即JSESSIONID)(即,考虑在许多用户利用相同RestTemplate的服务器上使用RestTemplate)。

如果要做到这一点,你可以使用这样的配置是:

RestTemplate rest = new RestTemplate(); 

// initialize the RequestFactory to allow cookies 
HttpClient httpClient = HttpClientBuilder.create().build(); 
ClientHttpRequestFactory factory = new HttpComponentsClientHttpRequestFactory(httpClient); 
rest.setRequestFactory(factory); 

MultiValueMap<String, String> map = new LinkedMultiValueMap<String, String>(); 
map.add("username", "user"); 
map.add("password", "password"); 

String result = rest.postForObject("http://localhost:8080/login", map, String.class); 

String hello = rest.postForObject("http://localhost:8080/", map, String.class); 

assertThat(hello).isEqualTo("Hello"); 

要使用此代码,你需要确保你的HttpClient在classpath。例如,下面可能是在你的pom.xml如果你正在使用Maven:

<dependency> 
    <groupId>org.apache.httpcomponents</groupId> 
    <artifactId>httpclient</artifactId> 
    <version>4.5</version> 
</dependency> 

显然,你需要确保你的HttpClient包含的版本,为您的依赖工作。

+0

httpclient已被弃用,你是否有点更新?我发现http://stackoverflow.com/questions/29058727/i-need-an-option-to-httpclient-in-android-to-send-data-to-php-as-it-is-deprecate,但我找不到如何为Rest设置数据。 –

+0

如果你在最新的httpclient(4.5),它不应该被弃用 –

+0

但httpclient有一个httpcore的要求,它不是4.5 ...另外,当我使用这些库时,出现其他错误,所以我使用了android-httpclient。然后,我也遇到了一些问题,最后,我用了一些我在文章底部提到的内容。如果您知道如何优化它,请告诉我。 –