Question

我正在使用Spring，Spring Security，BlazeDS，Flex和spring-flex。

我知道我可以调用channelSet.login()和channelSet.logout()挂钩到Spring Security进行身份验证。 channelSet.authenticated显然只知道当前的Flex会话，因为它始终以 false 开头，直到您致电channelSet.login()。

我想做什么：

从Flex检查以了解用户是否已在会话中。
如果是这样，我想要他们的用户名和角色。

更新
我只是想在下面的brd6644的答案中添加我使用的解决方案的详细信息，这样对于那些查看此内容的人来说这可能会更容易。我使用this StackOverflow回答来使SecurityContext注入。我不会在这个代码中重写该答案的代码，所以请查看SecurityContextFacade。

securityServiceImpl.java

public class SecurityServiceImpl implements SecurityService {
    private SecurityContextFacade securityContextFacade;

    @Secured({"ROLE_PEON"})
    public Map<String, Object> getUserDetails() {
        Map<String,Object> userSessionDetails = new HashMap<String, Object>();

        SecurityContext context = securityContextFacade.getContext();
        Authentication auth = context.getAuthentication();
        UserDetails userDetails = (UserDetails) auth.getPrincipal();

        ArrayList roles = new ArrayList();
        GrantedAuthority[] grantedRoles = userDetails.getAuthorities();
        for (int i = 0; i < grantedRoles.length; i++) {
            roles.add(grantedRoles[i].getAuthority());
        }

        userSessionDetails.put("username", userDetails.getUsername());
        userSessionDetails.put("roles", roles);
        return userSessionDetails;
    }
}

securityContext.xml

<security:http auto-config="true">
    <!-- Don't authenticate Flex app -->
    <security:intercept-url pattern="/flexAppDir/**" access="IS_AUTHENTICATED_ANONYMOUSLY" />
    <!-- Don't authenticate remote calls -->
    <security:intercept-url pattern="/messagebroker/amfsecure" access="IS_AUTHENTICATED_ANONYMOUSLY" />
</security:http>

<security:global-method-security secured-annotations="enabled" />

<bean id="securityService" class="ext.domain.project.service.SecurityServiceImpl">
    <property name="securityContextFacade" ref="securityContextFacade" />
</bean>
<bean id="securityContextFacade" class="ext.domain.spring.security.SecurityContextHolderFacade" />

flexContext.xml

<flex:message-broker>
    <flex:secured />
</flex:message-broker>

<flex:remoting-destination ref="securityService" />
<security:http auto-config="true" session-fixation-protection="none"/>

FlexSecurityTest.mxml

<mx:Application ... creationComplete="init()">

    <mx:Script><![CDATA[
        [Bindable]
        private var userDetails:UserDetails; // custom VO to hold user details

        private function init():void {
            security.getUserDetails();
        }

        private function showFault(e:FaultEvent):void {
            if (e.fault.faultCode == "Client.Authorization") {
                Alert.show("You need to log in.");
                // show the login form
            } else {
                // submit a ticket
            }
        }
        private function showResult(e:ResultEvent):void {
            userDetails = new UserDetails();
            userDetails.username = e.result.username;
            userDetails.roles = e.result.roles;
            // show user the application
        }
    ]]></mx:Script>

    <mx:RemoteObject id="security" destination="securityService">
        <mx:method name="getUserDetails" fault="showFault(event)" result="showResult(event)" />
    </mx:RemoteObject>

    ...
</mx:Application>

Answer 1

如果使用Spring Blazeds integration，则可以使用org.springframework.flex.security.AuthenticationResultUtils实现getUserDetails方法。

public Map<String, Object> getUserDetails() {  
 return AuthenticationResultUtils.getAuthenticationResult();
}

Answer 2

我会写一个安全的Spring服务方法，它返回当前用户的角色信息。让Flex应用程序在应用程序启动时调用它。如果由于安全性错误而收到FaultEvent，则提示用户进行身份验证并使用ChannelSet.login（）。

Answer 3

看到这个博客，我在Spring之前有一个flex模块，这很好地解决了这个问题。希望它能为您提供一些可能有帮助的宝石。

GridShore Blog

如何检查Spring Security以进行用户身份验证并从Flex获取角色？

3 个答案: