如何使用WSS4J拦截器在Web服务方法中获取经过身份验证的用户

Question

我在Spring中托管了Apache CXF服务.我正在使用WSS4J拦截器来验证用户名/密码安全性以访问服务器.验证工作正常,如果我从SoapUI发送错误的凭据我不能按预期使用该服务.如果我发送正确的凭据,服务没有问题.这是我在spring上下文文件中的配置.

<bean id="myPasswordCallback" class="cu.datys.sias.custom.ServerPasswordCallback"/>

<jaxws:endpoint id="siasEndpoint"
                implementor="#siasImpl"
                address="/sias">
    <jaxws:features>
        <!-- Soporte WS-Addressing -->
        <!--<wsa:addressing xmlns:wsa="http://cxf.apache.org/ws/addressing" addressingRequired="true" usingAddressingAdvisory="true" allowDuplicates="true"/>-->
    </jaxws:features>
    <jaxws:inInterceptors>
        <bean class="org.apache.cxf.ws.security.wss4j.WSS4JInInterceptor">
            <constructor-arg>
                <map>
                    <entry key="action" value="UsernameToken" />
                    <entry key="passwordType" value="PasswordText" />
                    <entry key="passwordCallbackRef"
                           value-ref="myPasswordCallback" />
                </map>
            </constructor-arg>
        </bean>
    </jaxws:inInterceptors>
</jaxws:endpoint>

现在我需要能够在我的服务方法中访问我的经过身份验证的用户,如下所示:

@WebResult(name = "UpdatePatternResponse", targetNamespace = "http://test.com/schemas/xsd/myservice/", partName = "UpdatePatternResponse")
@WebMethod(operationName = "UpdatePattern", action = "UpdatePattern")
@Generated(value = "org.apache.cxf.tools.wsdlto.WSDLToJava", date = "2015-02-19T12:49:59.491-05:00")
public test.com.schemas.xsd.myservice.UpdatePatternResponse updatePattern(
    @WebParam(partName = "UpdatePatternRequest", name = "UpdatePatternRequest", targetNamespace = "http://test.com/schemas/xsd/myservice/")
    test.com.schemas.xsd.myservice.UpdatePatternRequest updatePatternRequest
) throws SIASFaultMessage{
    .
    .
    User myAuthenticatedUser = //HOW TO GET THE USER???
    .....
    .
    .
    .
}

如何在Apache CXF服务方法中获取经过身份验证的用户？

Answer 1

我发现这是与这样一个重要主题有关的唯一问题,所以我想分享我为CXF 3.1.x发现的内容.

基本思路与@ alfredo-a相同.这是代码:

Message message=PhaseInterceptorChain.getCurrentMessage();
SecurityContext context=message.get(SecurityContext.class);
String userName=context.getUserPrincipal().getName();

我希望这对某人有帮助.

干杯!