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Apache CXF-无法满足任何其他政策选择

问题内容：

我正在尝试创建第三方WS的客户端。我的应用程序在JBoss AS 6（及其Apache CXF
2.3.1堆栈）上运行。我通过wsconsume（wsdl2java）生成了客户端代码。当我尝试连接到WS时，出现异常：

No assertion builder for type http://schemas.microsoft.com/ws/06/2004/policy/http}BasicAuthentication registered. 
Exception in thread "main" org.apache.cxf.ws.policy.PolicyException: None of the policy alternatives can be satisfied.

WSDL的Auth部分如下所示：

<wsp:Policy wsu:Id="abc_ssl_policy">
    <wsp:ExactlyOne>
        <wsp:All>
            <http:BasicAuthentication
                xmlns:http="http://schemas.microsoft.com/ws/06/2004/policy/http" />
            <sp:TransportBinding
                xmlns:sp="http://schemas.xmlsoap.org/ws/2005/07/securitypolicy">
                <wsp:Policy>
                    <sp:TransportToken>
                        <wsp:Policy>
                            <sp:HttpsToken RequireClientCertificate="false" />
                        </wsp:Policy>
                    </sp:TransportToken>
                    <sp:AlgorithmSuite>
                        <wsp:Policy>
                            <sp:Basic256 />
                        </wsp:Policy>
                    </sp:AlgorithmSuite>
                    <sp:Layout>
                        <wsp:Policy>
                            <sp:Strict />
                        </wsp:Policy>
                    </sp:Layout>
                </wsp:Policy>
            </sp:TransportBinding>
        </wsp:All>
    </wsp:ExactlyOne>
</wsp:Policy>

客户代码：

@WebServiceClient(name = "Abc", 
              wsdlLocation = "https://hiddendomain.com/abc/abc.svc?wsdl",
              targetNamespace = "http://tempuri.org/")                  
public class Abc extends Service {

public final static URL WSDL_LOCATION;

public final static QName SERVICE = new QName("http://tempuri.org/", "Abc");
public final static QName AbcSsl = new QName("http://tempuri.org/", "abc_ssl");
static {

    Authenticator.setDefault(new Authenticator() {
        @Override
        protected PasswordAuthentication getPasswordAuthentication() {
            return new PasswordAuthentication("user", "pas".toCharArray());
        }

    });

    URL url = null;
    try {
        url = new URL("https://hiddendomain.com/abc/abc.svc?wsdl");

    } catch (MalformedURLException e) {
        java.util.logging.Logger.getLogger(DistrInfo.class.getName())
            .log(java.util.logging.Level.INFO, 
                 "Can not initialize the default wsdl from {0}", "...");
    }
    WSDL_LOCATION = url;
}

我尝试获取导管时抛出异常：

    Client client = ClientProxy.getClient(port);
    HTTPConduit con = (HTTPConduit) client.getConduit(); <- exception

我怀疑这是由于非标准的MS政策所致，我需要适当的Intercerptor来处理此政策，但是有人可以向我展示一种方法吗？

我什至没有，我应该将HTTPS凭据放在auth（我无法获取管道）中


问题答案：

当我使用以下代码时，问题消失了：

import org.apache.cxf.endpoint.Client;
import org.apache.cxf.frontend.ClientProxy;
import org.apache.cxf.jaxws.JaxWsProxyFactoryBean;
import org.apache.cxf.transport.http.HTTPConduit;

...

JaxWsProxyFactoryBean factory = new JaxWsProxyFactoryBean();

//factory.getInInterceptors().add(new LoggingInInterceptor());
//factory.getOutInterceptors().add(new LoggingOutInterceptor());

factory.setServiceClass(IAbc.class);
factory.setAddress("https://hiddendomain.com/abc/abc.svc/soap"); <- must be /soap there, otherwise 404

IAbc info = (IAbc) factory.create();

Client client = ClientProxy.getClient(info);
HTTPConduit http = (HTTPConduit) client.getConduit();

http.getAuthorization().setUserName("user");
http.getAuthorization().setPassword("pass");

String abc = info.abc();