Question

我在.NET项目中使用AXIS服务参考时遇到了一些问题。我今天简单地在VS2010中添加了引用，但是提供的Method都没有返回Type或Params。我只是看了生成的WSDL，它看起来像这样:(简短的这个Schema，也许缺少一些消息元素，但让我们关注参数的东西）

<?xml version="1.0" encoding="utf-8"?>
<wsdl:definitions xmlns:soapenc="http://schemas.xmlsoap.org/soap/encoding/" xmlns:intf="urn:api.broadmail.de/soap11/RpcSession" xmlns:apachesoap="http://xml.apache.org/xml-soap" xmlns:wsdlsoap="http://schemas.xmlsoap.org/wsdl/soap/" xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns:tns2="http://lang.java" xmlns:impl="urn:api.broadmail.de/soap11/RpcSession" targetNamespace="urn:api.broadmail.de/soap11/RpcSession" xmlns:wsdl="http://schemas.xmlsoap.org/wsdl/">
  <wsdl:types>
    <xsd:schema xmlns="http://www.w3.org/2001/XMLSchema" targetNamespace="urn:api.broadmail.de/soap11/RpcSession">
      <xsd:import namespace="http://lang.java" />
      <xsd:import namespace="http://schemas.xmlsoap.org/soap/encoding/" />
      <xsd:complexType name="WebserviceException">
        <xsd:sequence>
          <xsd:element name="cause" nillable="true" type="xsd:anyType" />
          <xsd:element name="message" nillable="true" type="xsd:string" />
        </xsd:sequence>
      </xsd:complexType>
    </xsd:schema>
  </wsdl:types>
  <wsdl:message name="loginRequest">
    <wsdl:part name="in0" type="xsd:long" />
    <wsdl:part name="in1" type="xsd:string" />
    <wsdl:part name="in2" type="xsd:string" />
  </wsdl:message>
  <wsdl:portType name="SessionWebservice">
    <wsdl:operation name="login" parameterOrder="in0 in1 in2">
      <wsdl:input name="loginRequest" message="impl:loginRequest" />
      <wsdl:output name="loginResponse" message="impl:loginResponse" />
      <wsdl:fault name="WebserviceException" message="impl:WebserviceException" />
    </wsdl:operation>
  </wsdl:portType>
  <wsdl:binding name="RpcSessionSoapBinding" type="impl:SessionWebservice">
    <wsdlsoap:binding transport="http://schemas.xmlsoap.org/soap/http" style="rpc" />
     <wsdl:operation name="login">
      <wsdlsoap:operation soapAction="" />
      <wsdl:input name="loginRequest">
        <wsdlsoap:body use="encoded" namespace="urn:api.broadmail.de/soap11/RpcSession" encodingStyle="http://schemas.xmlsoap.org/soap/encoding/" />
      </wsdl:input>
      <wsdl:output name="loginResponse">
        <wsdlsoap:body use="encoded" namespace="urn:api.broadmail.de/soap11/RpcSession" encodingStyle="http://schemas.xmlsoap.org/soap/encoding/" />
      </wsdl:output>
      <wsdl:fault name="WebserviceException">
        <wsdlsoap:fault use="encoded" name="WebserviceException" namespace="urn:api.broadmail.de/soap11/RpcSession" encodingStyle="http://schemas.xmlsoap.org/soap/encoding/" />
      </wsdl:fault>
    </wsdl:operation>
    </wsdl:binding>
  <wsdl:service name="SessionWebserviceService">
    <wsdl:port name="RpcSession" binding="impl:RpcSessionSoapBinding">
      <wsdlsoap:address location="https://api.broadmail.de/soap11/RpcSession" />
    </wsdl:port>
  </wsdl:service>
</wsdl:definitions>

之后我查看了我生成的代理类“reference.cs”，通知方法中没有参数。

namespace OptivoInterfaceApp.broadmailSession {


    [System.CodeDom.Compiler.GeneratedCodeAttribute("System.ServiceModel", "4.0.0.0")]
    [System.ServiceModel.ServiceContractAttribute(Namespace="urn:api.broadmail.de/soap11/RpcSession", ConfigurationName="broadmailSession.SessionWebservice")]
    public interface SessionWebservice {


        [System.ServiceModel.OperationContractAttribute()]
        void login();

    }

    [System.CodeDom.Compiler.GeneratedCodeAttribute("System.ServiceModel", "4.0.0.0")]
    public interface SessionWebserviceChannel : OptivoInterfaceApp.broadmailSession.SessionWebservice, System.ServiceModel.IClientChannel {
    }

    [System.Diagnostics.DebuggerStepThroughAttribute()]
    [System.CodeDom.Compiler.GeneratedCodeAttribute("System.ServiceModel", "4.0.0.0")]
    public partial class SessionWebserviceClient : System.ServiceModel.ClientBase<OptivoInterfaceApp.broadmailSession.SessionWebservice>, OptivoInterfaceApp.broadmailSession.SessionWebservice {

        public SessionWebserviceClient() {
        }

        public SessionWebserviceClient(string endpointConfigurationName) : 
                base(endpointConfigurationName) {
        }

        public SessionWebserviceClient(string endpointConfigurationName, string remoteAddress) : 
                base(endpointConfigurationName, remoteAddress) {
        }

        public SessionWebserviceClient(string endpointConfigurationName, System.ServiceModel.EndpointAddress remoteAddress) : 
                base(endpointConfigurationName, remoteAddress) {
        }

        public SessionWebserviceClient(System.ServiceModel.Channels.Binding binding, System.ServiceModel.EndpointAddress remoteAddress) : 
                base(binding, remoteAddress) {
        }

        public void login() {
            base.Channel.login();
        }
    }
}

我是否必须手动将params添加到我的代理类中，或者我还有其他错误吗？我只是尝试手动添加参数，但在方法调用上我只得到一个异常。

我的任何提示？

Answer 1

也许你应该使用WRAPPED风格。

更多信息：http://axis.apache.org/axis/java/user-guide.html#ServiceStylesRPCDocumentWrappedAndMessage

.NET连接到AXIS服务。方法没有params或Datatypes

1 个答案: