如何在MessageContract类型中使用XML属性?

时间:2014-07-28 15:20:04

标签: c# xml wcf soap

预先注意:我无法更改进入的SOAP请求的格式,因为它们是由国际标准(weeeeeeeee)修复的。

我有一个SOAP请求进入我的WCF服务,看起来像这样:

<s:Body>
    <Request version="1.0">
        <data someOtherVersion="1.1">
            ...
        </data>
    </Request>
</s:Body>

到目前为止,我们一直在使用System.ServiceModel.Channels.Message个对象,这有点痛苦。我们正试图使用​​看起来像这样的强类型:

[MessageContract(IsWrapped = false)]
public class Request
{
    [MessageBodyMember]
    [XmlAttribute("version")]
    public string Version;

    [MessageBodyMember]
    [XmlElement("data")]
    public SomeOtherType Data;
}

[MessageContract(IsWrapped = false)]
public class Response
{
    [MessageBodyMember]
    [XmlAttribute("version")]
    public string Version;

    [MessageBodyMember]
    [XmlElement("data")]
    public SomeOtherType ResponseData;
}

[ServiceContract]
[XmlSerializerFormat]
public interface Service
{
    [OperationContract(Action = "request", ReplyAction = "response")]
    Response ServiceOperation(Request req);
}

不幸的是,当我们尝试启动时,我们收到一条错误消息“System.ServiceModel.dll中发生了'System.InvalidOperationException'类型的未处理异常

其他信息:XmlSerializer属性System.Xml.Serialization.XmlAttributeAttribute在版本中无效。当IsWrapped为false时,MessageContract中仅支持XmlElement,XmlArray,XmlArrayItem和XmlAnyElement属性。“

有趣的是,将“IsWrapped”设置为true会产生相同的错误。有没有办法在消息协定类型中序列化XML属性,或者使用包装器是我们唯一的选择吗?

2 个答案:

答案 0 :(得分:1)

不幸的是,我发现实现此目标的唯一方法是使用包装器类

[MessageContract(IsWrapped = false)]
public class Response
{
    [MessageBodyMember(Name = "Response", Namespace = "Http://example.org/ns1")]
    public ResponseBody Body { get; set; }

    public Response(){}


    public Response(ResponseBody body)
    {
        Body = body;
    }
}

[XmlType(AnonymousType = true, Namespace = "Http://example.org/ns1")]
public class ResponseBody
{
    [XmlAttribute(AttributeName = "version")]
    public string Version { get; set; }

    [XmlElement(ElementName = "data", Namespace = "Http://example.org/ns1")]
    [MessageBodyMember]
    public SomeOtherType ResponseData { get; set; }
}

答案 1 :(得分:-3)

尝试使用XmlElement(typeof(data))