Question

我构建了一个WCF Rest服务来为另一个进程提供数据。假设他的名字是GetData。这个提供了具有以下结构的xml响应：

<?xml version="1.0" encoding="utf-8"?>
<GetDataResponse xmlns="http://tempuri.org/" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
  <GetDataResult>
    <DataMessage>   
      <a></a>
      <b></b>
      <c></c>
    </DataMessage>
  </GetDataResult>
</GetDataResponse>

服务界面：

    [XmlSerializerFormat]
    [OperationContract(Name = "GetData")]
    [WebInvoke(Method = "GET",
               ResponseFormat = WebMessageFormat.Xml,
               BodyStyle = WebMessageBodyStyle.Wrapped,
               UriTemplate = "Data/{Param}")]
    List<DataMessage> GetData(string Params);

我希望在保存它之后反序列化xml，遵循DataMessage类。所以，我想有这个架构：

<?xml version="1.0" encoding="utf-8"?>
<DataMessages xmlns="http://tempuri.org/" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
    <DataMessage>
      <a></a>
      <b></b>
      <c></c>
    </DataMessage>
</DataMessages>

如何定义服务响应模式以使其具有这样的效果？

谢谢。

Answer 1

您可以使用System.Xml.Serialization命名空间中的某些属性来定义映射到您拥有的架构的对象图。下面的代码就是这样。

public class StackOverflow_7905186
{
    [XmlType(TypeName = "DataMessage", Namespace = "http://tempuri.org/")]
    public class DataMessage
    {
        public string a;
        public string b;
        public string c;
    }
    [XmlRoot(ElementName = "DataMessages", Namespace = "http://tempuri.org/")]
    public class DataMessages
    {
        [XmlElement(ElementName = "DataMessage")]
        public List<DataMessage> Messages;
    }
    [ServiceContract]
    public class Service
    {
        [XmlSerializerFormat]
        [OperationContract(Name = "GetData")]
        [WebGet(ResponseFormat = WebMessageFormat.Xml,
                BodyStyle = WebMessageBodyStyle.Bare,
                UriTemplate = "Data/{Param}")]
        [return: MessageParameter(Name = "DataMessages")]
        public DataMessages GetData(string Param)
        {
            return new DataMessages
            {
                Messages = new List<DataMessage>
                {
                    new DataMessage
                    {
                        a = "1",
                        b = "2",
                        c = "3",
                    }
                }
            };
        }
    }
    public static void Test()
    {
        string baseAddress = "http://" + Environment.MachineName + ":8000/Service";
        WebServiceHost host = new WebServiceHost(typeof(Service), new Uri(baseAddress));
        host.Open();
        Console.WriteLine("Host opened");

        WebClient c = new WebClient();
        Console.WriteLine(c.DownloadString(baseAddress + "/Data/foo"));

        Console.Write("Press ENTER to close the host");
        Console.ReadLine();
        host.Close();
    }
}

定义WCF XML响应模式

1 个答案: