POX SVCUtil生成的类转换

Question

我使用SVCUTIL从XSD生成一个类。 我在弄清楚如何获取传入的请求对象以及从对象中检索“ MsgType”值时遇到困难。

我认为通过这样做，我可以使用以下命令简单地访问数据：

request.Request.MsgType

但是，这不是那么简单。 “请求”给我的唯一选项是：等于GetHashCode GetSchema GetType节点ReadXML ToString WriteXML

为了访问MsgType，我需要对序列化对象进行某种类型的转换吗？

public ServiceProviderTic callRequestFunc(ServiceProviderTic request) {
      //How do I get request.Request.MsgType Value?
}

生成类中的根元素：

using System.Runtime.Serialization;

[assembly: System.Runtime.Serialization.ContractNamespaceAttribute("", ClrNamespace="")]

[System.Diagnostics.DebuggerStepThroughAttribute()]
[System.CodeDom.Compiler.GeneratedCodeAttribute("System.Runtime.Serialization", "3.0.0.0")]
[System.Runtime.Serialization.DataContractAttribute(Name="RequestType", Namespace="")]
[System.Runtime.Serialization.KnownTypeAttribute(typeof(ResponseType))]
[System.Runtime.Serialization.KnownTypeAttribute(typeof(DateTimeInfoType))]
[System.Runtime.Serialization.KnownTypeAttribute(typeof(OriginType))]
[System.Runtime.Serialization.KnownTypeAttribute(typeof(LocaleInfoType))]
[System.Runtime.Serialization.KnownTypeAttribute(typeof(ProductType))]
[System.Runtime.Serialization.KnownTypeAttribute(typeof(ValueType))]
[System.Runtime.Serialization.KnownTypeAttribute(typeof(AuthInfoType))]
public partial class RequestType : object, System.Runtime.Serialization.IExtensibleDataObject
{

private RequestType.MsgTypeType MsgTypeField;

[System.Runtime.Serialization.DataMemberAttribute(IsRequired=true)]
public RequestType.MsgTypeType MsgType
{
    get
    {
        return this.MsgTypeField;
    }
    set
    {
        this.MsgTypeField = value;
    }
}

 [System.Runtime.Serialization.DataContractAttribute(Name="RequestType.MsgTypeType", Namespace="")]
public enum MsgTypeType : int
{
    [System.Runtime.Serialization.EnumMemberAttribute()]
    act = 0
}
}

[System.Diagnostics.DebuggerStepThroughAttribute()]
[System.CodeDom.Compiler.GeneratedCodeAttribute("System.Runtime.Serialization", "3.0.0.0")]
[System.Xml.Serialization.XmlSchemaProviderAttribute("ExportSchema")]
[System.Xml.Serialization.XmlRootAttribute(IsNullable=false)]
public partial class ServiceProviderTic : object, System.Xml.Serialization.IXmlSerializable
{

private System.Xml.XmlNode[] nodesField;

private static System.Xml.XmlQualifiedName typeName = new System.Xml.XmlQualifiedName("ServiceProviderTic", "");

public System.Xml.XmlNode[] Nodes
{
    get
    {
        return this.nodesField;
    }
    set
    {
        this.nodesField = value;
    }
}

public void ReadXml(System.Xml.XmlReader reader)
{
    this.nodesField = System.Runtime.Serialization.XmlSerializableServices.ReadNodes(reader);
}

public void WriteXml(System.Xml.XmlWriter writer)
{
    System.Runtime.Serialization.XmlSerializableServices.WriteNodes(writer, this.Nodes);
}

public System.Xml.Schema.XmlSchema GetSchema()
{
    return null;
}

public static System.Xml.XmlQualifiedName ExportSchema(System.Xml.Schema.XmlSchemaSet schemas)
{
    System.Runtime.Serialization.XmlSerializableServices.AddDefaultSchema(schemas, typeName);
    return typeName;
}

XML：

<ServiceProviderTic>
<Request>
<MsgType>act</MsgType>

XSD模式

<xs:schema xmlns:xs="http://www.w3.org/2001/XMLSchema" elementFormDefault="qualified" attributeFormDefault="unqualified">
<xs:element name="ServiceProvideTic" nillable="false">
<xs:annotation>
<xs:documentation></xs:documentation>
</xs:annotation>
<xs:complexType>
<xs:sequence>
<xs:element name="Version" type="xs:string" nillable="false"/>
<xs:choice>
<xs:element name="Request" type="RequestType" nillable="false"/>
<xs:element name="Response" type="ResponseType" nillable="false"/>
</xs:choice>
</xs:sequence>
</xs:complexType>
</xs:element>
<xs:complexType name="RequestType">
<xs:annotation>
<xs:documentation> Request Information</xs:documentation>
</xs:annotation>
<xs:sequence>
<xs:element name="MsgType" nillable="false">
<xs:simpleType>
<xs:restriction base="xs:string">
<xs:enumeration value="act"/>

Answer 1

在此处查看有关创建支持XML和JSON的REST Web服务的一些好的建议（请参阅WebHttpBehavior类）

除此之外，我不确定您的XSD是什么。

Answer 2

经过几天的学习并弄清楚了如何不用旧的方式访问数据。 这是我想出的：

Microsoft提供了XSD和SVCUTIL，用于将xsd转换为类并使它们可序列化。 我之所以被困在这个项目上是因为类型复杂，我之前从未做过。 我用了：

命令提示符：XSD.exe ServiceProviderTic.xsd / CLASSES

哪个生成了ServiceProviderTic.cs

我创建了一个Web服务：

接口：

[OperationContract]
[WebInvoke(Method = "POST", UriTemplate="/MyService", RequestFormat = WebMessageFormat.Xml, ResponseFormat = WebMessageFormat.Xml, BodyStyle = WebMessageBodyStyle.Bare)]
    XElement callRequestFunc(XElement request);

类：

public XElement callRequestFunc(XElement request)
    {
        ServiceProviderTic requestSer = Utility.DeserializeData(request);

        if (requestSer.Item.GetType() == typeof(RequestType))
        {
            RequestType reqObj = (RequestType)requestSer.Item;
            string datapiece = reqObj.MsgType.ToString();
        }

        XElement responseSer = Utility.SerializeData(requestSer);

        return responseSer;
    }
}

XElement帮助我接受普通的旧xml（POX）并使用普通的旧xml进行响应。 以下是对我的xelement进行序列化和反序列化的帮助函数。 我还包括了多余的代码，这些代码删除了不需要的名称空间。

public class Utility
{

    public static ServiceProviderTic DeserializeData(XElement request)
    {
        var ser = new XmlSerializer(typeof(ServiceProviderTic));
        return (ServiceProviderTic)ser.Deserialize(request.CreateReader());
    }

    public static XElement SerializeData(ServiceProviderTic response)
    {
        using (var memoryStream = new MemoryStream())
        {
            using (TextWriter streamWriter = new StreamWriter(memoryStream))
            {
                var xmlSerializer = new XmlSerializer(typeof(ServiceProviderTic));
                xmlSerializer.Serialize(streamWriter, response);
                return Utility.RemoveAllNamespaces(XElement.Parse(Encoding.ASCII.GetString(memoryStream.ToArray())));
            }
        }
    }

    public static XElement RemoveAllNamespaces(XElement source)
    {
        return !source.HasElements
                   ? new XElement(source.Name.LocalName)
                   {
                       Value = source.Value
                   }
                   : new XElement(source.Name.LocalName, source.Elements().Select(el => RemoveAllNamespaces(el)));
    }
}

希望这对以后的人有所帮助！