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我使用 SVCUTIL 从 XSD 生成一个类。我很难弄清楚如何获取传入的请求对象并从对象中检索“MsgType”值。

我想通过这样做，我可以简单地使用以下方法访问数据：

request.Request.MsgType

然而，事情并没有这么简单。'request' 给我的唯一选项是： Equals GetHashCode GetSchema GetType Nodes 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 架构

<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"/>
4

2 回答 2

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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，我不确定你的 XSD 是什么。

于 2013-01-07T20:29:36.017 回答
0

经过几天的学习和弄清楚如何在不使用旧时尚方式的情况下访问数据。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。我被困在这个项目上的原因是因为复杂的类型，我以前从未这样做过。我用了：

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

其中生成了 ServiceProviderTic.cs

我创建了一个网络服务：

界面：

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

班级：

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 帮助我接受普通旧 xml (POX) 并以普通旧 xml 响应。下面是序列化和反序列化我的 xelement 的辅助函数。我还包含了额外的代码，这些代码删除了我不需要的命名空间。

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)));
    }
}

我希望这对将来的人有所帮助！

于 2013-01-10T00:41:00.987 回答