3

我想使用 XML 属性而不是 xmlrequest 和 xmlresponse 中的元素,我实现了 IXmlSerializable 如下:

public partial class Id : IXmlSerializable
{

    /// <remarks/>
    [XmlAttribute]
    public string lmsId;

    /// <remarks/>

    [XmlAttribute]
    public string unitId;

    /// <remarks/>

    [XmlAttribute]
    public string lmsPresId;

    /// <remarks/>

    [XmlAttribute]
    public string callId;

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

    public void ReadXml(System.Xml.XmlReader reader)
    {
        //implement if remote callers are going to pass your object in
    }

    public void WriteXml(System.Xml.XmlWriter writer)
    {
        writer.WriteAttributeString("lmsId", lmsId.ToString());
        writer.WriteAttributeString("unitId", unitId.ToString());
        writer.WriteAttributeString("lmsPresId", lmsPresId.ToString());
        writer.WriteAttributeString("callId", callId.ToString());
    }
}

但我的服务帮助显示请求和响应为未知


我什至尝试了 XmlSerializerFormat

[System.ServiceModel.MessageContract]
public partial class Id
{

    /// <remarks/>
    [XmlAttribute, MessageBodyMember]
    public string lmsId;

    /// <remarks/>

    [XmlAttribute, MessageBodyMember]
    public string unitId;

    /// <remarks/>

    [XmlAttribute, MessageBodyMember]
    public string lmsPresId;

    /// <remarks/>

    [XmlAttribute, MessageBodyMember]
    public string callId;
}

然后我将 xml 中的所有内容作为 xmlElements 而不是 XmlAttributes 但我将所有数据作为元素而不是 Xmlattributes 获取。

4

0 回答 0