C# 从SOAP响应获取附加数据

我正在从一个web服务接收下面的SOAP响应

<soap:Body>
<SubscriptionQueryResponse>
    <Subscriptions>
        <Subscription></Subscription>
        <Subscription></Subscription>
    </Subscriptions>
    <TotalCount>2</TotalCount>
    <ResultCount>2</ResultCount>
    <StartingSequence>0</StartingSequence>
</SubscriptionQueryResponse>
</soap:Body>

正如您所看到的，我已经返回订阅了。订阅列表位于返回数组的索引0中。我还想将Total Count作为返回数组的索引1返回，但return-XmlArray语句只返回订阅并自动强制转换它们

编辑：至少，我希望原始XML没有序列化，我可以自己解析它

---

注意：忽略SoapDocumentMethod中的&&&。我删除了它们的值以供显示。

我的问题的解决方案是不让框架序列化xml

[SoapDocumentMethod( "&&&", RequestElementName = "SubscriptionQueryRequest", RequestNamespace = "&&&", ResponseNamespace = "&&&", Use = SoapBindingUse.Literal, ParameterStyle = SoapParameterStyle.Wrapped )]
[return: XmlAnyElement]
public List<XmlNode> SubscriptionQuery(string SubscriberId, int SortType, bool SortDescending, string Service, string ReferenceID, string SubscriptionName, int StartingSequence, int ResultCount )
{
    object[] results = this.Invoke( "SubscriptionQuery", new object[] {
            SubscriberId, 
            SortType, 
            SortDescending,
            Service,
            ReferenceID,
            SubscriptionName, 
            StartingSequence,
            ResultCount
    } );
    return ( (List<XmlNode>)( results[0] ) );
}

这会将响应中的所有节点强制转换为一个列表，然后我可以获取响应中所需的每个节点，并自己对它们进行反序列化，而不是让框架进行反序列化

[SoapDocumentMethod( "&&&", RequestElementName = "SubscriptionQueryRequest", RequestNamespace = "&&&", ResponseNamespace = "&&&", Use = SoapBindingUse.Literal, ParameterStyle = SoapParameterStyle.Wrapped )]
[return: XmlAnyElement]
public List<XmlNode> SubscriptionQuery(string SubscriberId, int SortType, bool SortDescending, string Service, string ReferenceID, string SubscriptionName, int StartingSequence, int ResultCount )
{
    object[] results = this.Invoke( "SubscriptionQuery", new object[] {
            SubscriberId, 
            SortType, 
            SortDescending,
            Service,
            ReferenceID,
            SubscriptionName, 
            StartingSequence,
            ResultCount
    } );
    return ( (List<XmlNode>)( results[0] ) );
}