在C＃中，如何将我的派生对象序列化到此XML？

Question

我正在尝试将一些XML发送给第三方，而我需要发送的元素的格式如下所示：

<CONTACT>
    <CONTACTMETHODS>
        <PHONE number='202-555-1234' />
        <EMAIL address='myname@gmail.com' />
    </CONTACTMETHODS>
</CONTACT>

我在C＃中有以下模型，我正在尝试将其序列化为XML字符串：

[XmlRoot("CONTACT")]
[XmlInclude(typeof(Phone))]
[XmlInclude(typeof(Email))]
public class Contact
{
    [XmlArray("CONTACTMETHODS")]
    public List<ContactMethod> ContactMethods { get; set; }
}

public abstract class ContactMethod
{
}

[XmlRoot("PHONE")]
public class Phone : ContactMethod
{
    [XmlAttribute("number")]
    public string Number { get; set; }
}

[XmlRoot("EMAIL")]
public class Email : ContactMethod
{
    [XmlAttribute("address")]
    public string Address { get; set; }
}

发送到字符串的XML是：

<CONTACT>
    <CONTACTMETHODS>
        <ContactMethod xsi:type="Phone" number="202-555-1234" />
        <ContactMethod xsi:type="Email" address="myname@gmail.com" />
    </CONTACTMETHODS>
</CONTACT>

如何获得序列化程序来创建所需的XML？

编辑：根据要求，以下是序列化对象的代码：

protected string ObjToXmlString<T>(T obj) where T : class
{
    var stringwriter = new System.IO.StringWriter();
    var serializer = new XmlSerializer(typeof(T));
    serializer.Serialize(stringwriter, obj);
    var returnXml = stringwriter.ToString();

    return returnXml;
}

Answer 1

我以错误的方式思考它。 我不是通过电话和电子邮件派生的抽象类，而是这样做的：

[XmlRoot("CONTACT")]
[XmlInclude(typeof(CR.Models.XactAnalysis.Phone))]
[XmlInclude(typeof(CR.Models.XactAnalysis.Email))]
public class Contact
{
    [XmlArray("CONTACTMETHODS")]
    public List<ContactMethod> ContactMethods { get; set; }
}

public class ContactMethod
{
    [XmlElement("PHONE")]
    public Phone Phone { get; set; }

    [XmlElement("EMAIL")]
    public Email Email { get; set; }
}

[XmlRoot("PHONE")]
public class Phone
{
    [XmlAttribute("number")]
    public string Number { get; set; }
}

[XmlRoot("EMAIL")]
public class Email
{
    [XmlAttribute("address")]
    public string Address { get; set; }
}

似乎工作顺畅。