Question

I am designing my class model. The serialized message of the class model needs to be in this format:

<?xml version="1.0" encoding="UTF-8" ?>
<Request>
  <Name>TesterScript</Name>
  <ID>CD_20110628133820576</ID>
  <Type>
    <ItemId>191_20110628T133821</ItemId>
    <ShopId>MyBCShop</ShopId>
    <MessageXml>
    <ChildMessage>
      This is my message
    </ChildMessage>
    </MessageXml>
  </Type>
  <SentTime>2011-06-30T15:27:06-07:00</SentTime>
</Request>

How would I design the classes?
Also what should be the best way to serialize the suggested class model to above XML message? I am thinking of using:

// Serialize the request
XmlSerializer xs = new XmlSerializer(typeof(Request));
StringWriter sw = new StringWriter();
xs.Serialize(sw, dispatchRequest);
string xml = sw.ToString();
return new xml;

Is this the most suitable way?

Answer 1

I hope my solution is useful ….

The xml content (i think there is something wrong in your post…)

<?xml version="1.0" encoding="utf-8" ?>
<Request>
  <Name>TesterScript</Name>
  <ID>CD_20110628133820576</ID>
  <Type>
    <ItemId>191_20110628T133821</ItemId>
    <ShopId>BARCGB2L</ShopId>
    <MessageXml>
      <ChildMessage>
        This is my message
      </ChildMessage>
    </MessageXml>
  </Type>
  <SentTime>2011-06-30T15:27:06-07:00</SentTime>
</Request>

The class :

[XmlRoot("Request")]
public class SampleClass
{
    public string Name { get; set; }
    public string ID { get; set; }
    [XmlElement("Type")]
    public SubClass SC { get; set; }
    public string SentTime { get; set; }

    public class SubClass
    {
        public string ItemId { get; set; }
        public string ShopId { get; set; }
        [XmlElement("MessageXml")]
        public Sub2Class SC2 { get; set; }

        public class Sub2Class
        {
            public string ChildMessage { get; set; }
        }
    }
}

The Deserialize method:

public static T DeserializeForXml<T>(string filePath)
    {
        XmlSerializer selializer = new XmlSerializer(typeof(T));
        using (Stream fs = new FileStream(filePath, FileMode.Open, FileAccess.Read, FileShare.Read))
        {
            return (T)selializer.Deserialize(fs);
        }
    }

How to use ?

SampleClass sc = Utility.DeserializeForXml<SampleClass>("test.xml");