How to serialize a class property to be a part of xml root?

Question

I have the following xml

<OTA_HotelResNotifRS xmlns:i="http://www.w3.org/2001/XMLSchema-instance" xmlns="http://schemas.datacontract.org/2004/07/Connect.Domain.OTA_2014B.Reservations.OTA_HotelResNotifRS">
    <HotelReservations>
        <HotelReservation>
            <ResGlobalInfo>
                <HotelReservationIDs>
                    <OTA_HotelResNotifRSHotelReservationsHotelReservationResGlobalInfoHotelReservationID>
                        <ResID_Source>some_source</ResID_Source>
                        <ResID_Type>0</ResID_Type>
                        <ResID_Value>51550</ResID_Value>
                    </OTA_HotelResNotifRSHotelReservationsHotelReservationResGlobalInfoHotelReservationID>
                </HotelReservationIDs>
            </ResGlobalInfo>
        </HotelReservation>
    </HotelReservations>
    <Success i:nil="true" />
    <Target i:nil="true" />
    <TimeStamp>0001-01-01T00:00:00</TimeStamp>
    <Version>0</Version>
</OTA_HotelResNotifRS>

With the following c# code.

    [System.ComponentModel.DesignerCategoryAttribute("code")]
    [System.Xml.Serialization.XmlTypeAttribute(AnonymousType = true, Namespace = 

    "http://www.opentravel.org/OTA/2003/05")]
    [System.Xml.Serialization.XmlRootAttribute(Namespace = "http://www.opentravel.org/OTA/2003/05", IsNullable = false)]
    public partial class OTA_HotelResNotifRS

{   
    /// <remarks/>
    public OTA_HotelResNotifRSHotelReservations HotelReservations
    {
        get; set;
    }

    /// <remarks/>
    [System.Xml.Serialization.XmlAttributeAttribute()]
    public System.DateTime TimeStamp
    {
        get; set;
    }
}

How can I place timestamp, target and version in root element? I tried to add XmlRootAttribute instead of XmlAttribute, but I am getting an error.

Answer 1

If you want to make TimeStamp and Version attributes, just add a XmlAttribute.

[XmlAttribute]
public DateTime TimeStamp { get; set; }

[XmlAttribute]
public byte Version { get; set; }

To make the Target property as attribute, it must be a simple type, like int, DateTime, etc.

[XmlAttribute]
public int Target { get; set; }

If it is complex type like object, etc, then will have to leave it as element.

[XmlElement]
public object Target { get; set; }

---

I copied your xml to the clipboard.

In the Visual Studio menu choose Edit > Paste Special > Paste XML As Classes. Were generated a set of classes.

I have added the attribute [XmlAttribute] before the properties TimeStamp and Version. That's all, I have not made any other changes.

Execute this code:

var xs = new XmlSerializer(typeof(OTA_HotelResNotifRS));
OTA_HotelResNotifRS ota;

using (var fs = new FileStream("in.xml", FileMode.Open))
    ota = (OTA_HotelResNotifRS)xs.Deserialize(fs);

using (var fs = new FileStream("out.xml", FileMode.Create))
    xs.Serialize(fs, ota);

In the end, I got the following xml (I formatted the attributes for easier reading and skipped inner nodes):

<?xml version="1.0"?>
<OTA_HotelResNotifRS xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
                     xmlns:xsd="http://www.w3.org/2001/XMLSchema"
                     TimeStamp="0001-01-01T00:00:00"
                     Version="0"
                     xmlns="http://schemas.datacontract.org/2004/07/Connect.Domain.OTA_2014B.Reservations.OTA_HotelResNotifRS">
  <HotelReservations>
    ...
  </HotelReservations>
  <Success xsi:nil="true" />
  <Target xsi:nil="true" />
</OTA_HotelResNotifRS>

TimeStamp and Version became attributes of the root element.