WebApi/DataBind: How Databind XML and class model?

Question

I am trying to databind in a web api next xml: application/xml to a class

<COMPANY>
  <PROPERTIES>
    <DATASOURCE>UXXI</DATASOURCE>
    <DATETIME>2012-07-11T11:18:12</DATETIME>
  </PROPERTIES>
  <PERSON recstatus="1">
    <SOURCEDID>
      <SOURCE>U123XXI</SOURCE>
      <ID>usuario26</ID>
    </SOURCEDID>
    <USERID password="123456">usuario26</USERID>
    <NAME>
      <FN>Usuario Prueba 26</FN>
      <PI>
        <FAMILY>Prueba 26</FAMILY>
        <GIVEN>Usuario</GIVEN>
        <EMAIL>[email protected]</EMAIL>
      </PI>
    </NAME>
  </PERSON>
</COMPANY>

To class:

public class COMPANY {
    public class PROPERTIES {
        public string DATASOURCE { get; set; }
        public string MYDATE { get; set; }
    }

    public class PERSON {
        public string recstatus { get; set; }

        public class SOURCEID {
            public string SOURCE { get; set; }
            public string ID { get; set; }
        }

        public class USER {
            public string password { get; set; }
            public string USERID { get; set; }
        }

        public class NAME {
            public string FN {set; get;}
            public class PI {
                public string FAMILY { set; get; }
                public string GIVEN { set; get; }

                public string EMAIL { set; get; }
            }
        }
    }
}

In controller I have :

[HttpPut]
public HttpResponseMessage GetPerson(COMPANY req) {
...
}

How can I databind it?

I use PostMan to send request

I am getting null in the controller because I think I have to databind class with XML.

Nasreddine · Accepted Answer · 2016-06-01 10:18:40Z

First let's annotate the classes/member with the different ElementName/AttributeName:

[XmlRoot(ElementName = "PROPERTIES")]
public class Properties
{
    [XmlElement(ElementName = "DATASOURCE")]
    public string Datasource { get; set; }
    [XmlElement(ElementName = "DATETIME")]
    public string Datetime { get; set; }
}

[XmlRoot(ElementName = "SOURCEDID")]
public class Sourcedid
{
    [XmlElement(ElementName = "SOURCE")]
    public string Source { get; set; }
    [XmlElement(ElementName = "ID")]
    public string ID { get; set; }
}

[XmlRoot(ElementName = "USERID")]
public class UserId
{
    [XmlAttribute(AttributeName = "password")]
    public string Password { get; set; }
    [XmlText]
    public string Text { get; set; }
}

[XmlRoot(ElementName = "PI")]
public class PI
{
    [XmlElement(ElementName = "FAMILY")]
    public string Family { get; set; }
    [XmlElement(ElementName = "GIVEN")]
    public string Given { get; set; }
    [XmlElement(ElementName = "EMAIL")]
    public string Email { get; set; }
}

[XmlRoot(ElementName = "NAME")]
public class Name
{
    [XmlElement(ElementName = "FN")]
    public string FN { get; set; }
    [XmlElement(ElementName = "PI")]
    public PI PI { get; set; }
}

[XmlRoot(ElementName = "PERSON")]
public class Person
{
    [XmlElement(ElementName = "SOURCEDID")]
    public Sourcedid Sourcedid { get; set; }
    [XmlElement(ElementName = "USERID")]
    public UserId Userid { get; set; }
    [XmlElement(ElementName = "NAME")]
    public Name Name { get; set; }
    [XmlAttribute(AttributeName = "recstatus")]
    public string Recstatus { get; set; }
}

[XmlRoot(ElementName = "COMPANY")]
public class Company
{
    [XmlElement(ElementName = "PROPERTIES")]
    public Properties Properties { get; set; }
    [XmlElement(ElementName = "PERSON")]
    public Person Person { get; set; }
}

Then we'll tell WebA PI that we want to use the XmlSerializer instead of the DataContractSerializer: Inside the Application_start() method in your Global.asax file add this line:

GlobalConfiguration.Configuration.Formatters.XmlFormatter.UseXmlSerializer = true;

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