How can I convert object to strongly typed XML

Question

Is there a way to convert an object to XML but each element contains an attribute (let's be like this xt:type="int").
I know I can do it manually using reflection and iterate properties and ....
I'm asking if there's a way to generate it using pre-made library or something.
What I'm doing now is:

XmlDocument doc = new XmlDocument();
XmlSerializer ser = new XmlSerializer(document.GetType());
string result = string.Empty;
using (MemoryStream memStm = new MemoryStream())
 {
   ser.Serialize(memStm, document);
   memStm.Position = 0;
   result = new StreamReader(memStm).ReadToEnd();
 }

Because later I need to read it back to an object. All this I want to do it programmatically, not using XSD tool.

Thanks

UPDATE 1:
What I want to get looks something like this:

<note>
   <propertyName1 xs:type="string">value1</to>
   <propertyName2 xs:type="int">10</to>
   <propertyName2 xs:type="datetime">04-06-2015 01:10:00</to>
</note>

The most important is the attribute xs:type.

I'm using as you see "XmlSerializer " but how to write the type attribute in the xml element. — Dabbas, Jun 04 '15 at 08:14
let me know if this helps:https://msdn.microsoft.com/en-us/library/system.xml.serialization.xmlelementattribute.type(v=vs.110).aspx — Amit Kumar Ghosh, Jun 04 '15 at 08:19
If I decorated the property with 'XmlAttributeAttribute' then property will be as an Attribute not as Element and it'll not event add the type to the XML result. — Dabbas, Jun 04 '15 at 09:19
I don't understand what you want to achieve, can you post an example of an object and of its expected xml? — Matteo Umili, Jun 04 '15 at 10:15

Benoit Gidon · Accepted Answer · 2015-06-04T11:07:56.133

1

If you use System.XML.Linq, you can add an XAttribute with whatever you want in it.

Lets say you have a Personn P with string Name and int Age

XElement e = new XElement( "Personne" );
XElement age = new XElement("Age",Personne.Age);
age.add(new XAttribute("xs:type", typeOf(Personne.Age)));
e.Add(age);
XElement name = new XElement("Name",Personne.Name);
name.add(new XAttribute("xs:type", typeOf(Personne.Name)));
e.Add(name);`

You will receive as XML:

<Personne>
<Age type="int">(the age of the personne)</age>
<Name type="string">(he name of the personne)</Name>
<Personne>

with "automated" process:

XElement p = new XElement( "Personne" );

foreach(var property in Personn.GetType().GetProperties()) {
XElement e = new XElement(prop.Name, prop.GetValue(P, null));
e.Add(new XAttribute("xs:type", typeOf(Personne.prop)));
p.Add(e);
}

EDITED: added "xs:" as you need it

edited Jun 04 '15 at 11:07

answered Jun 04 '15 at 10:24

Benoit Gidon

79
10

So I need to do it manually, there's no automation for this process – Dabbas Jun 04 '15 at 10:44
Well assuming that you can loop through all properties of an object: See http://stackoverflow.com/questions/737151/how-to-get-the-list-of-properties-of-a-class So you don't have do it manually, do you want me to edit the above sample to make it "automated" by this process? – Benoit Gidon Jun 04 '15 at 10:49
It's a deep traverse in the object so it's not easy I think to do it manually :) Any way thanks – Dabbas Jun 04 '15 at 10:55
1

I'll redo the sample for you – Benoit Gidon Jun 04 '15 at 11:02
@Dabbas Done , hope it will help – Benoit Gidon Jun 04 '15 at 13:16

How can I convert object to strongly typed XML

1 Answers1