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How to deserialize a given xml document [duplicate]

How do I Deserialize this XML document:
<?xml version="1.0" encoding="utf-8"?>
<Cars>
<Car>
<StockNumber>1020</StockNumber>
<Make>Nissan</Make>
<Model>Sentra</Model>
</Car>
<Car>
<StockNumber>1010</StockNumber>
<Make>Toyota</Make>
<Model>Corolla</Model>
</Car>
<Car>
<StockNumber>1111</StockNumber>
<Make>Honda</Make>
<Model>Accord</Model>
</Car>
</Cars>
I have this:
[Serializable()]
public class Car
{
[System.Xml.Serialization.XmlElementAttribute("StockNumber")]
public string StockNumber{ get; set; }
[System.Xml.Serialization.XmlElementAttribute("Make")]
public string Make{ get; set; }
[System.Xml.Serialization.XmlElementAttribute("Model")]
public string Model{ get; set; }
}
.
[System.Xml.Serialization.XmlRootAttribute("Cars", Namespace = "", IsNullable = false)]
public class Cars
{
[XmlArrayItem(typeof(Car))]
public Car[] Car { get; set; }
}
.
public class CarSerializer
{
public Cars Deserialize()
{
Cars[] cars = null;
string path = HttpContext.Current.ApplicationInstance.Server.MapPath("~/App_Data/") + "cars.xml";
XmlSerializer serializer = new XmlSerializer(typeof(Cars[]));
StreamReader reader = new StreamReader(path);
reader.ReadToEnd();
cars = (Cars[])serializer.Deserialize(reader);
reader.Close();
return cars;
}
}
that don't seem to work :-(

How about you just save the xml to a file, and use xsd to generate C# classes?
Write the file to disk (I named it foo.xml)
Generate the xsd: xsd foo.xml
Generate the C#: xsd foo.xsd /classes
Et voila - and C# code file that should be able to read the data via XmlSerializer:
XmlSerializer ser = new XmlSerializer(typeof(Cars));
Cars cars;
using (XmlReader reader = XmlReader.Create(path))
{
cars = (Cars) ser.Deserialize(reader);
}
(include the generated foo.cs in the project)

Here's a working version. I changed the XmlElementAttribute labels to XmlElement because in the xml the StockNumber, Make and Model values are elements, not attributes. Also I removed the reader.ReadToEnd(); (that function reads the whole stream and returns a string, so the Deserialize() function couldn't use the reader anymore...the position was at the end of the stream). I also took a few liberties with the naming :).
Here are the classes:
[Serializable()]
public class Car
{
[System.Xml.Serialization.XmlElement("StockNumber")]
public string StockNumber { get; set; }
[System.Xml.Serialization.XmlElement("Make")]
public string Make { get; set; }
[System.Xml.Serialization.XmlElement("Model")]
public string Model { get; set; }
}
[Serializable()]
[System.Xml.Serialization.XmlRoot("CarCollection")]
public class CarCollection
{
[XmlArray("Cars")]
[XmlArrayItem("Car", typeof(Car))]
public Car[] Car { get; set; }
}
The Deserialize function:
CarCollection cars = null;
string path = "cars.xml";
XmlSerializer serializer = new XmlSerializer(typeof(CarCollection));
StreamReader reader = new StreamReader(path);
cars = (CarCollection)serializer.Deserialize(reader);
reader.Close();
And the slightly tweaked xml (I needed to add a new element to wrap <Cars>...Net is picky about deserializing arrays):
<?xml version="1.0" encoding="utf-8"?>
<CarCollection>
<Cars>
<Car>
<StockNumber>1020</StockNumber>
<Make>Nissan</Make>
<Model>Sentra</Model>
</Car>
<Car>
<StockNumber>1010</StockNumber>
<Make>Toyota</Make>
<Model>Corolla</Model>
</Car>
<Car>
<StockNumber>1111</StockNumber>
<Make>Honda</Make>
<Model>Accord</Model>
</Car>
</Cars>
</CarCollection>

You have two possibilities.
Method 1. XSD tool
Suppose that you have your XML file in this location C:\path\to\xml\file.xml
Open Developer Command Prompt
You can find it in Start Menu > Programs > Microsoft Visual Studio 2012 > Visual Studio Tools
Or if you have Windows 8 can just start typing Developer Command Prompt in Start screen
Change location to your XML file directory by typing cd /D "C:\path\to\xml"
Create XSD file from your xml file by typing xsd file.xml
Create C# classes by typing xsd /c file.xsd
And that's it! You have generated C# classes from xml file in C:\path\to\xml\file.cs
Method 2 - Paste special
Required Visual Studio 2012+
Copy content of your XML file to clipboard
Add to your solution new, empty class file (Shift+Alt+C)
Open that file and in menu click Edit > Paste special > Paste XML As Classes
And that's it!
Usage
Usage is very simple with this helper class:
using System;
using System.IO;
using System.Web.Script.Serialization; // Add reference: System.Web.Extensions
using System.Xml;
using System.Xml.Serialization;
namespace Helpers
{
internal static class ParseHelpers
{
private static JavaScriptSerializer json;
private static JavaScriptSerializer JSON { get { return json ?? (json = new JavaScriptSerializer()); } }
public static Stream ToStream(this string #this)
{
var stream = new MemoryStream();
var writer = new StreamWriter(stream);
writer.Write(#this);
writer.Flush();
stream.Position = 0;
return stream;
}
public static T ParseXML<T>(this string #this) where T : class
{
var reader = XmlReader.Create(#this.Trim().ToStream(), new XmlReaderSettings() { ConformanceLevel = ConformanceLevel.Document });
return new XmlSerializer(typeof(T)).Deserialize(reader) as T;
}
public static T ParseJSON<T>(this string #this) where T : class
{
return JSON.Deserialize<T>(#this.Trim());
}
}
}
All you have to do now, is:
public class JSONRoot
{
public catalog catalog { get; set; }
}
// ...
string xml = File.ReadAllText(#"D:\file.xml");
var catalog1 = xml.ParseXML<catalog>();
string json = File.ReadAllText(#"D:\file.json");
var catalog2 = json.ParseJSON<JSONRoot>();

The following snippet should do the trick (and you can ignore most of the serialization attributes):
public class Car
{
public string StockNumber { get; set; }
public string Make { get; set; }
public string Model { get; set; }
}
[XmlRootAttribute("Cars")]
public class CarCollection
{
[XmlElement("Car")]
public Car[] Cars { get; set; }
}
...
using (TextReader reader = new StreamReader(path))
{
XmlSerializer serializer = new XmlSerializer(typeof(CarCollection));
return (CarCollection) serializer.Deserialize(reader);
}

See if this helps:
[Serializable()]
[System.Xml.Serialization.XmlRootAttribute("Cars", Namespace = "", IsNullable = false)]
public class Cars
{
[XmlArrayItem(typeof(Car))]
public Car[] Car { get; set; }
}
.
[Serializable()]
public class Car
{
[System.Xml.Serialization.XmlElement()]
public string StockNumber{ get; set; }
[System.Xml.Serialization.XmlElement()]
public string Make{ get; set; }
[System.Xml.Serialization.XmlElement()]
public string Model{ get; set; }
}
And failing that use the xsd.exe program that comes with visual studio to create a schema document based on that xml file, and then use it again to create a class based on the schema document.

I don't think .net is 'picky about deserializing arrays'. The first xml document is not well formed.
There is no root element, although it looks like there is. The canonical xml document has a root and at least 1 element (if at all). In your example:
<Root> <-- well, the root
<Cars> <-- an element (not a root), it being an array
<Car> <-- an element, it being an array item
...
</Car>
</Cars>
</Root>

try this block of code if your .xml file has been generated somewhere in disk and if you have used List<T>:
//deserialization
XmlSerializer xmlser = new XmlSerializer(typeof(List<Item>));
StreamReader srdr = new StreamReader(#"C:\serialize.xml");
List<Item> p = (List<Item>)xmlser.Deserialize(srdr);
srdr.Close();`
Note: C:\serialize.xml is my .xml file's path. You can change it for your needs.

For Beginners
I found the answers here to be very helpful, that said I still struggled (just a bit) to get this working. So, in case it helps someone I'll spell out the working solution:
XML from Original Question. The xml is in a file Class1.xml, a path to this file is used in the code to locate this xml file.
I used the answer by #erymski to get this working, so created a file called Car.cs and added the following:
using System.Xml.Serialization; // Added
public class Car
{
public string StockNumber { get; set; }
public string Make { get; set; }
public string Model { get; set; }
}
[XmlRootAttribute("Cars")]
public class CarCollection
{
[XmlElement("Car")]
public Car[] Cars { get; set; }
}
The other bit of code provided by #erymski ...
using (TextReader reader = new StreamReader(path))
{
XmlSerializer serializer = new XmlSerializer(typeof(CarCollection));
return (CarCollection) serializer.Deserialize(reader);
}
... goes into your main program (Program.cs), in static CarCollection XCar() like this:
using System;
using System.IO;
using System.Xml.Serialization;
namespace ConsoleApp2
{
class Program
{
public static void Main()
{
var c = new CarCollection();
c = XCar();
foreach (var k in c.Cars)
{
Console.WriteLine(k.Make + " " + k.Model + " " + k.StockNumber);
}
c = null;
Console.ReadLine();
}
static CarCollection XCar()
{
using (TextReader reader = new StreamReader(#"C:\Users\SlowLearner\source\repos\ConsoleApp2\ConsoleApp2\Class1.xml"))
{
XmlSerializer serializer = new XmlSerializer(typeof(CarCollection));
return (CarCollection)serializer.Deserialize(reader);
}
}
}
}
Hope it helps :-)

Kevin's anser is good, aside from the fact, that in the real world, you are often not able to alter the original XML to suit your needs.
There's a simple solution for the original XML, too:
[XmlRoot("Cars")]
public class XmlData
{
[XmlElement("Car")]
public List<Car> Cars{ get; set; }
}
public class Car
{
public string StockNumber { get; set; }
public string Make { get; set; }
public string Model { get; set; }
}
And then you can simply call:
var ser = new XmlSerializer(typeof(XmlData));
var data = (XmlData)ser.Deserialize(XmlReader.Create(PathToCarsXml));

One liner:
var object = (Cars)new XmlSerializer(typeof(Cars)).Deserialize(new StringReader(xmlString));

Try this Generic Class For Xml Serialization & Deserialization.
public class SerializeConfig<T> where T : class
{
public static void Serialize(string path, T type)
{
var serializer = new XmlSerializer(type.GetType());
using (var writer = new FileStream(path, FileMode.Create))
{
serializer.Serialize(writer, type);
}
}
public static T DeSerialize(string path)
{
T type;
var serializer = new XmlSerializer(typeof(T));
using (var reader = XmlReader.Create(path))
{
type = serializer.Deserialize(reader) as T;
}
return type;
}
}

How about a generic class to deserialize an XML document
//++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
// Generic class to load any xml into a class
// used like this ...
// YourClassTypeHere InfoList = LoadXMLFileIntoClass<YourClassTypeHere>(xmlFile);
using System.IO;
using System.Xml.Serialization;
public static T LoadXMLFileIntoClass<T>(string xmlFile)
{
T returnThis;
XmlSerializer serializer = new XmlSerializer(typeof(T));
if (!FileAndIO.FileExists(xmlFile))
{
Console.WriteLine("FileDoesNotExistError {0}", xmlFile);
}
returnThis = (T)serializer.Deserialize(new StreamReader(xmlFile));
return (T)returnThis;
}
This part may, or may not be necessary. Open the XML document in Visual Studio, right click on the XML, choose properties. Then choose your schema file.

The idea is to have all level being handled for deserialization
Please see a sample solution that solved my similar issue
<?xml version="1.0" ?>
<TRANSACTION_RESPONSE>
<TRANSACTION>
<TRANSACTION_ID>25429</TRANSACTION_ID>
<MERCHANT_ACC_NO>02700701354375000964</MERCHANT_ACC_NO>
<TXN_STATUS>F</TXN_STATUS>
<TXN_SIGNATURE>a16af68d4c3e2280e44bd7c2c23f2af6cb1f0e5a28c266ea741608e72b1a5e4224da5b975909cc43c53b6c0f7f1bbf0820269caa3e350dd1812484edc499b279</TXN_SIGNATURE>
<TXN_SIGNATURE2>B1684258EA112C8B5BA51F73CDA9864D1BB98E04F5A78B67A3E539BEF96CCF4D16CFF6B9E04818B50E855E0783BB075309D112CA596BDC49F9738C4BF3AA1FB4</TXN_SIGNATURE2>
<TRAN_DATE>29-09-2015 07:36:59</TRAN_DATE>
<MERCHANT_TRANID>150929093703RUDZMX4</MERCHANT_TRANID>
<RESPONSE_CODE>9967</RESPONSE_CODE>
<RESPONSE_DESC>Bank rejected transaction!</RESPONSE_DESC>
<CUSTOMER_ID>RUDZMX</CUSTOMER_ID>
<AUTH_ID />
<AUTH_DATE />
<CAPTURE_DATE />
<SALES_DATE />
<VOID_REV_DATE />
<REFUND_DATE />
<REFUND_AMOUNT>0.00</REFUND_AMOUNT>
</TRANSACTION>
</TRANSACTION_RESPONSE>
The above XML is handled in two level
[XmlType("TRANSACTION_RESPONSE")]
public class TransactionResponse
{
[XmlElement("TRANSACTION")]
public BankQueryResponse Response { get; set; }
}
The Inner level
public class BankQueryResponse
{
[XmlElement("TRANSACTION_ID")]
public string TransactionId { get; set; }
[XmlElement("MERCHANT_ACC_NO")]
public string MerchantAccNo { get; set; }
[XmlElement("TXN_SIGNATURE")]
public string TxnSignature { get; set; }
[XmlElement("TRAN_DATE")]
public DateTime TranDate { get; set; }
[XmlElement("TXN_STATUS")]
public string TxnStatus { get; set; }
[XmlElement("REFUND_DATE")]
public DateTime RefundDate { get; set; }
[XmlElement("RESPONSE_CODE")]
public string ResponseCode { get; set; }
[XmlElement("RESPONSE_DESC")]
public string ResponseDesc { get; set; }
[XmlAttribute("MERCHANT_TRANID")]
public string MerchantTranId { get; set; }
}
Same Way you need multiple level with car as array
Check this example for multilevel deserialization

If you're getting errors using xsd.exe to create your xsd file, then use the XmlSchemaInference class as mentioned on msdn. Here's a unit test to demonstrate:
using System.Xml;
using System.Xml.Schema;
[TestMethod]
public void GenerateXsdFromXmlTest()
{
string folder = #"C:\mydir\mydata\xmlToCSharp";
XmlReader reader = XmlReader.Create(folder + "\some_xml.xml");
XmlSchemaSet schemaSet = new XmlSchemaSet();
XmlSchemaInference schema = new XmlSchemaInference();
schemaSet = schema.InferSchema(reader);
foreach (XmlSchema s in schemaSet.Schemas())
{
XmlWriter xsdFile = new XmlTextWriter(folder + "\some_xsd.xsd", System.Text.Encoding.UTF8);
s.Write(xsdFile);
xsdFile.Close();
}
}
// now from the visual studio command line type: xsd some_xsd.xsd /classes

You can just change one attribute for you Cars car property from XmlArrayItem to XmlElment. That is, from
[System.Xml.Serialization.XmlRootAttribute("Cars", Namespace = "", IsNullable = false)]
public class Cars
{
[XmlArrayItem(typeof(Car))]
public Car[] Car { get; set; }
}
to
[System.Xml.Serialization.XmlRootAttribute("Cars", Namespace = "", IsNullable = false)]
public class Cars
{
[XmlElement("Car")]
public Car[] Car { get; set; }
}

My solution:
Use Edit > Past Special > Paste XML As Classes to get the class in your code
Try something like this: create a list of that class (List<class1>), then use the XmlSerializer to serialize that list to a xml file.
Now you just replace the body of that file with your data and try to deserialize it.
Code:
StreamReader sr = new StreamReader(#"C:\Users\duongngh\Desktop\Newfolder\abc.txt");
XmlSerializer xml = new XmlSerializer(typeof(Class1[]));
var a = xml.Deserialize(sr);
sr.Close();
NOTE: you must pay attention to the root name, don't change it. Mine is "ArrayOfClass1"

Parsing XML from string C#

I am trying to deserialize a string, response.Content, with this XML
<?xml version="1.0" encoding="utf-8"?><root><uri><![CDATA[http://api.bart.gov/api/stn.aspx?cmd=stns]]></uri><stations><station><name>12th St. Oakland City Center</name><abbr>12TH</abbr><gtfs_latitude>37.803664</gtfs_latitude><gtfs_longitude>-122.271604</gtfs_longitude><address>1245 Broadway</address><city>Oakland</city><county>alameda</county><state>CA</state><zipcode>94612</zipcode></station>
I am using this code to deserialize it:
var serializer = new XmlSerializer(typeof(Stations), new XmlRootAttribute("root"));
Stations result;
using (TextReader reader = new StringReader(response.Content))
{
result = (Stations)serializer.Deserialize(reader);
}
I then have the Stations class declared here
[XmlRoot]
public class Stations
{
[XmlElement]
public string name;
}
However, my name is null. Any idea why?

While using XmlSerializer you should imitate all the xml structure with your classes.
[XmlRoot(ElementName = "root")]
public class Root
{
[XmlArray(ElementName = "stations"), XmlArrayItem(ElementName = "station")]
public Station[] Stations { get; set; }
}
public class Station
{
[XmlElement(ElementName = "name")]
public string Name { get; set; }
}
Then you can deserialize your data in that way.
var data = ""; //your xml goes here
var serializer = new XmlSerializer(typeof(Root));
using (var reader = new StringReader(data))
{
var root = (Root)serializer.Deserialize(reader);
}

Stations is a list of Station objects. Stations does not have an element called Name, only Station does.
You should probably do something like
public Station[] Stations
in the root-class.
Then define a new class called Station with a Name property.

Stations should not be a class, it should be a collection of Station elements.

How to serialize the base class with derived classes

.
Hello,
I have this sample code :
public class Vehicule
{
public string Name { get; set; }
public Brand Brand { get; set; }
}
public class Car : Vehicule
{
public string Matriculation { get; set; }
}
public class Brand
{
public string Name { get; set; }
}
public class Renault : Brand
{
public string Information { get; set; }
}
If I create this instance :
var car = new Car { Name = "Clio", Matriculation = "XXX-XXX", Brand = new Renault { Name = "Renault", Information = "Contact Infos" } };
When I serialize this object like that :
var serializer = new XmlSerializer(typeof(Car), new Type[] { typeof(Renault)});
serializer.Serialize(wr, car);
I obtain this :
<?xml version="1.0" encoding="utf-8"?>
<Car xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
<Name>Clio</Name>
<Brand xsi:type="Renault">
<Name>Renault</Name>
<Information>Contact Infos</Information>
</Brand>
<Matriculation>XXX-XXX</Matriculation>
</Car>
But, in my project, I don't have to have informations on derived classes, I would like only elements of base classes from this instance like this :
var serializer = new XmlSerializer(typeof(Vehicule));
serializer.Serialize(wr, car);
The Xml :
<?xml version="1.0" encoding="utf-8"?>
<Vehicule xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
<Name>Clio</Name>
<Brand>
<Name>Renault</Name>
</Brand>
</Vehicule>
Can you please, help me to obtain the good Xml (only with base type Vehicule and Brand) ?
Many thanks

You can't magically serialize a derived class as it's base because
"...Serialization checks type of instance by calling Object.getType()
method. This method always returns the exact type of object."
http://bytes.com/topic/net/answers/809946-how-force-serialize-base-type
The solution here, if you really need to only serialize the base class is to implement the IXmlSerializable interface and create your own custom serializer.
IXmlSerializable:
http://msdn.microsoft.com/en-us/library/system.xml.serialization.ixmlserializable(v=vs.110).aspx
One more thought. If you can work around the limitation of outputting the extra XML elements, you are able to serialize the derived class using only the base object by either 1) using XmlIncludeAttributes on the base class to tell it which types to expect or 2) using the XmlSerializer constructor overload that takes a list of types.
Edit:
After thinking about this a little more, a workaround would be that you would add a Clone() method onto your base object, then serialize the clone of the base.
LinqPad code:
public class Vehicule
{
public string Name { get; set; }
public Brand Brand { get; set; }
public Vehicule Clone()
{
return new Vehicule { Name = this.Name, Brand = new Brand { Name = this.Brand.Name } };
}
}
public class Car : Vehicule
{
public string Matriculation { get; set; }
}
public class Brand
{
public string Name { get; set; }
}
public class Renault : Brand
{
public string Information { get; set; }
}
void Main()
{
var car = new Car { Name = "Clio", Matriculation = "XXX-XXX", Brand = new Renault { Name = "Renault", Information = "Contact Infos" } };
var vehicle = car as Vehicule;
var serializer = new System.Xml.Serialization.XmlSerializer(typeof(Vehicule));
XmlWriterSettings settings = new XmlWriterSettings
{
Encoding = new UnicodeEncoding(false, false),
Indent = false,
OmitXmlDeclaration = false
};
using(StringWriter textWriter = new StringWriter())
using(XmlWriter xmlWriter = XmlWriter.Create(textWriter, settings)) {
serializer.Serialize(xmlWriter, vehicle.Clone());
textWriter.ToString().Dump();
}
}

This is one of the issues with inheritance, and another reason to favor composition imho.
I ran into the same issue on a mobile app where I had a Contact class that derives from ContactSummary. The repository returns Contact instances, but in lots of cases I only wanted the ContactSummary going over the wire to save on message sizes and data usage etc. The default Xml and Json serialisers would only work when the derived class was attributed with the [KnownType()] of the base class, but this still meant all those extra properties going over the wire.
Using inheritance it is problematic to achieve a viable solution, and I didn't want to resort to custom serialisers, and if the solution is to pollute the DTO with copy constructors and clone properties, then why not change the DTO to use composition instead?
If you have control over your DTOs, then restructuring them to use composition rather than inheritance may be the answer. In my example it was fairly simple...
public class ContactSummary
{
public string Name { get; set;}
public string Phone { get; set; }
}
public class Contact
{
public ContactSummary Summary { get; set; }
// ... other properties here
}
In your example, Car would need to contain a reference to Vehicle not inherit from it - something like...
[KnowTypes(typeof(Renault))]
public class Vehicle
{
public string Name { get; set; }
public Brand Brand { get; set; }
}
public class Car
{
public Vehicle Vehicle { get; set; }
public string Matriculation { get; set; }
}
Then when you want the 'base' type in your example, simply serialise Car.Vehicle.

I had the same problem and I got around it by re-mapping the inheriting class into the base class using AutoMapper:
MapperConfiguration config = new MapperConfiguration(cfg => cfg.CreateMap<Inheriting, Base>());
IMapper mapper = config.CreateMapper();
var baseObj = mapper.Map<Base>(InheritingObj);
There is not much you can customize on XmlSerializer out-of-the-box options.

How to serialize a class to an xml structure?

How could I serialize my below class {0} to the below xml {1}.
So, the class name, property names should match to the xml.
{0}:
[Serializable]
public class ProfileSite
{
[XmlAttribute("profileId")]
public int ProfileId { get; set; }
[XmlAttribute("siteId")]
public int SiteId { get; set; }
public Link[] Links { get; set; }
public XElement Deserialize()
{
}
}
{1}:
<profileSite profileId="" siteId="">
<links>
<link>
<originalUrl></originalUrl>
<isCrawled></isCrawled>
<isBroken></isBroken>
<isHtmlPage></isHtmlPage>
<firstAppearedLevel></firstAppearedLevel>
</link>
</links>
</profileSite>
Many thanks,

[XmlRoot("profileSite")]
public class ProfileSite
{
[XmlAttribute("profileId")]
public int ProfileId { get; set; }
[XmlAttribute("siteId")]
public int SiteId { get; set; }
[XmlArray("links"), XmlArrayItem("link")]
public Link[] Links { get; set; }
}
then:
var ser = new XmlSerializer(typeof(ProfileSite));
var site = (ProfileSite) ser.Deserialize(source);

The first step is to mark up your class with the relevant Xml... attributes which control the sreialization and whether to have attributes or elements. Your requirement basically changes the case, and has properties of the main object as attributes and properties of all child Link objects as elements:
[XmlRoot("profileSite")]
public class ProfileSite
{
[XmlAttribute("profileId")]
public int ProfileId { get; set; }
[XmlAttribute("siteId")]
public int SiteId { get; set; }
[XmlArray("links"), XmlArrayItem("link")]
public Link[] Links { get; set; }
}
public class Link
{
[XmlElement("originalUrl")]
public string OriginalUrl{get;set;}
// You other props here much like the above
}
Then to serialize it use XmlSerializer.Serialize there are many overloads taking varios places to output the result. For testing you can use the Console.Out.
XmlSerializer serializer = new XmlSerializer(typeof(ProfileSite));
serializer.Serialize(Console.Out, obj);
You may want to add an empty namespace manager, which stops the ugly extra xmlns attributes:
XmlSerializerNamespaces ns = new XmlSerializerNamespaces();
ns.Add("","");
XmlSerializer serializer = new XmlSerializer(typeof(ProfileSite));
serializer.Serialize(Console.Out, obj,ns);
Output of the above using this example object:
var obj = new ProfileSite{
ProfileId=1,
SiteId=2,
Links = new[]{
new Link{OriginalUrl="www.google.com" },
new Link{OriginalUrl="www.foo.com" }
}};
is this:
<?xml version="1.0" encoding="utf-8"?>
<profileSite profileId="1" siteId="2">
<links>
<link>
<originalUrl>www.google.com</originalUrl>
</link>
<link>
<originalUrl>www.foo.com</originalUrl>
</link>
</links>
</profileSite>
Finally, here's a working example for you to play around with: http://rextester.com/XCJHD55693

Xml serialization - Change the serialization type of xml

I have a struct like this:
public struct Vehicles
{
public string Name { get; set; }
public string Count { get; set; }
public List<Car> Cars { get; set; }
}
public struct Car
{
public string Name { get; set; }
public int Count { get; set; }
public List<Tire> Tires { get; set; }
}
public struct Tire
{
public string Brand { get; set; }
public int Count { get; set; }
public int UniqueCount { get; set; }
public List<Dimension> Dimensions { get; set; }
}
public struct Dimension
{
public string Size { get; set; }
public int AlternateSize { get; set; }
}
When I serialize "Vehicles" it is like:
<root xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns:xsd="http://www.w3.org /2001/XMLSchema">
<Vehicles>
<Name>SuperVehicles</Name>
<Cars>
<Car>
<Name>BMW</Name>
<Count>29</Count>
<Tires>
<Tire>
<Name>DMZ</Name>
<Count>26</Count>
<UniqueCount>24</UniqueCount>
<Dimensions>
<Dimension>
<Size>70x570</Size>
<AlternateSize>70x580</AlternateSize>
</Dimension>
<Dimension>
<Size>60x570</Size>
<AlternateSize>60x580</AlternateSize>
</Dimension>
<Dimension>
<Size>50x570</Size>
<AlternateSize>50x580</AlternateSize>
</Dimension>
</Dimensions>
</Tire>
</Tires>
</Car>
</Cars>
</Vehicles>
</root>
Now the problem is, I want to serialize it like this:
<root>
<vehicles vehicleName="superVehicles" vehicleCount="50" carName="BMW"
carCount="25" tireBrand="kamu" tireCount="15" tireUniqueCount="15"
dimensionSize="70x570" dimensionAlternateSize="70x580" />
<vehicles vehicleName="superVehicles" vehicleCount="35" carName="MERCEDES"
carCount="22" tireBrand="kamu" tireCount="12" tireUniqueCount="12"
dimensionSize="60x570" dimensionAlternateSize="60x580" />
<vehicles vehicleName="superVehicles" vehicleCount="35" carName="PORSCHE"
carCount="22" tireBrand="kamu" tireCount="12" tireUniqueCount="12"
dimensionSize="60x570" dimensionAlternateSize="60x580" />
</root>
Do I have to change the structure and avoid the groupings or is there any way to create a schema for xml serialization to gather this result.
Summary:
I get all the child items in a new tag when I serialize the root struct to xml but I need to take them as properties of an instance that create only the count of root (Vehicles in this situation) element of rows to xml.

You need to do manual serialization.
Here is how you can implement this using System.Xml.Linq :
var xmlElementsVehicles = new[]{
new XElement("vehicles ", new object[]
{
new XElement("vehicleName", "superVehicles"),
new XElement("vehicleCount", 35),
new XElement("carName", "PORSCHE"),
new XElement("carCount", 2)
}),
new XElement("vehicles ", new object[]
{
new XElement("vehicleName", "superVehicles"),
new XElement("vehicleCount", 35),
new XElement("carName", "PORSCHE"),
new XElement("carCount", 2)
})
};
var root = new XElement("root", xmlElementsVehicles );
var myXml = new XDocument(new XDeclaration("1.0", "utf-8", "yes"), root);
using (var xmlWriter = XmlWriter.Create(stream))
{
myXml.Save(xmlWriter);
}

To use XmlSerializer the model must roughly be the same as the layout; a few things can change (names, etc). However, your model is nothing like the XML. Three options, then:
create a second DTO model that looks like the XML (you can use xsd.exe on the sample XML to automate this), and use XmlSerializer
don't use XmlSerializer, but build the XML somehow else (XmlDocument or XDocument would be the obvious two, or XmlWriter if the size is very large)
use something like xslt to reshape the XML after writing
There is nothing "easy" can be done to make XmlSerializer write that model into your desired XML.