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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"

How to rename the serialization root object name in a class?

(First, feel free to edit my title, I really don't find a better one for my problem)
I got my root class:
[XmlRoot("ProductList")]
public class Product
{
[XmlElement("Property1")]
public string property1 { get; set; }
[XmlElement("Property2")]
public Property2 property2 { get; set; }
[XmlArray("Property3Array")]
[XmlArrayItem("Property3ArrayItem")]
public List<Property3> property3{ get; set; }
}
I'm serializing the List of Products like this:
public void Execute(IJobExecutionContext context)
{
var products = _productionService.GetAllProducts();
XmlSerializer xs = new XmlSerializer(typeof(List<Product>));
using (StreamWriter sw = new StreamWriter("products.xml"))
{
xs.Serialize(sw, products);
}
}
The serialization is working properly BUT, in my products.xml file, the root node is:
<ArrayOfProduct xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
...
</ArrayOfProduct>
But I want the root element of the list to be named ProductList
I tried with [XmlRoot("ProductList")] But that doesn't work.
So, how can I rename the name of the root xml object of a List<Class> ?

Use a constructor overload that accepts an XmlRootAttribute.
var xs = new XmlSerializer(typeof(List<Product>), new XmlRootAttribute("ProductList"));
See this fiddle for a working demo.

Serializing list of nullable CDATA

I am trying to deserialize a XML file:
<?xml version="1.0" encoding="UTF-8"?>
<Foos xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
<FooList>
<Foo>
<Bar>bar value</Bar>
<Stack />
</Foo>
<Foo>
<Bar>bar value</Bar>
<Stack><![CDATA[This is some cdata, where <br> html code is left as is..]]></Stack>
</Foo>
</FooList>
</Foos>
Into the following class
[XmlRoot("Foos")]
public class Foos
{
[XmlArray("FooList")]
[XmlArrayItem("Foo")]
public List<Foo> FooList { get; set; }
}
public class Foo
{
public string Bar { get; set; }
[XmlElement("Stack", typeof(XmlCDataSection))]
public XmlCDataSection Stack {get; set; }
}
The problem is, that the resulting FooList of Foos only contains one (1) element, which is the first Foo block, with the Slack property set to null. If i add a CDATA value to stack, like with the second Foo block, then i will end up with a list of both elements.
For some reason, the deserialization stops, after reaching a CDATA value which is null.
I have tried creating a private string and creating the setter and getter of Stack, such that is uses the private property to store the CDATA string. This doesn't change anything.
Any suggestions?
Thank you.

Get to a cdata section through a string. Object model:
[Serializable]
[XmlRoot("Foos", Namespace="", IsNullable=false)]
public partial class Foos {
[XmlArrayItem ("Foo", IsNullable=false)] public Foo[] FooList { get; set; } }
[Serializable] public partial class Foo {
public string Bar { get; set; }
[XmlIgnore] public string Stack { get; set; }
[XmlElement("Stack", IsNullable=true)]
public XmlCDataSection StackCDATA {
get { return new XmlDocument().CreateCDataSection(Stack ?? String.Empty); }
set {
Stack = value == null
? String.Empty
: ( String.IsNullOrWhiteSpace(value.Value)
? String.Empty
: value.Value); } }
public override string ToString() { return String.Format("Bar:{0} Stack:{1}", Bar, Stack); } }
The right way to specify a null element is <Stack xsi:nil='true' />. With this addition, your deserialization will be flawless. However, on the first occurrence of <Stack />, the deserializer will encounter an unknown node. Workaround in case you are missing this attribute, subscribe to UnknownElement:
var fooSrlz = new XmlSerializer(typeof(Foo), new XmlRootAttribute { ElementName = "Foo", IsNullable = true} );
var foosMissedByDeserializer = new List<CData.Foo>();
srlz.UnknownElement += (/*object*/ sender, /*XmlElementEventArgs */ e) => {
using (TextReader tr = new StringReader(e.Element.OuterXml)) {
var foo = (CData.Foo)fooSrlz.Deserialize(tr);
foosMissedByDeserializer.Add (foo);
} };
Merge the deserialization result with the foosMissedByDeserializer above.

Deserialize XML to object (need to return a list of objects)

Started practicing with XML and C# and I have an error message of "There is an error in XML document (3,2)". After looking at the file, I can't see anything wrong with it (Mind you, I probably missed something since I'm a noob). I'm using a Console Application for C# right now. I'm trying to return a list of Adventurers and just a side note, the GEAR element is optional. Here is what I have so far:
XML File - Test1
<?xml version="1.0" encoding="utf-8"?>
<Catalog>
<Adventurer>
<ID>001</ID>
<Name>John Smith</Name>
<Address>123 Fake Street</Address>
<Phone>123-456-7890</Phone>
<Gear>
<Attack>
<Item>
<IName>Sword</IName>
<IPrice>15.00</IPrice>
</Item>
<Item>
<IName>Wand</IName>
<IPrice>20.00</IPrice>
</Item>
</Attack>
<Defense>
<Item>
<IName>Shield</IName>
<IPrice>5.00</IPrice>
</Item>
</Defense>
</Gear>
</Adventurer>
<Adventurer>
<ID>002</ID>
<Name>Guy noone likes</Name>
<Address>Some Big House</Address>
<Phone>666-666-6666</Phone>
<Gear></Gear>
</Adventurer>
</Catalog>
C# Classes
public class Catalog
{
List<Adventurer> Adventurers { get; set; }
}
public class Adventurer
{
public int ID { get; set; }
public string Name { get; set; }
public string Address { get; set; }
public string Phone { get; set; }
public Gear Gear { get; set; }
}
public class Gear
{
public List<Item> Attack { get; set; }
public List<Item> Defense { get; set; }
}
public class Item
{
public string IName { get; set; }
public decimal IPrice { get; set; }
}
Serialize Function - Where the Problem Occurs at Line 5
Catalog obj = null;
string path = #"C:\Users\Blah\Desktop\test1.xml";
XmlSerializer serializer = new XmlSerializer(typeof(Catalog));
StreamReader reader = new StreamReader(path);
obj = (Catalog)serializer.Deserialize(reader);
reader.Close();
Console.ReadLine();

The issue is the list of Adventurers in Catalog:
<?xml version="1.0" encoding="utf-8"?>
<Catalog>
<Adventurers> <!-- you're missing this -->
<Adventurer>
</Adventurer>
...
<Adventurer>
</Adventurer>
</Adventurers> <!-- and missing this -->
</Catalog>
You don't have the wrapping element for the Adventurers collection.
EDIT: By the way, I find the easiest way to build the XML structure and make sure it's compatible is to create the object(s) in C#, then run through the built-in XmlSerializer and use its XML output as a basis for any XML I create rather than forming it by hand.

First, "Adventurers" property is not public, it's inaccessible, I think that the best way to find the error is to serialize your object and then compare the result with your xml file.

Your XML doesn't quite line up with your objects... namely these two...
public string City { get; set; }
and
<Address>123 Fake Street</Address>
Change City to Address or vice versa and it should fix the problem.
Edit: Got this to work in a test project, combination of all our answers...
Add <Adventurers> tag after <Catalog> (and </Adventurers> before </Catalog>) and change
List<Adventurer> Adventurers { get; set; }
to
public List<Adventurer> Adventurers { get; set; }
and it works properly for me.

I was able to get your xml to deserialize with a couple minor changes (namely the public qualifier on Adventurer).
using System;
using System.Collections.Generic;
using System.Text;
using System.IO;
using System.Xml.Serialization;
namespace TempSandbox
{
[XmlRoot]
public class Catalog
{
[XmlElement("Adventurer")]
public List<Adventurer> Adventurers;
private readonly static Type[] myTypes = new Type[] { typeof(Adventurer), typeof(Gear), typeof(Item) };
private readonly static XmlSerializer mySerializer = new XmlSerializer(typeof(Catalog), myTypes);
public static Catalog Deserialize(string xml)
{
return (Catalog)Utils.Deserialize(mySerializer, xml, Encoding.UTF8);
}
}
[XmlRoot]
public class Adventurer
{
public int ID;
public string Name;
public string Address;
public string Phone;
[XmlElement(IsNullable = true)]
public Gear Gear;
}
[XmlRoot]
public class Gear
{
public List<Item> Attack;
public List<Item> Defense;
}
[XmlRoot]
public class Item
{
public string IName;
public decimal IPrice;
}
}
I'm using [XmlElement("Adventurer")] because the xml element names don't exactly match the class property names.
NOTE: I'm using a generic deserialization utility i already had on hand .

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.