Unable to deserialize property with specific name - c#

I'm using DataContractSerializer in C# to deserialize xml file into object, and everything was working fine :)
My issue started when I add a new property to my class and xml file. My new property name is EncryptionKey every time I deserialize file, property value remain null, but when i change the name of xml element and property to anEncryptionKey it deserialized correctly without changing anything else in the code.
Actually I try a lot of options for property name like EncryptKey, Encrypt and a lot more but I ended up with anEncryptionKey.
Maybe there is some constraint on properties names or something like that or I just need more caffeine to figure it out.
Xml file:
<?xml version="1.0" encoding="utf-8" ?>
<KioskSettings xmlns="http://schemas.datacontract.org/2004/07/Proxies" xmlns:i="http://www.w3.org/2001/XMLSchema-instance">
<ID>20198</ID>
<HeartBeatInterval>1</HeartBeatInterval>
<ServerURL></ServerURL>
<EncryptionKey>abcd</EncryptionKey>
</KioskSettings>
Code i use to deserialize file into object:
private KioskSettings ReadEngineSettingsFromSimulatorXmlFile()
{
KioskSettings engineSettings = new KioskSettings();
DataContractSerializer serializer = new DataContractSerializer(typeof(KioskSettings));
FileStream fs = new FileStream(#"c:\simulation.xml", FileMode.Open);
XmlDictionaryReader reader = XmlDictionaryReader.CreateTextReader(fs, new XmlDictionaryReaderQuotas());
while (reader.Read())
{
switch (reader.NodeType)
{
case XmlNodeType.Element:
if (serializer.IsStartObject(reader))
{
engineSettings = (KioskSettings)serializer.ReadObject(reader);
}
break;
}
}
fs.Flush(); fs.Close();
reader.Close();
reader = null;
serializer = null;
return engineSettings;
}
KioskSettings class:
public class KioskSettings
{
public string ID { get; set; }
public int HeartBeatInterval {get; set;}
public string ServerURL {get; set;}
public string EncryptionKey { get; set; }
}
Note: Above code may contains some syntax errors because I modify it to make it short.


You need to ensure your XML nodes are in alphabetical order:
<?xml version="1.0" encoding="utf-8" ?>
<KioskSettings xmlns="http://schemas.datacontract.org/2004/07/Proxies" xmlns:i="http://www.w3.org/2001/XMLSchema-instance">
<EncryptionKey>abcd</EncryptionKey>
<HeartBeatInterval>1</HeartBeatInterval>
<ID>20198</ID>
<ServerURL></ServerURL>
</KioskSettings>
Serialize a KioskSettings object with the following code, and ensure your XML takes the same form:
public static string DataContractSerializeObject<T>(T objectToSerialize)
{
using (MemoryStream memStm = new MemoryStream())
{
var serializer = new DataContractSerializer(typeof(T));
serializer.WriteObject(memStm, objectToSerialize);
memStm.Seek(0, SeekOrigin.Begin);
using (var streamReader = new StreamReader(memStm))
{
string result = streamReader.ReadToEnd();
return result;
}
}
}
If you need to preserve a specific order then specify the DataMember attribute on your class properties - Data Member Order e.g.
[DataContract]
public class KioskSettings
{
[DataMember(Order = 1)]
public string ID { get; set; }
[DataMember(Order = 2)]
public int HeartBeatInterval { get; set; }
[DataMember(Order = 3)]
public string ServerURL { get; set; }
[DataMember(Order = 4)]
public string EncryptionKey { get; set; }
}

Related

Reading an XML and Parsing as Class and Properties [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"

Trouble deserializing XML into a List<T>

Here's the XML file I'm trying to deserialize:
<?xml version="1.0" encoding="utf-8"?>
<EntityType>
<Name>SomeName</Name>
<Components>
<ComponentAssembly>Some assembly name</ComponentAssembly>
</Components>
</EntityType>
And here is the Data contract I am using to deserialize it:
using System.Collections.Generic;
using System.Runtime.Serialization;
namespace GameUtilities.Entities.DataContracts
{
[DataContract(Name="EntityType",Namespace="")]
public class EntityTypeData
{
[DataMember(IsRequired=true,Order = 0)]
public string Name { get; private set; }
[DataMember(IsRequired=false,Order=1)]
public List<ComponentEntry> Components { get; private set; }
public EntityTypeData(string name, List<ComponentEntry> components = null)
{
Name = name;
if(components == null)
{
Components = new List<ComponentEntry>();
}
else
{
Components = components;
}
}
}
[DataContract]
public class ComponentEntry
{
[DataMember(IsRequired = true, Order = 0)]
public string ComponentAssembly { get; private set; }
public ComponentEntry(string componentAssembly)
{
ComponentAssembly = componentAssembly;
}
}
}
Deserializing it works correctly, but the Components list is always empty, no matter how many entrys I put inside the tags. I have tried marking the [DataMemeber] attribute for Components as "IsRequired=true", and deserialization still completes without error, but the List is not getting populated. Can you see any issues with my data contract that would make this fail?
EDIT: As a test, I ran an object using the Data Contract above through a serializer to see what XML got spat out. Here's what I saw:
<EntityType xmlns:i="http://www.w3.org/2001/XMLSchema-instance">
<Name>TESTNAME</Name>
<Components xmlns:a="http://schemas.datacontract.org/2004/07/GameUtilities.Entities.DataContracts">
<a:ComponentEntry>
<a:ComponentAssembly>ONE</a:ComponentAssembly>
</a:ComponentEntry>
<a:ComponentEntry>
<a:ComponentAssembly>TWO</a:ComponentAssembly>
</a:ComponentEntry>
<a:ComponentEntry>
<a:ComponentAssembly>THREE</a:ComponentAssembly>
</a:ComponentEntry>
</Components>
Here is the serialization code I used:
public static void SerializeObject<T>(string path, T obj)
{
FileStream fs = new FileStream(path,FileMode.Create);
XmlDictionaryWriter writer = XmlDictionaryWriter.CreateTextWriter(fs);
DataContractSerializer ser = new DataContractSerializer(typeof(T));
//Serialize the data to a file
ser.WriteObject(writer, obj);
writer.Close();
fs.Close();
}
As you can see, there is a separate ComponentEntry being created for each ComponentAssembly that is listed. Is there a way to get rid of that and just get the ComponentAssembly?

Easiest solution would be to define a collection data contract instead. I don't think there is a way to have the ComponentEntry as a separate type, when using DataContractSerializer.
[DataContract(Name="EntityType",Namespace="")]
public class EntityTypeData
{
[DataMember(IsRequired=true,Order=0)]
public string Name { get; private set; }
[DataMember(IsRequired=false,Order=1)]
public ComponentList Components { get; private set; }
public EntityTypeData(string name, IEnumerable<string> components = null)
{
Name = name;
Components = new ComponentList();
if(components != null)
{
Components.AddRange(components);
}
}
}
[CollectionDataContract(ItemName = "ComponentAssembly", Namespace="")]
public class ComponentList : List<string> {}
var ser = new DataContractSerializer(typeof(EntityTypeData));
var entity = (EntityTypeData) ser.ReadObject(stream);

Deserialize xml to IList c#

I am trying to deserialize some xml into an IList, but I am having problems. This is what I have done so far:
The XML:
<?xml version="1.0" encoding="utf-8"?>
<Animals>
<Animal>
<Name>Cow</Name>
<Color>Brown</Color>
</Animal>
</Animals>
The Model:
[XmlRoot("Animals")]
public class Model
{
[XmlElement("Animal")]
public IList<Animal> AnimalList { get; set; }
}
public class Animal
{
[XmlElement("Name")]
public string Name{ get; set; }
[XmlElement("Color")]
public string Color{ get; set; }
}
Deserialization:
FileStream fs = new FileStream("file.xml", FileMode.Open);
XmlReader xml = XmlReader.Create(fs);
XmlSerializer ser = new XmlSerializer(typeof(List<Model>));
var list = (List<Model>)ser.Deserialize(xml);
I get an invalid operation exception when running the code above. What am I doing wrong?
Thanks,
James Ford

Try that:
// Create a new XmlSerializer instance with the type of the test class
XmlSerializer SerializerObj = new XmlSerializer(typeof(List<Model>));
// Create a new file stream for reading the XML file
FileStream ReadFileStream = new FileStream(#"C:\file.xml", FileMode.Open, FileAccess.Read, FileShare.Read);
// Load the object saved above by using the Deserialize function
List<Model> LoadedObj = (List<Model>)SerializerObj.Deserialize(ReadFileStream);
// Cleanup
ReadFileStream.Close();

The problem is that you are using an IList<Animal>. You need to use a List<Animal> so that it knows the specific type to use.
EDIT: Using the following code in LINQPad works perfectly. Only difference is I am loading the XML via string instead of file, but even when I change to a file it works fine. I just added the using for System.Xml.Serialization.
void Main()
{
string xml = #"<?xml version=""1.0"" encoding=""utf-8""?>
<Animals>
<Animal>
<Name>Cow</Name>
<Color>Brown</Color>
</Animal>
</Animals>";
XmlReader reader = XmlReader.Create(new StringReader(xml));
XmlSerializer ser = new XmlSerializer(typeof(Model));
var list = (Model)ser.Deserialize(reader);
list.Dump();
}
// Define other methods and classes here
[XmlRoot("Animals")]
public class Model
{
[XmlElement("Animal")]
public List<Animal> AnimalList { get; set; }
}
public class Animal
{
[XmlElement("Name")]
public string Name{ get; set; }
[XmlElement("Color")]
public string Color{ get; set; }
}

I think you need to change your XmlSerializer to this:
XmlSerializer ser = new XmlSerializer(typeof(Model));
Before you were trying to serialize a list of Models, you want to serialize a XML file into a Model, which contains a list of stuff.
Also, you need to change your ObjectList definition to
public List<Animal> AnimalList { get; set; }

Element was not expected While Deserializing an Array with XML Serialization

OK. I'm trying to work on communicating with the Pivotal Tracker API, which only returns data in an XML format. I have the following XML that I'm trying to deserialize into my domain model.
<?xml version="1.0" encoding="UTF-8"?>
<stories type="array" count="2" total="2">
<story>
<id type="integer">2909137</id>
<project_id type="integer">68153</project_id>
<story_type>bug</story_type>
<url>http://www.pivotaltracker.com/story/show/2909137</url>
<current_state>unscheduled</current_state>
<description></description>
<name>Test #2</name>
<requested_by>Anthony Shaw</requested_by>
<created_at type="datetime">2010/03/23 20:05:58 EDT</created_at>
<updated_at type="datetime">2010/03/23 20:05:58 EDT</updated_at>
</story>
<story>
<id type="integer">2909135</id>
<project_id type="integer">68153</project_id>
<story_type>feature</story_type>
<url>http://www.pivotaltracker.com/story/show/2909135</url>
<estimate type="integer">-1</estimate>
<current_state>unscheduled</current_state>
<description></description>
<name>Test #1</name>
<requested_by>Anthony Shaw</requested_by>
<created_at type="datetime">2010/03/23 20:05:53 EDT</created_at>
<updated_at type="datetime">2010/03/23 20:05:53 EDT</updated_at>
</story>
</stories>
My 'story' object is created as follows:
public class story
{
public int id { get; set; }
public int estimate { get; set; }
public int project_id { get; set; }
public string story_type { get; set; }
public string url { get; set; }
public string current_state { get; set; }
public string description { get; set; }
public string name { get; set; }
public string requested_by { get; set; }
public string labels { get; set; }
public string lighthouse_id { get; set; }
public string lighthouse_url { get; set; }
public string owned_by { get; set; }
public string accepted_at { get; set; }
public string created_at { get; set; }
public attachment[] attachments { get; set; }
public note[] notes { get; set; }
}
When I execute my deserialization code, I receive the following exception:
Exception:
There is an error in XML document (2, 2).
Inner Exception:
<stories xmlns=''> was not expected.
I can deserialize the individual stories just fine, I just cannot deserialize this xml into an array of 'story' objects
And my deserialization code (value is a string of the xml)
var byteArray = Encoding.ASCII.GetBytes(value);
var stream = new MemoryStream(byteArray);
var deserializedObject = new XmlSerializer(typeof (story[])).Deserialize(stream)
Does anybody have any ideas?

Let me offer a more concise solution. Set your deserialization up to look like this:
var deserializedObject = new XmlSerializer(typeof(story[]), new XmlRootAttribute("stories")).Deserialize(stream);
By specifying that second parameter in the XmlSerializer, you can avoid having to stub out that class. It lets the serializer know what the root element's name is.
For this to work, the name of the class that represents the array-element type must exactly match the XML name, e.g. class story {}, <story>. You can get around this (and I'd recommend it as a best practice anyway) by specifying the XmlType:
[XmlType("story")]
public class Story
{
...
}
I prefer to do this as it frees me from being stuck with the XML type name.

The problem is that you have no property named "stories". The XML Serializer has no idea what to do with the stories element when it sees it.
One thing you could try is to create a "stories" class:
public class stories : List<story> {}
and use
var byteArray = Encoding.ASCII.GetBytes(value);
stories deserializedObject = null;
using (var stream = new MemoryStream(byteArray))
{
var storiesSerializer = new XmlSerializer(typeof (stories));
deserializedObject = (stories)storiesSerializer .Deserialize(stream);
}

Try something like
public class stories
{
[XmlElement("story")]
public story[] storyarray { get; set; }
}
...
var byteArray = Encoding.ASCII.GetBytes(value);
XmlSerializer serializer = new XmlSerializer(typeof(stories));
stories myStories = null;
using (var stream = new MemoryStream(byteArray))
{
myStories = (stories)serializer.Deserialize(stream);
}
foreach (story stor in myStories.storyarray)
Console.WriteLine(stor.story_type);
Edit: Updated code sample to use using statement based on feedback.

XMSerializer expects an XML Namespace with which to understand your XML from.
xmlns="http://schemas.microsoft.com"
... ought to do. See the XML sample at the bottom of this page.

I would recommend that you generate an XSD from some sample XML you get from the web service. Then, with that XSD, you can generate the classes that have all the proper serialization attributes affixed to them.
To generate a schema (unless you prefer to write your own), open the sample XML file in Visual Studio, and select the XML -> Create Schema menu option. Save that XSD.
To generate the classes, run the XSD command from the VS command prompt. If you run it without parameters, it'll show you the command-line parameters you can use.
Now you can create a type-safe XML serializer.

How to Deserialize XML document

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"

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