Develop Reference

C# XML Serialization Observable Collection

I'm trying to deserialize a bunch of data that was serialized by an old version of the code. When the data was serialized the classes had a different structure from the current class structure. To keep this data working in my new code, I hade to add the old classes structure to the code just for import this serialized data. I'm calling this classes as 'class'_oldVersions. To deserialize, I'm using this code:
className_oldVersions temp_className = new className_oldVersions();
XmlSerializer testSerializer = new XmlSerializer(typeof(className_oldVersions),
new XmlRootAttribute { ElementName = "className" });
temp_className = (ObservedData_OldVersions)testSerializer.Deserialize(ms_MemoryStream);
This code works fine, and I can deserialize the data using a diffent class name from the original. My problem is when I try to use this same procedure to deserialize an observable collection.
I created a code that reproduce my problem. In this code I serialize an observable collection of class OptimizationVariables and I would like to deserialize to an observable collection of class OptimizationVariablies_NewClass that has similar structure from the original one.
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.IO;
using System.Threading.Tasks;
using System.Xml.Serialization;
using System.Collections.ObjectModel;
public class OptimizationVariables
{
public string VariableName { get; set; }
}
public class OptimizationVariables_NewClass
{
public string VariableName { get; set; }
}
public class ModelsCollection
{
private ModelsCollection()
{
}
private ObservableCollection<OptimizationVariables> m_optimizationVariables =
new ObservableCollection<OptimizationVariables>();
public ObservableCollection<OptimizationVariables> OptimizationVariables
{
get { return m_optimizationVariables; }
set { m_optimizationVariables = value; }
}
private ObservableCollection<OptimizationVariables_NewClass> m_optimizationVariables_NewClass =
new ObservableCollection<OptimizationVariables_NewClass>();
public ObservableCollection<OptimizationVariables_NewClass> OptimizationVariables_NewClass
{
get { return m_optimizationVariables_NewClass; }
set { m_optimizationVariables_NewClass = value; }
}
}
class Program
{
static void Main(string[] args)
{
//Here I serialize an ObservableCollection of 2 OptimizationVariables instances
Serialize();
//Here I deserialize for the same class and works fine
Deserialize();
//Here I try to deserialize to a new class with same structure, but different name. I a have an error.
Deserialize2NewClass();
}
static void Serialize()
{
MemoryStream ms;
ObservableCollection<OptimizationVariables> OptimizationVariables2Serialize = new ObservableCollection<OptimizationVariables>();
OptimizationVariables opt_var1 = new OptimizationVariables();
opt_var1.VariableName = "Variable Name 1";
OptimizationVariables2Serialize.Add(opt_var1);
OptimizationVariables opt_var2 = new OptimizationVariables();
opt_var1.VariableName = "Variable Name 2";
OptimizationVariables2Serialize.Add(opt_var1);
ms = new MemoryStream();
XmlSerializer serializer = new XmlSerializer(typeof(ObservableCollection<OptimizationVariables>));
serializer.Serialize(ms, OptimizationVariables2Serialize);
TextWriter sw = new StreamWriter("XML_File_x64.bin");
sw.WriteLine(Convert.ToBase64String(ms.ToArray()));
sw.Close();
}
static void Deserialize()
{
byte[] memoryData;
MemoryStream ms;
TextReader sw = new StreamReader("XML_File_x64.bin");
memoryData = Convert.FromBase64String(sw.ReadLine());
ms = new MemoryStream(memoryData);
ObservableCollection<OptimizationVariables> OptimizationVariablesDeserialized = new ObservableCollection<OptimizationVariables>();
XmlSerializer deserializer = new XmlSerializer(typeof(ObservableCollection<OptimizationVariables>));
OptimizationVariablesDeserialized = (ObservableCollection<OptimizationVariables>)deserializer.Deserialize(ms);
Console.Write(OptimizationVariablesDeserialized.Count());
sw.Close();
}
static void Deserialize2NewClass()
{
byte[] memoryData;
MemoryStream ms;
TextReader sw = new StreamReader("XML_File_x64.bin");
memoryData = Convert.FromBase64String(sw.ReadLine());
ms = new MemoryStream(memoryData);
ObservableCollection<OptimizationVariables_NewClass> OptimizationVariablesDeserialized = new ObservableCollection<OptimizationVariables_NewClass>();
XmlSerializer deserializer = new XmlSerializer(typeof(ObservableCollection<OptimizationVariables_NewClass>));
OptimizationVariablesDeserialized = (ObservableCollection<OptimizationVariables_NewClass>)deserializer.Deserialize(ms);
Console.Write(OptimizationVariablesDeserialized.Count());
sw.Close();
}
}

The problem in your code is that the class OptimizationVariables name is saved in the xml data when serializing. but for deserialization it expects a OptimizationVariables_NewClass name which is not within the xml file.
consider the fact that not only the variable names, but also the class names will be used when serializing a class or struct...
So, just changed your code to make it save the data in readable xml file, then replacing "OptimizationVariables" with "OptimizationVariables_NewClass" solved the error:
using System;
using System.Linq;
using System.IO;
using System.Xml.Serialization;
using System.Collections.ObjectModel;
public class OptimizationVariables
{
public string VariableName { get; set; }
}
public class OptimizationVariables_NewClass
{
public string VariableName { get; set; }
}
public class ModelsCollection
{
private ModelsCollection()
{
}
private ObservableCollection<OptimizationVariables> m_optimizationVariables =
new ObservableCollection<OptimizationVariables>();
public ObservableCollection<OptimizationVariables> OptimizationVariables
{
get { return m_optimizationVariables; }
set { m_optimizationVariables = value; }
}
private ObservableCollection<OptimizationVariables_NewClass> m_optimizationVariables_NewClass =
new ObservableCollection<OptimizationVariables_NewClass>();
public ObservableCollection<OptimizationVariables_NewClass> OptimizationVariables_NewClass
{
get { return m_optimizationVariables_NewClass; }
set { m_optimizationVariables_NewClass = value; }
}
}
class Program
{
static void Main(string[] args)
{
Serialize();
Deserialize();
Deserialize2NewClass();
}
static void Serialize()
{
ObservableCollection<OptimizationVariables> OptimizationVariables2Serialize = new ObservableCollection<OptimizationVariables>();
OptimizationVariables opt_var1 = new OptimizationVariables();
opt_var1.VariableName = "Variable Name 1";
OptimizationVariables2Serialize.Add(opt_var1);
OptimizationVariables opt_var2 = new OptimizationVariables();
opt_var1.VariableName = "Variable Name 2";
OptimizationVariables2Serialize.Add(opt_var1);
TextWriter writer = new StreamWriter("XML_File.xml");
XmlSerializer serializer = new XmlSerializer(typeof(ObservableCollection<OptimizationVariables>));
serializer.Serialize(writer, OptimizationVariables2Serialize);
writer.Close();
}
static void Deserialize()
{
TextReader sw = new StreamReader("XML_File.xml");
ObservableCollection<OptimizationVariables> OptimizationVariablesDeserialized = new ObservableCollection<OptimizationVariables>();
XmlSerializer deserializer = new XmlSerializer(typeof(ObservableCollection<OptimizationVariables>));
OptimizationVariablesDeserialized = (ObservableCollection<OptimizationVariables>)deserializer.Deserialize(sw);
Console.Write(OptimizationVariablesDeserialized.Count());
sw.Close();
}
static void Deserialize2NewClass()
{
TextReader sw = new StreamReader("XML_File.xml");
var str = sw.ReadToEnd();
sw.Close();
str = str.Replace("OptimizationVariables", "OptimizationVariables_NewClass");
var stream = new StringReader(str);
ObservableCollection<OptimizationVariables_NewClass> OptimizationVariablesDeserialized = new ObservableCollection<OptimizationVariables_NewClass>();
XmlSerializer deserializer = new XmlSerializer(typeof(ObservableCollection<OptimizationVariables_NewClass>));
OptimizationVariablesDeserialized = (ObservableCollection<OptimizationVariables_NewClass>)deserializer.Deserialize(stream);
Console.Write(OptimizationVariablesDeserialized.Count());
}
}
now it works fine!

How can I serialize a class to XML using Windows Phone 8 SDK?

I've got one attribute (_XMLPlaylist) that I want to serialize in a XML-file likeshown in the code:
private void btn_Save_Click(object sender, RoutedEventArgs e)
{
_Playlist.Pl_Name = tb_Name.Text.ToString();
_XMLPlaylist.Playlists.Add(_Playlist);
IsolatedStorageFile isoStore = IsolatedStorageFile.GetUserStoreForApplication();
using (IsolatedStorageFileStream isoStream = new IsolatedStorageFileStream(StudiCast.Resources.AppResources.Playlists, FileMode.CreateNew, isoStore))
{
using (StreamWriter writer = new StreamWriter(isoStream))
{
var ser = new XmlSerializer(typeof(XMLPlaylist));
ser.Serialize(writer, _XMLPlaylist);
writer.Close();
}
isoStream.Close();
}
}
The Type XMLPlaylist looks as follows:
class XMLPlaylist
{
public XMLPlaylist()
{
Playlists = new List<Playlist>();
}
public List<Playlist> Playlists;
}
And the class Playlist like that:
class Playlist
{
public Playlist()
{
Casts = new List<Cast>();
}
public string Pl_Name;
public List<Cast> Casts;
}
'Cast' owns two strings. In .NET 4 I used the keyword [Serializable] in front of the class's name but there isn't the [Serializable]-attribute anymore.
Need fast Help please!
Edit:
Error at 'var ser = new XmlSerializer(typeof(XMLPlaylist));':
In System.InvalidOperationException an unhandled error of type "System.Xml.Serialization.ni.dll" occurs.

XmlSerializer can serialize only public classes - make your class XMLPlaylist public (also all properties/classes you want to serialize - so Playlist should be public as well).

XmlSerializer deserialize class hierarchy

I got some issues with the XmlSerializer in .NET.
Here I have a small example I built up just right now. (also available # gist https://gist.github.com/2d84be9041a3f9c06237)
using System.IO;
using System.Xml.Serialization;
namespace XmlSerializingSample
{
internal class Program
{
private static void Main(string[] args)
{
var specialType = new SpecialType()
{
Id = 1,
Name = "test"
};
var serializer = new XmlSerializer(typeof (SpecialType));
var des = new XmlSerializer(typeof (BaseType));
using (var memeStream = new MemoryStream())
{
serializer.Serialize(memeStream, specialType);
memeStream.Flush();
memeStream.Seek(0, SeekOrigin.Begin);
var instance = des.Deserialize(memeStream); // Here it throws the exception
}
}
}
[XmlInclude(typeof(SpecialType))]
[XmlType("baseType")]
public class BaseType
{
public long Id { get; set; }
}
[XmlRoot("special")]
public class SpecialType : BaseType
{
public string Name { get; set; }
}
}
In line 24 of the code I get an InvalidOperationException stating "{"<special xmlns=''> wurde nicht erwartet."}" [yes, it's german]
All Posts I found stated, after adding XmlIncludeAttribute on the base type being deserialized, this should work. Did I forget sth.?
Regards,
MacX

The problem is that your Serializer is serializing the SpecialType with a root element as so:
<special ...>
<Id>...
But then you try to Deserialize it using var des = new XmlSerializer(typeof (BaseType)); it knows about both types but it doesn't know how to handle the root element in the xml.
If you want this to work, you need to also set the root element of the base type to serialize as special. In other words, you would need to do this:
[XmlInclude(typeof(SpecialType))]
[XmlType("baseType")]
[XmlRoot("special")]
public class BaseType
{
public long Id { get; set; }
}
That way, the deserializer knows how to handle special as the root element.
I don't think there are other simple alternatives to make this work out of the box.
Update
This is another alternative using the XmlAttributeOverrides class.
LinqPad code:
void Main()
{
var specialType = new SpecialType()
{
Id = 1,
Name = "test"
};
var serializer = new XmlSerializer(typeof (SpecialType));
XmlAttributeOverrides attrOverrides = new XmlAttributeOverrides();
XmlAttributes attrs = new XmlAttributes();
// Create an XmlRootAttribute to override.
XmlRootAttribute attr = new XmlRootAttribute();
attr.ElementName = "special";
// Add the XmlRootAttribute to the collection of objects.
attrs.XmlRoot=attr;
attrOverrides.Add(typeof(BaseType), attrs);
var des = new XmlSerializer(typeof (BaseType), attrOverrides);
using (var memeStream= new MemoryStream())
{
serializer.Serialize(memeStream, specialType);
memeStream.Flush();
memeStream.Seek(0, SeekOrigin.Begin);
var instance = des.Deserialize(memeStream);
}
}
[XmlInclude(typeof(SpecialType))]
[XmlType("baseType")]
public class BaseType
{
public long Id { get; set; }
}
[XmlRoot("special")]
public class SpecialType : BaseType
{
public string Name { get; set; }
}

Model to XML file mapping - How to find location of deserialized object in soucre XML File.

Which is the best way to map my model to XML file using c# serializer. I mean that if for example I select an deserialized object I could be able to find the xml source text in XML file.

I got a working sample for you and you can explore further on it.
using System;
using System.Linq;
using System.Xml;
using System.Xml.Linq;
using System.Xml.Serialization;
using System.Collections.Generic;
using System.IO;
namespace ConsoleApplication5
{
public class Person
{
public int Age { get; set; }
public string Name { get; set; }
public int XMLLine { get; set; }
}
public class Persons : List<Person> { }
class Program
{
static void Main(string[] args)
{
//create your objects
Person p = new Person();
p.Age = 35;
p.Name = "Arnold";
Person p2 = new Person();
p2.Age = 36;
p2.Name = "Tom";
Persons ps = new Persons();
ps.Add(p);
ps.Add(p2);
//Serialize them to XML
XmlSerializer xs = new XmlSerializer(typeof(Persons));
XDocument d = new XDocument();
using (XmlWriter xw = d.CreateWriter())
xs.Serialize(xw, ps);
//print xml
//System.Diagnostics.Debug.WriteLine(d.ToString());
// it will produce following xml. You can save it to file.
//I have saved it to variable xml for demo
string xml = #"<ArrayOfPerson>
<Person>
<Age>35</Age>
<Name>Arnold</Name>
<XMLLine>0</XMLLine>
</Person>
<Person>
<Age>36</Age>
<Name>Tom</Name>
<XMLLine>0</XMLLine>
</Person>
</ArrayOfPerson>";
XDocument xdoc = XDocument.Parse(xml, LoadOptions.SetLineInfo);
// A little trick to get xml line
xdoc.Descendants("Person").All(a => { a.SetElementValue("XMLLine", ((IXmlLineInfo)a).HasLineInfo() ? ((IXmlLineInfo)a).LineNumber : -1); return true; });
//deserialize back to object
Persons pplz = xs.Deserialize((xdoc.CreateReader())) as Persons;
pplz.All(a => { Console.WriteLine(string.Format("Name {0} ,Age{1} ,Line number of object in XML File {2}", a.Name, a.Age, a.XMLLine)); return true; });
Console.ReadLine();
}
}
}
and It will give your results like
Name Arnold ,Age35 ,Line number of object in XML File 2
Name Tom ,Age36 ,Line number of object in XML File 7

You can try this extension method:
public static string ToXml<T>(this object obj)
{
using (var memoryStream = new MemoryStream())
{
using (TextWriter streamWriter = new StreamWriter(memoryStream))
{
var xmlSerializer = new XmlSerializer(typeof(T));
xmlSerializer.Serialize(streamWriter, obj);
return Encoding.ASCII.GetString(memoryStream.ToArray());
}
}
}
public static void ToXmlFile<T>(this object obj, string fileName)
{
using (TextWriter streamWriter = new StreamWriter(fileName))
{
var xmlSerializer = new XmlSerializer(typeof(T));
xmlSerializer.Serialize(streamWriter, obj);
}
}
USAGE:
// you will get this on a string variable
var xmlString = yourModel.ToXml<YourModel>();
// you will save our object in a file.
yourModel.ToXmlFile<YourModel>(#"C:\yourModelDump.xml");
Please be noted to add SerializableAttribute on your class
[Serializable]
public class YourModel
{
//...
}
This should do it

Mapping a class and xml file using serialization

If I create a class in C#, how can I serialize/deserialize it to a file? Is this somethat that can be done using built in functionality or is it custom code?

XmlSerializer; note that the exact xml names can be controlled through various attributes, but all you really need is:
a public type
with a default constructor
and public read/write members (ideally properties)
Example:
using System;
using System.Xml;
using System.Xml.Serialization;
public class Person {
public string Name { get; set; }
}
static class Program {
static void Main() {
Person person = new Person { Name = "Fred"};
XmlSerializer ser = new XmlSerializer(typeof(Person));
// write
using (XmlWriter xw = XmlWriter.Create("file.xml")) {
ser.Serialize(xw, person);
}
// read
using (XmlReader xr = XmlReader.Create("file.xml")) {
Person clone = (Person) ser.Deserialize(xr);
Console.WriteLine(clone.Name);
}
}
}

You need to use class XmlSerializer. Main methods are Serialize and Deserialize. They accept streams, text readers\writers and other classes.
Code sample:
public class Program
{
public class MyClass
{
public string Name { get; set; }
}
static void Main(string[] args)
{
var myObj = new MyClass { Name = "My name" };
var fileName = "data.xml";
var serializer = new XmlSerializer(typeof(MyClass));
using (var output = new XmlTextWriter(fileName, Encoding.UTF8))
serializer.Serialize(output, myObj);
using (var input = new StreamReader(fileName))
{
var deserialized = (MyClass)serializer.Deserialize(input);
Console.WriteLine(deserialized.Name);
}
Console.WriteLine("Press ENTER to finish");
Console.ReadLine();
}
}