Develop Reference

C# XmlSerializer Serialize the same class with different namespaces

Suppose I have a class:
using System.Xml;
using System.Xml.Serialization;
class Foo {
[XmlElement(ElementName="Bar")]
public string Bar {get; set;}
}
And now I want to serialize it. But. I want to specify different namespaces,
On one occasion I want my XML look like this:
<Foo xmlns:v1="http://site1">
<v1:Bar />
</Foo>
And on the other one - like this:
<Foo xmlns:v2="http://site2">
<v2:Bar />
</Foo>
I know I need to specify the namespace in the XmlElement attribute, but this is exactly what I want to avoid.
I could, of course, make 2 different classes which are identical except the field attributes, but this feels wrong somehow.
Is there any way I could force the XmlSerializer use the namespace I choose at runtime?

Yes; XmlAttributeOverrides:
static void Main()
{
var obj = new Foo { Bar = "abc" };
GetSerializer("http://site1").Serialize(Console.Out, obj);
Console.WriteLine();
GetSerializer("http://site2").Serialize(Console.Out, obj);
}
static XmlSerializer GetSerializer(string barNamespace)
{
var ao = new XmlAttributeOverrides();
var a = new XmlAttributes();
a.XmlElements.Add(new XmlElementAttribute { Namespace = barNamespace });
ao.Add(typeof(Foo), nameof(Foo.Bar), a);
return new XmlSerializer(typeof(Foo), ao);
}
However!!!
When you do this, it generates an additional assembly in memory each time; you must cache and re-use the serializer instances - usually via a concurrent dictionary or similar. For example:
private static readonly ConcurrentDictionary<string, XmlSerializer>
s_serializersByNamespace = new();
static XmlSerializer GetSerializer(string barNamespace)
{
if (!s_serializersByNamespace.TryGetValue(barNamespace, out var serializer))
{
lock (s_serializersByNamespace)
{
// double-checked, avoid dups
if (!s_serializersByNamespace.TryGetValue(barNamespace, out serializer))
{
var ao = new XmlAttributeOverrides();
var a = new XmlAttributes();
a.XmlElements.Add(new XmlElementAttribute { Namespace = barNamespace });
ao.Add(typeof(Foo), nameof(Foo.Bar), a);
serializer = new XmlSerializer(typeof(Foo), ao);
s_serializersByNamespace[barNamespace] = serializer;
}
}
}
return serializer;
}
Note: if you want the specific xmlns control too, that's XmlSerializerNamespaces:
var obj = new Foo { Bar = "abc" };
var ns = new XmlSerializerNamespaces();
ns.Add("v1", "http://site1");
GetSerializer("http://site1").Serialize(Console.Out, obj, ns);
Console.WriteLine();
ns = new XmlSerializerNamespaces();
ns.Add("v2", "http://site2");
GetSerializer("http://site2").Serialize(Console.Out, obj, ns);

Add root element attribute

I'm serializing a class like below
XmlSerializerNamespaces namespaces = new XmlSerializerNamespaces();
namespaces.Add(string.Empty, string.Empty);
StringWriter sw = new StringWriter();
XmlSerializer serializer1 = new XmlSerializer(typeof(List<student>), new XmlRootAttribute("Response"));
XmlTextWriter xmlWriter = new XmlTextWriter(sw);
serializer1.Serialize(xmlWriter, ls, namespaces);
sw.ToString()
The result string below
<?xml version="1.0" encoding="utf-16"?>
<Response><student><name>xxx</name></student></Response>
but, How can i add an attribute to the root element(Response)?
like below one
<?xml version="1.0" encoding="utf-16"?>
<Response status="1"><student><name>xxx</name></student></Response>

You just need to mark that property of the class with XmlAttribute, i.e.
class MyClass{
[XmlAttribute("status")]
public string ErrorStatus { get; set; }
}
Edit:
Just realised you are serializing the list directly. Put your list inside a parent class, Response, and add the above attribute to this Response class, then serialise the Response object.
Hope this helps.

You can create another object that contains the list, and then create a property to add the attribute to the root node.
The trick is to preface the list in this new class with an explicit type assignment to the Student type to avoid having your list nested within another parent node.
[XmlType(TypeName = "Response")]
public class ResponseObject
{
[XmlAttribute("status")]
public string file { get; set; }
[XmlElement("Student", Type = typeof(Student))]
public List<Student> studentList { get; set; }
}
Your code would then look like the following
XmlSerializerNamespaces namespaces = new XmlSerializerNamespaces();
namespaces.Add(string.Empty, string.Empty);
StringWriter sw = new StringWriter();
XmlSerializer serializer1 = new XmlSerializer(typeof(ResponseObject));
XmlTextWriter xmlWriter = new XmlTextWriter(sw);
//Creating new object and assign the existing list and status
ResponseObject resp = new ResponseObject();
resp.studentList = ls;
resp.status = 1;
//Serialize with the new object
serializer1.Serialize(xmlWriter, resp, namespaces);
sw.ToString()

How to serialize to xml using Linq

How can i serialize Instance of College to XML using Linq?
class College
{
public string Name { get; set; }
public string Address { get; set; }
public List<Person> Persons { get; set; }
}
class Person
{
public string Gender { get; set; }
public string City { get; set; }
}

You can't serialize with LINQ. You can use XmlSerializer.
XmlSerializer serializer = new XmlSerializer(typeof(College));
// Create a FileStream to write with.
Stream writer = new FileStream(filename, FileMode.Create);
// Serialize the object, and close the TextWriter
serializer.Serialize(writer, i);
writer.Close();

Not sure why people are saying you can't serialize/deserialize with LINQ. Custom serialization is still serialization:
public static College Deserialize(XElement collegeXML)
{
return new College()
{
Name = (string)collegeXML.Element("Name"),
Address = (string)collegeXML.Element("Address"),
Persons = (from personXML in collegeXML.Element("Persons").Elements("Person")
select Person.Deserialize(personXML)).ToList()
}
}
public static XElement Serialize(College college)
{
return new XElement("College",
new XElement("Name", college.Name),
new XElement("Address", college.Address)
new XElement("Persons", (from p in college.Persons
select Person.Serialize(p)).ToList()));
);
Note, this probably isn't the greatest approach, but it's answering the question at least.

You can't use LINQ. Look at the below code as an example.
// This is the test class we want to
// serialize:
[Serializable()]
public class TestClass
{
private string someString;
public string SomeString
{
get { return someString; }
set { someString = value; }
}
private List<string> settings = new List<string>();
public List<string> Settings
{
get { return settings; }
set { settings = value; }
}
// These will be ignored
[NonSerialized()]
private int willBeIgnored1 = 1;
private int willBeIgnored2 = 1;
}
// Example code
// This example requires:
// using System.Xml.Serialization;
// using System.IO;
// Create a new instance of the test class
TestClass TestObj = new TestClass();
// Set some dummy values
TestObj.SomeString = "foo";
TestObj.Settings.Add("A");
TestObj.Settings.Add("B");
TestObj.Settings.Add("C");
#region Save the object
// Create a new XmlSerializer instance with the type of the test class
XmlSerializer SerializerObj = new XmlSerializer(typeof(TestClass));
// Create a new file stream to write the serialized object to a file
TextWriter WriteFileStream = new StreamWriter(#"C:\test.xml");
SerializerObj.Serialize(WriteFileStream, TestObj);
// Cleanup
WriteFileStream.Close();
#endregion
/*
The test.xml file will look like this:
<?xml version="1.0"?>
<TestClass xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
<SomeString>foo</SomeString>
<Settings>
<string>A</string>
<string>B</string>
<string>C</string>
</Settings>
</TestClass>
*/
#region Load the object
// Create a new file stream for reading the XML file
FileStream ReadFileStream = new FileStream(#"C:\test.xml", FileMode.Open, FileAccess.Read, FileShare.Read);
// Load the object saved above by using the Deserialize function
TestClass LoadedObj = (TestClass)SerializerObj.Deserialize(ReadFileStream);
// Cleanup
ReadFileStream.Close();
#endregion
// Test the new loaded object:
MessageBox.Show(LoadedObj.SomeString);
foreach (string Setting in LoadedObj.Settings)
MessageBox.Show(Setting);

you have to use the XML serialization
static public void SerializeToXML(College college)
{
XmlSerializer serializer = new XmlSerializer(typeof(college));
TextWriter textWriter = new StreamWriter(#"C:\college.xml");
serializer.Serialize(textWriter, college);
textWriter.Close();
}

You can use that if you needed XDocument object after serialization
DataClass dc = new DataClass();
XmlSerializer x = new XmlSerializer(typeof(DataClass));
MemoryStream ms = new MemoryStream();
x.Serialize(ms, dc);
ms.Seek(0, 0);
XDocument xDocument = XDocument.Load(ms); // Here it is!

I'm not sure if that is what you want, but to make an XML-Document out of this:
College coll = ...
XDocument doc = new XDocument(
new XElement("College",
new XElement("Name", coll.Name),
new XElement("Address", coll.Address),
new XElement("Persons", coll.Persons.Select(p =>
new XElement("Person",
new XElement("Gender", p.Gender),
new XElement("City", p.City)
)
)
)
);

Xml Serialization without XML Declaration [duplicate]

How do I serialize an XML-serializable object to an XML fragment (no XML declaration nor namespace references in the root element)?

Here is a hack-ish way to do it without having to load the entire output string into an XmlDocument:
using System;
using System.Text;
using System.Xml;
using System.Xml.Serialization;
public class Example
{
public String Name { get; set; }
static void Main()
{
Example example = new Example { Name = "Foo" };
XmlSerializer serializer = new XmlSerializer(typeof(Example));
XmlSerializerNamespaces emptyNamespace = new XmlSerializerNamespaces();
emptyNamespace.Add(String.Empty, String.Empty);
StringBuilder output = new StringBuilder();
XmlWriter writer = XmlWriter.Create(output,
new XmlWriterSettings { OmitXmlDeclaration = true });
serializer.Serialize(writer, example, emptyNamespace);
Console.WriteLine(output.ToString());
}
}

You should be able to just serialize like you usually do, and then use the Root property from the resulting document.
You may need to clear the attributes of the element first.

By the way this is awesome.
I implemented this code to make it easy to work with xml fragments as classes quickly and then you can just replace the node when finished. This makes the transition between code and xml ultra-easy.
First create some extension methods.
public static class SerializableFragmentExtensions
{
public static XElement ToElement(this ISerializableFragment iSerializableFragment)
{
var serializer = new XmlSerializer(iSerializableFragment.GetType());
var emptyNamespace = new XmlSerializerNamespaces();
emptyNamespace.Add(String.Empty, String.Empty);
var output = new StringBuilder();
var writer = XmlWriter.Create(output,
new XmlWriterSettings { OmitXmlDeclaration = true });
serializer.Serialize(writer, iSerializableFragment, emptyNamespace);
return XElement.Parse(output.ToString(), LoadOptions.None);
}
public static T ToObject<T>(this XElement xElement)
{
var serializer = new XmlSerializer(typeof (T));
var reader = xElement.CreateReader();
var obj = (T) serializer.Deserialize(reader);
return obj;
}
}
Next Implement the required interface (marker interface--I know you are not supposed to but I think this is the perfect reason to it.)
public interface ISerializableFragment
{
}
Now all you have to do is decorate any Serializable class, you want to convert to an XElement Fragment, with the interface.
[Serializable]
public class SomeSerializableClass : ISerializableFragment
{
[XmlAttribute]
public string SomeData { get; set; }
}
Finally test the code.
static void Main(string[] args)
{
var someSerializableClassObj = new SomeSerializableClass() {SomeData = "Testing"};
var element = someSerializableClass.ToElement();
var backToSomeSerializableClassObj = element.ToObject<SomeSerializableClass>();
}
Thanks again for this amazingly useful code.

How to rename <ArrayOf> XML attribute that generated after serializing List of objects

I am serializing List of objects List<TestObject>
, and XmlSerializer generates <ArrayOfTestObject> attribute, I want rename it or remove it.
Can it be done with creating new class that encapsulated List as field?
[XmlRoot("Container")]
public class TestObject
{
public TestObject() { }
public string Str { get; set; }
}
List<TestObject> tmpList = new List<TestObject>();
TestObject TestObj = new TestObject();
TestObj.Str = "Test";
TestObject TestObj2 = new TestObject();
TestObj2.Str = "xcvxc";
tmpList.Add(TestObj);
tmpList.Add(TestObj2);
XmlWriterSettings settings = new XmlWriterSettings();
settings.OmitXmlDeclaration = true;
settings.Indent = true;
XmlSerializer serializer = new XmlSerializer(typeof(List<TestObject>));
using (XmlWriter writer = XmlWriter.Create(#"C:\test.xml", settings))
{
XmlSerializerNamespaces namespaces = new XmlSerializerNamespaces();
namespaces.Add(string.Empty, string.Empty);
serializer.Serialize(writer, tmpList, namespaces);
}
<ArrayOfTestObject>
<TestObject>
<Str>Test</Str>
</TestObject>
<TestObject>
<Str>xcvxc</Str>
</TestObject>
</ArrayOfTestObject>

The most reliable way is to declare an outermost DTO class:
[XmlRoot("myOuterElement")]
public class MyOuterMessage {
[XmlElement("item")]
public List<TestObject> Items {get;set;}
}
and serialize that (i.e. put your list into another object).
You can avoid a wrapper class, but I wouldn't:
class Program
{
static void Main()
{
XmlSerializer ser = new XmlSerializer(typeof(List<Foo>),
new XmlRootAttribute("Flibble"));
List<Foo> foos = new List<Foo> {
new Foo {Bar = "abc"},
new Foo {Bar = "def"}
};
ser.Serialize(Console.Out, foos);
}
}
public class Foo
{
public string Bar { get; set; }
}
The problem with this is that when you use custom attributes you need to be very careful to store and re-use the serializer, otherwise you get lots of dynamic assemblies loaded into memory. This is avoided if you just use the XmlSerializer(Type) constructor, as it caches this internally automatically.

Change the following line from:
XmlSerializer serializer = new XmlSerializer(typeof(List<TestObject>));
To:
XmlRootAttribute root = new XmlRootAttribute("TestObjects");
XmlSerializer serializer = new XmlSerializer(typeof(List<TestObject>), root);
It should work.

You can add an additional parameter to the XmlSerializer constructor to essentially name the root element.
XmlSerializer xsSubmit = new XmlSerializer(typeof(List<DropDownOption>), new XmlRootAttribute("DropDownOptions"));
This would result in the following structure:
<DropDownOptions xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
<DropDownOption>
<ID>1</ID>
<Description>foo</Description>
</DropDownOption>
<DropDownOption>
<ID>2</ID>
<Description>bar</Description>
</DropDownOption>
</DropDownOptions>

Create another class like :
[XmlRoot("TestObjects")]
public class TestObjects: List<TestObject>
{
}
Then apply below code while sealizing :
XmlSerializer serializer = new XmlSerializer(typeof(TestObjects));
MemoryStream memStream = new MemoryStream();
serializer.Serialize(memStream, tmpList);