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XML Serialization and Deserialization: Part-1

, 10 Dec 2013
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XML serialization and deserialization using C#.

Introduction  

The article talks about serialization of objects in XML format and deserialization of an XML file back to an object. Serialization is a process by which an object's state is transformed in some serial data format, such as XML or binary format. Deserialization on the other hand is used to convert the byte of data, such as XML or binary data, to object type. Serialization is the process of converting an object into a form that can be readily transported. For example, you can serialize an object and transport it over the Internet using HTTP between a client and a server. On the other end, deserialization reconstructs the object from the stream. XML serialization results in strongly typed classes with public properties and fields that are converted to a serial format (in this case, XML) for storage or transport. 

Let's start with a basic example. Here is a simple class the need to be serialized :   

public class AddressDetails
{
    public int HouseNo { get; set; }
    public string StreetName { get; set; }
    public string City { get; set; }
    private string PoAddress { get; set; }
}     

The following points should be noted while creating a class for serialization:

  1. XML serialization only serializes public fields and properties. 
  2. XML serialization does not include any type information. 
  3. We need to have a default/ non-parameterised constructor in order to serialize an object. 
  4. ReadOnly properties are not serialized.    

Code to serialize the above class:  

public static void Main(string[] args) 
{ 
    AddressDetails details = new AddressDetails();
    details.HouseNo = 4;
    details.StreeName = "Rohini";
    details.City = "Delhi";
    Serialize(details);
}   
static public void Serialize(AddressDetails details)
{ 
    XmlSerializer serializer = new XmlSerializer(typeof(AddressDetails)); 
    using (TextWriter writer = new StreamWriter(@"C:\Xml.xml"))
    {
        serializer.Serialize(writer, details); 
    } 
}

XmlSerializer (located in the System.Xml.Serialization namespace) class is used to serialize and deserialize. The class method Serialize is called. Since we have to serialize in a file we create a "TextWriter". Since TextWriter implements IDisposable, we used using so that we need not close the writer.  

The output after the serialization is : 

<?xml version="1.0" encoding="utf-8"?> 

<AddressDetails>
  <HouseNo>4</HouseNo>
  <StreetName>Rohini</StreetName>
  <City>Delhi</City>
</AddressDetails>

Here in the XML we can see that the Head tag of the XML created is same as that of the class name and the subtag names are same as the properties in class AddressDetails. Each public property is displayed in the form of Tags in the XML created. We can observe here that only public fields are displayed here.

XML Serialization and Attributes

Some common attributes that are available while Serialization are: 

  • XmlAttribute: This member will be serialized as an XML attribute 
  • XmlElement: The field will be serialized as an XML element  
  • XmlIgnore: Field will be ignored while Serialization
  • XmlRoot: Represent XML document's root Element   

Use of XmlElement: 

Further if we need to have different Tag name in XML from the class property Name. We can introduce the XmlElement attribute to it in the class structure.    

public class AddressDetails
{ 
    [XmlElement("Number")]
    public int HouseNo { get; set; }
    [XmlElement("Street")] 
    public  string StreetName { get; set; } 
    [XmlElement("CityName")]
}

[XmlElement("Number")] specifies that the property HouseNo will be serialized with the Tag name "Number" in the Xml File. It help us to map between the XML Tag name and the class Property Name. The resultant XML string with the Custom tag name is given below : 

<AddressDetails> 
  <Number>4</Number>
  <Street>Rohini</Street>
  <CityName>Delhi</CityName>
</AddressDetails>

Use of XmlAttribute

If we want that the property HouseNo should occur as the attribute for the Tag AddressDetails then we should use XmlAttribute. XmlAttribute serializes the object property as the attribute for the parent tag. The following code illustrates the functionality:

public class AddresssDetails
{ 
    [XmlAttribute]("Number")]
    public int HouseNo { get; set; }
    [XmlElement("Street")] 
    public  string StreetName { get; set; } 
    [XmlElement("CityName")]
    public string City {get; set;}
}

The XML serialized output for the code will be : 

<AddressDetails Number="4">
  <Street>Rohini</Street>
  <CityName>Delhi</CityName>
</AddressDetails>

Notice here, Since the class property HouseNo is specified as XMLAttribute therefore this property is an Attribute for the parent tag AddressDetails.

Use to XmlIgnore  

By default, all public fields and public read/write properties are serialized by the XmlSerializer . That is, the value of each public field or property is persisted as an XML element or XML attribute in an XML-document instance. In order to override this property apply XmlIgnore attribute to it. This will remove the element from the XML. The code below explains the following:

public class AddressDetails
{
    [XmlElement("Number")]
    public int HouseNo;
    [XmlElement("Street")]
    public string StreetName;
    [XmlIgnore]
    public string City;
}

Here we can see that the property City contains XmlIgnore attribute. The resultant XML created wont contain the City tag in it.

<AddressDetails>
    <Number>4</Number>
    <Street>ABC</Street>
</AddressDetails>

Notice here that the property City is not serialized because of the attribute XmlIgnore placed on it.

Use of XmlRoot 

Every XML has a root element. By default the name of the root element is same as the name of the class that is serialized. In order to give a custom name to the root element of XML, we use XmlRoot attribute. Implementation of this attribute is provided below:

[XmlRoot("Root")]
public class AddressDetails
{
    [XmlElement("Number")]
    public int HouseNo;
    [XmlElement("Street")]
    public string StreetName;
    [XmlElement("CityName")]
    public string City;
} 

Here we can see that the attribute XmlRoot is placed over AddressDetails class. This will now override the default serialization behavior which takes xml tag root name same as the class name. The XML will now have "Root" as the root tag. 

<Root>
  <HouseNo>4</HouseNo>
  <StreetName>Rohini</StreetName>
  <City>Delhi</City> 
</Root>   

Notice here that the root tag here is now "Root" and not the Class name. 

Object List Serialization

Now let's try to serialize a list of AddressDetails object to XML file:  

public static void Main(string[] args) 
{ 
    List<AddressDetails> AddressList = new List<AddressDetails>();
    AddressDetails detail1 = new AddressDetails();
    detail1.HouseNo ="4";
    detail1.StreetName = "ABC";
    detail1.City = "Delhi";
 
    AddressDetails detail2 = new AddressDetails(); 
    detail2.HouseNo ="3";
    detail2.StreetName = "ABCD";
    detail2.City = "New Delhi";
 
    AddressList.Add(detail1); 
    AddressList.Add(detail2); 
    Serialize(AddressList); 
}   
public void Serialize(List<AddressDetails> list)
{
    XmlSerializer serializer = new XmlSerializer(typeof(List<AddressDetails>));
    using ( TextWriter writer = new StreamWriter( @"C:\Xml.txt")
    {
        serializer.Serialize(writer, list) 
    }  
}

The XML output for the above execution will be :  

<ArrayOfAddressDetails>
  <AddressDetails>
    <Number>4</Number>
    <Street>ABC</Street>
    <CityName>Delhi</CityName>
  </AddressDetails>
  <AddressDetails>
    <Number>3</Number>
    <Street>ABCD</Street>
    <CityName>New Delhi</CityName>
  </AddressDetails>
</ArrayOfAddressDetails>

Notice that the XML produced gives a list of AddressDetails object. 

Serialization of classes containing other class objects

If we have a class structure such that a class contains an object of other class and we want to include that class object also for serialization. Let's see the following example :

public class PersonalDetails
{
    public string Name { get; set; }
    public int Age { get; set; }
    public Address address;
}
public class Address
{
    public int HouseNo { get; set; }
    public string StreetName { get; set; }
    public string City { get; set; }
}

This is how the PersonalDetails class will be serialized: 

<PersonalDetails>
  <Name>Mayank</Name>
  <Age>24</Age>
  <Address>
    <HouseNo>4</HouseNo>
    <StreetName>Rohini</StreetName>
    <City>Delhi</City>
  </Address>
</PersonalDetails>

To add more complexity to it lets try creating the following XML structure : 

<PersonalDetails>
  <Name>Mayank</Name>
  <Age>24</Age>
  <Address HouseNo="4">
    <StreetName>Rohini</StreetName>
    <City>Delhi</City>
  </Address>
</PersonalDetails>

Observe that the difference over here is that we need to have "HouseNo" as the attribute for the Address Tag.  Lets see what change will be made in class in order to create this structure:

public class PersonalDetails
{
    public string Name { get; set; }
    public int Age { get; set; }
    public Address address;
    public PersonalDetails()
    {
        Name = "Mayank";
        Age = 24;
        address = new Address();
    }
    
}
public class Address
{
    [XmlAttribute("HouseNo")]
    public int HouseNo { get; set; }
    public string StreetName { get; set; }
    public string City { get; set; }
    public Address()
    {
        HouseNo = 8;
        StreetName = "Rohini";
        City = "Delhi";
    }
}

As per the requirement we wanted to have "HouseNo" as XML attribute instead of the normal XMLElement. Therefore we introduce "XmlAttribute" on property.

Let's try creating the following XML Structure:

<PersonalDetails>
  <Name>Mayank</Name>
  <Age>24</Age>
  <address HouseNo="8">Rohini</address>
</PersonalDetails>

The difference over here is that we require the StreetName as innertext of the XML node "address". So in order to create such structure we have another attribute XmlText. This help us to add the particular property as innertext for a tag. 

So the Code for creating such a structure is :   

public class PersonalDetails
{
    public string Name { get; set; }
    public int Age { get; set; }
    public Address address;
}
public class Address
{
    [XmlAttribute("HouseNo")]
    public int HouseNo { get; set; }
    [XmlText]
    public string StreetName { get; set; }
}

The "XmlText" attribute here adds StreetName as InnerText to the tag "address"  

Deserialization of XML

See the following article for details on deserialization :  XML Serialization and Deserialization (Part-2) .

Conclusion

Serialization is a very efficient way to convert the object to XML. This save lots of saving time and effort.

License

This article, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

About the Author

mayankgupta688

United States United States
No Biography provided

Comments and Discussions

 
GeneralMy vote of 2 PinmemberMic10-Nov-12 5:36 
GeneralMy vote of 5 PinmemberSavalia Manoj M7-Nov-12 1:08 
GeneralMy vote of 5 PinmemberHernán Hegykozi6-Nov-12 19:08 
GeneralMy vote of 5 PinmemberVelrats6-Nov-12 1:15 
GeneralMy vote of 3 Pinmembertlhintoq5-Nov-12 10:27 
Questioneh. 3 or 4 out of 5 Pinmembertlhintoq5-Nov-12 10:26 
GeneralMy vote of 5 Pinmemberkbe4-Nov-12 23:30 
Generalmy vote of 5 PinmemberWen Hao2-Nov-12 18:03 
Before this, I know nothing about serialization on xml.
But after reading your article, I know how to do it. Laugh | :laugh:
Looking forward for your next article on deserialization on xml.
GeneralRe: my vote of 5 Pinmembermayankgupta6884-Nov-12 19:53 
GeneralRe: my vote of 5 PinmemberShobhana.n4-Sep-13 9:00 
Questionvoynguyen Pinmemberhayvevoianh1-Nov-12 18:25 
GeneralMy vote of 4 PinmemberChelo19741-Nov-12 14:39 
QuestionSwitching Seamlessly Between Serialization PinmemberMic1-Nov-12 6:53 
AnswerRe: Switching Seamlessly Between Serialization PinmemberMic10-Nov-12 5:33 
QuestionWhat about "Yet Another XML Serialization Library for the .NET Framework" ? Pinmemberandre123451-Nov-12 6:08 
AnswerRe: What about "Yet Another XML Serialization Library for the .NET Framework" ? PinmemberRui Jarimba1-Nov-12 11:41 
GeneralMy vote of 3 PinmemberRui Jarimba1-Nov-12 2:53 
GeneralRe: My vote of 3 Pinmembermayankgupta6881-Nov-12 8:25 
GeneralMy vote of 5 PinmemberMashudu Nemukula30-Oct-12 20:25 
QuestionArticle not complete PinmemberMashudu Nemukula30-Oct-12 20:18 
AnswerRe: Article not complete Pinmembermayankgupta68830-Oct-12 20:23 
GeneralMy vote of 5 PinmemberRemi Lebrun30-Oct-12 11:57 
GeneralMy vote of 3 PinmemberDaniel Gidman30-Oct-12 10:42 
GeneralRe: My vote of 3 Pinmembermayankgupta68830-Oct-12 20:27 
GeneralMy vote of 5 PinmemberMember 770470529-Oct-12 22:00 

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