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 that needs 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:
- XML serialization only serializes
public
fields and properties. - XML serialization does not include any type information.
- We need to have a default/non-parameterised constructor in order to serialize an object.
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:
="1.0"="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 the same as that of the class name and the subtag names are the 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 during 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 during 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 helps 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 X
mlSerializer
. 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 won't 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 the 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 take a look at 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, let's 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. Let's 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 helps 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 time and effort.