xml

Question

http://www.codeproject.com/Articles/483055/XML-Serialization-and-Deserialization-Part-1

http://www.codeproject.com/Articles/487571/XML-Serialization-and-Deserialization-Part-2 I need all in C# form application

Iftikar Hussain · Answer

try like this public partial class Form1 : Form { public Form1() { InitializeComponent(); } private void button1_Click(object sender EventArgs e) { //Serialization of object to XML form AddressDetails details = new AddressDetails(); details.HouseNo = 4; details.StreetName = Rohini; details.City = Delhi; Serialize(details); MessageBox.Show(Completed); } private void button2_Click(object sender EventArgs e) { //Deserialization of XML to object form XmlSerializer deserializer = new XmlSerializer(typeof(AddressDetails)); TextReader reader = new StreamReader(@C:\Test\Xml.xml); object obj = deserializer.Deserialize(reader); AddressDetails XmlData = (AddressDetails)obj; reader.Close(); string data = string.Empty; data=String.Format(House No :{0} XmlData.HouseNo) + Environment.NewLine; data = data + String.Format(StreetName :{0} XmlData.StreetName) + Environment.NewLine; data = data + String.Format(City :{0} XmlData.City) + Environment.NewLine; MessageBox.Show(data); } public static void Serialize(AddressDetails details) { XmlSerializer serializer = new XmlSerializer(typeof(AddressDetails)); using (TextWriter writer = new StreamWriter(@C:\Test\Xml.xml)) { serializer.Serialize(writer details); } } } Regards Iftikar

om prakash · Answer

I need in windows form appliation

om prakash · Answer

I need in windows form

Iftikar Hussain · Answer

try like this Console.WriteLine(Serialization of object to XML form); Console.WriteLine(); AddressDetails details = new AddressDetails(); details.HouseNo = 4; details.StreetName = Rohini; details.City = Delhi; Serialize(details); Console.WriteLine(); Console.WriteLine(Completed); Console.WriteLine(===========================================); Console.WriteLine(Deserialization of XML to object form); Console.WriteLine(); XmlSerializer deserializer = new XmlSerializer(typeof(AddressDetails)); TextReader reader = new StreamReader(@C:\Test\Xml.xml); object obj = deserializer.Deserialize(reader); AddressDetails XmlData = (AddressDetails)obj; reader.Close(); Console.WriteLine(House No :{0} XmlData.HouseNo); Console.WriteLine(StreetName :{0} XmlData.StreetName); Console.WriteLine(City :{0} XmlData.City); Console.WriteLine(); Console.WriteLine(Completed); Console.ReadKey(); Regards Iftikar

om prakash · Answer

ifikhar thnx I got output.. but I want to do deserialize method how can I do......I want to complete my tasks

Iftikar Hussain · Answer

declare address details as public as like this public class AddressDetails Regards Iftikar

om prakash · Answer

ConsoleApplication6.AddressDetails is inaccessible due to its protection level. Only public types can be processed.

om prakash · Answer

ConsoleApplication6.AddressDetails is inaccessible due to its protection level. Only public types can be processed.

Iftikar Hussain · Answer

Hi What error message your getting? Your code seems to be correct. Regards Iftikar

om prakash · Answer

XmlSerializer serializer = new XmlSerializer ( typeof ( AddressDetails )); I have got error in this line............... the code is class file: Using System; using System.Collections.Generic; using System.Linq; using System.Text; using System.Threading.Tasks; namespace ConsoleApplication6 { class AddressDetails { public int HouseNo { get ; set ; } public string StreetName { get ; set ; } public string City { get ; set ; } private string PoAddress { get ; set ; } } } form: using System; using System.Collections.Generic; using System.Linq; using System.Text; using System.Threading.Tasks; using System.Xml.Serialization; using System.IO; using System.Xml; namespace ConsoleApplication6 { class Program { static void Main( string [] args) { { AddressDetails details = new AddressDetails (); details.HouseNo = 4; details.StreetName = Rohini ; details.City = Delhi ; Serialize(details); } } static public void Serialize( AddressDetails details) { XmlSerializer serializer = new XmlSerializer ( typeof ( AddressDetails )); using ( TextWriter writer = new StreamWriter ( @D:\Xml.xml )) { serializer.Serialize(writer details); } } } }