Deserializing Data

Problem

You have a DataSet that has been serialized and written to a file. You want to recreate the DataSet from this file.

Solution

Use the serializer object's Deserialize ( ) method and cast the result as a DataSet .

The sample code loads a file stream containing a previously serialized DataSet in a specified format and deserializes it to recreate the original DataSet .

The C# code is shown in Example 5-5.

Example 5-5. File: DeserializeForm.cs

// Namespaces, variables, and constants
using System;
using System.Windows.Forms;
using System.IO;
using System.Data;
using System.Runtime.Serialization;
using System.Runtime.Serialization.Formatters;
using System.Runtime.Serialization.Formatters.Binary;
using System.Runtime.Serialization.Formatters.Soap;
using System.Xml.Serialization;

private OpenFileDialog ofd;

// . . . 

private void goButton_Click(object sender, System.EventArgs e)
{
 // Create and open the stream for deserializing.
 Stream stream = null;
 try
 {
 stream = File.Open(fileNameTextBox.Text, FileMode.Open,
 FileAccess.Read);
 }
 catch(Exception ex)
 {
 MessageBox.Show(ex.Message, "Deserializing Data",
 MessageBoxButtons.OK, MessageBoxIcon.Error);
 return;
 }

 // Deserialize the DataSet from the stream.
 DataSet ds = null;
 try
 {
 if (xmlReadRadioButton.Checked)
 {
 ds = new DataSet( );
 ds.ReadXml(stream);
 }
 else if (xmlSerializerRadioButton.Checked)
 {
 XmlSerializer xs = new XmlSerializer(typeof(DataSet));
 ds = (DataSet)xs.Deserialize(stream);
 }
 else if(soapRadioButton.Checked)
 {
 SoapFormatter sf = new SoapFormatter( );
 ds = (DataSet)sf.Deserialize(stream);
 } 
 else if(binaryRadioButton.Checked)
 {
 BinaryFormatter bf = new BinaryFormatter( );
 ds = (DataSet)bf.Deserialize(stream);
 }
 }
 catch (System.Exception ex)
 {
 MessageBox.Show(ex.Message, "Deserializing Data",
 MessageBoxButtons.OK, MessageBoxIcon.Error);
 return;
 }
 finally
 {
 stream.Close( );
 }

 // Bind the DataSet to the grid.
 dataGrid.DataSource = ds.DefaultViewManager;

 MessageBox.Show("Deserialization complete.", "Deserializing Data",
 MessageBoxButtons.OK, MessageBoxIcon.Information);
}

private void fileDialogButton_Click(object sender, System.EventArgs e)
{
 // File dialog to save file
 if(xmlReadRadioButton.Checked xmlSerializerRadioButton.Checked)
 ofd.Filter = "XML files (*.xml)*.xml";
 else if(soapRadioButton.Checked)
 ofd.Filter = "SOAP files (*.soap)*.soap";
 else if(binaryRadioButton.Checked)
 ofd.Filter = "Binary files (*.bin)*.bin";

 ofd.Filter += "All files (*.*)*.*";
 ofd.FilterIndex = 0;

 if (ofd.ShowDialog( ) == DialogResult.OK)
 fileNameTextBox.Text = ofd.FileName;
}

Discussion

This sample deserializes any of the serialized DataSet objects from Recipe 5.4.

The sample allows the user to select a file and specify a serialization type. The appropriate serializing object is created and in the case of the XmlSerializer object, its type is specified in the constructor. The Deserialize( ) method of the serializer object is then used to deserialize the file stream into an object graph. This is then cast to a DataSet to complete the deserialization.

See the discussion in Recipe 5.4 for more information about the serialization and the formatter classes that can serialize ADO.NET objects.