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.