Adding image from source property of image control(WPF) at Access Database

Question

Hi to all
First and foremost, I apologize for my grammatical errors; my first language is Persian (Iran).
This is tested solution

Answer 1

I found the solution which is working

  //Global field  
  BitmapImage BM;  
  //Convert image to bytes by following method  
  private static byte[] ImageToBytes(BitmapImage image)  
  {  
         byte[] Data;  
         JpegBitmapEncoder JpegEncoder = new JpegBitmapEncoder();  
         JpegEncoder.Frames.Add(BitmapFrame.Create(image));  
         using (System.IO.MemoryStream MS = new System.IO.MemoryStream())  
         {  
             JpegEncoder.Save(MS);  
             Data = MS.ToArray();  
         }  
         return Data;  
  }  
  //Convert bytes to image by following method  
  private BitmapImage GetImageFromBytes(byte[] bytes)  
  {  
         System.IO.MemoryStream Stream = new System.IO.MemoryStream();  
         Stream.Write(bytes, 0, bytes.Length);  
         Stream.Position = 0;  
         System.Drawing.Image img = System.Drawing.Image.FromStream(Stream);  
         BitmapImage bitImage = new BitmapImage();  
         bitImage.BeginInit();  
         System.IO.MemoryStream MS = new System.IO.MemoryStream();  
         img.Save(MS, System.Drawing.Imaging.ImageFormat.Jpeg);  
         MS.Seek(0, System.IO.SeekOrigin.Begin);  
         bitImage.StreamSource = MS;  
         bitImage.EndInit();  
         return bitImage;  
  }  
  //Insert image in WPF control image  
  private void UploadButton_PreviewMouseLeftButtonUp(object sender, MouseButtonEventArgs e)  
  {  
         if(OpenFileDialog.ShowDialog()==System.Windows.Forms.DialogResult.OK)  
         {  
          var IMG = System.Drawing.Image.FromFile(OpenFileDialog.FileName);  
          BM = new BitmapImage(new Uri(OpenFileDialog.FileName));  
          BitmapImage BitMapImage = new BitmapImage();  
          BitMapImage.BeginInit();  
          System.IO.MemoryStream MemoryStream = new System.IO.MemoryStream();  
          IMG.Save(MemoryStream, System.Drawing.Imaging.ImageFormat.Bmp);  
          MemoryStream.Seek(0, System.IO.SeekOrigin.Begin);  
          BitMapImage.StreamSource = MemoryStream;  
          BitMapImage.EndInit();  
          MemberImage.Source = BitMapImage;  
          OpenFileDialog.Dispose();  
         }  
  }  
  //Now Insert Image and Name by The following code  
  private void Add_Button_PreviewMouseLeftButtonDown(object sender, MouseButtonEventArgs e)  
  {  
         OleDbConnect.Open();  
         byte[] Image_Bytes = ImageToBytes(BM);  
         OleDbParameter Parameter = new OleDbParameter();  
         Parameter.OleDbType = OleDbType.Binary;  
         Parameter.ParameterName = "Image";  
         Parameter.Value = Image_Bytes;  
         OleDbCommand OleDbCommand_Insert = new OleDbCommand("Insert Into         
         MemberTable(FirstName,MemberImage)values(" +  
         "'" + FirstName_TextBox.Text + "',[@](/users/na/?userId=120b56f4-bffd-0003-0000-000000000000))", OleDbConnect);  
         OleDbCommand_Insert.Parameters.Add(Parameter);  
         OleDbCommand_Insert.ExecuteScalar();  
         OleDbConnect.Close();  
  }  
  //Now retrieve Image and Name by The following code from DataBase  
  private void Search_Button_PreviewMouseLeftButtonDown(object sender, MouseButtonEventArgs e)  
  {  
         OleDbCommand OleDCmd = new OleDbCommand("Select  * From MemberTable Where FirstName='"        
         + FirstName_TextBox.Text + "'",OleDbConnect);  
         OleDCmd.CommandType = System.Data.CommandType.Text;  
         FirstName_TextBox.Text = null;  
         OleDbConnect.Open();  
         OleDbDataReader DataReader = OleDCmd.ExecuteReader();  
         while (DataReader.Read())  
         {  
          FirstName_TextBox.Text = DataReader[0].ToString();  
          BitmapImage BMP = GetImageFromBytes((byte[])DataReader[11]);  
          MemberImage.Source = BMP;  
         }  
  }