Open the image while click on the filename in datagridview

Question

Hi Friends In windows application I want to open the file while click the file name in gatagridview cell. DataGridView (database)having 4 fields 1) Image_Name 2) Project_Name 3) Received_Date 4) Image_Path when i click or press the down arrow on the datagridview the respective image should be open in picturebox according to image_path (i have populated the image location and saved in 4th column). Could you please anybody help me here Regards Gopi A

Dorababu Meka · Answer

Try this private void dataGridView1_CellMouseClick(object sender DataGridViewCellMouseEventArgs e) { DataGridViewImageCell img = new DataGridViewImageCell(); img = dataGridView1 .Item(ImageColumn e.RowIndex); PictureBox1.Image = byteArrayToImage(img.Value); } static private Image byteArrayToImage(byte[] byteImg) { MemoryStream ms= new MemoryStream(); Image imgcol= null; try { ms.Write( byteImg 0 byteImg .Length); imgcol = new Bitmap(ms); } finally { ms.Close(); } return imgcol ; }

Mohammad Kanderi · Answer

Hi Gopi Asai Try this code private void dataGridView1_CellClick(object sender DataGridViewCellEventArgs e) { pictureBox1.Image = Image.FromFile(dataGridView1.Rows[e.RowIndex].Cells[3].Value.ToString()); }