Вставить картинку в datagrid?

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Problem line 13 and 38..data column type is image in sql

Мне нужна классическая вставка в таблицу из picturebox

Что я уже пробовал:

   String cs = "Data Source=.\\SQLEXPRESS;Initial Catalog=bss;Integrated Security=True";
               if (string.IsNullOrEmpty(idTextBox.Text))
               {

                   using (SqlConnection openCon = new SqlConnection(cs))
                   {
                       string saveStaff = "INSERT into dbo.poslovni_partneri (Ime, data) VALUES (@Ime,@data)";

                       using (SqlCommand querySaveStaff = new SqlCommand(saveStaff))
                       {
                           querySaveStaff.Connection = openCon;
                           querySaveStaff.Parameters.Add("@Ime", SqlDbType.VarChar, 255).Value = imeTextBox.Text;
                           querySaveStaff.Parameters.Add("@data", SqlDbType.image).Value = pictureBox.Text;


                           openCon.Open();
                           querySaveStaff.ExecuteNonQuery();
                           openCon.Close();


                       }

                   }
               }

               else
               {


                   using (SqlConnection openCon = new SqlConnection(cs))
                   {
                       string saveStaff = "UPDATE  dbo.poslovni_partneri SET Ime=@Ime, data=@data WHERE id=" + idTextBox.Text;

                       using (SqlCommand querySaveStaff = new SqlCommand(saveStaff))
                       {
                           querySaveStaff.Connection = openCon;
                           querySaveStaff.Parameters.Add("@Ime", SqlDbType.VarChar, 255).Value = imeTextBox.Text;
                           querySaveStaff.Parameters.Add("@data", SqlDbType.image).Value = pictureBox.Text;

                           openCon.Open();
                           querySaveStaff.ExecuteNonQuery();
                           MessageBox.Show("Uspješno ste izmenili stavku!", "Informacija", MessageBoxButtons.OK, MessageBoxIcon.Information);
                           openCon.Close();
                       }
                   }
               }

/////////For upload picture in picture box use this code and it work...


   private void button4_Click(object sender, EventArgs e)
           {
               //Read image file
               using (OpenFileDialog ofd = new OpenFileDialog() { Filter = "JPEG|*.jpg", ValidateNames = true, Multiselect = false })
               {
                   if (ofd.ShowDialog() == DialogResult.OK)
                   {
                       fileName = ofd.FileName;
                       lblFilename.Text = fileName;
                       dataPictureBox1.Image = System.Drawing.Image.FromFile(fileName);
                   }
               }
           }

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