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SqlConnection cn= new SqlConnection(connectionString);
OpenFileDialog open = new OpenFileDialog();
open.Filter = "Image Files(*.jpeg;*.bmp;*.png;*.jpg)|*.jpeg;*.bmp;*.png;*.jpg";
if (open.ShowDialog() == DialogResult.OK)
{
textBoximage.Text = open.FileName;
}
cn.Open();
string image = textBoximage.Text;
Bitmap bmp = new Bitmap(image);
FileStream fs = new FileStream(image, FileMode.Open, FileAccess.Read);
byte[] bimage = new byte[fs.Length];
fs.Read(bimage, 0, Convert.ToInt32(fs.Length));
fs.Close();
SqlCommand cmd = new SqlCommand("insert into TableName(imgdata) values(@imgdata)",cn);
cmd.Parameters.AddWithValue("@imgdata",SqlDbType.Image).Value=bimage;
cmd.ExecuteNonQuery();
cn.Close();
}
cn.Open();
SqlDataAdapter da = new SqlDataAdapter(new SqlCommand("Select Picture From TableName where FieldName=condition ", cn));
DataSet ds = new DataSet();
da.Fill(ds);
byte[] myImage = new byte[0];
myImage = (byte[])ds.Tables[0].Rows[0]["Picture"];
MemoryStream stream = new MemoryStream(myImage);
pictureBox1.Image = Image.FromStream(stream);
cn.Close();
Я предполагаю, что вы использовали sql server в качестве базы данных.Следуйте приведенному ниже коду.
byte[] imag = File.ReadAllBytes("your image path");
SqlCommand cmd = new SqlCommand("INSERT INTO Table(Image) VALUES(@Image)", your connection);
SqlCommand.Parameters.AddWithValue("@Image", imag);
cmd.ExecuteNonQuery();