- Data can't be inserted in database
- protected void Button2_Click(object sender, EventArgs e)
- {
- SqlConnection conn = new SqlConnection(ConfigurationManager.ConnectionStrings["Connection"].ConnectionString);
- SqlCommand cmd = new SqlCommand("Insert into CarTab(Brand,Model,Plate,Color,Service) Values (@brand,@model,@plate,@color,@year,@service)",conn);
- cmd.CommandType = CommandType.Text;
- cmd.Parameters.AddWithValue("@brand", Label1.Text);
- cmd.Parameters.AddWithValue("@model", Label2.Text);
- cmd.Parameters.AddWithValue("@plate", Label3.Text);
- cmd.Parameters.AddWithValue("@color", Label4.Text);
- cmd.Parameters.AddWithValue("@year", Label5.Text);
- cmd.Parameters.AddWithValue("@service", Label6.Text);
- conn.Open();
- cmd.ExecuteNonQuery();
- }
- SqlCommand cmd = new SqlCommand(
- "Insert into CarTab(Brand,Model,Plate,Color,Year,Service)
- Values (@brand,@model,@plate,@color,@year,@service)",
- conn);
- using(SqlConnection conn = new SqlConnection(ConfigurationManager.ConnectionStrings["Connection"].ConnectionString))
- {
- using(SqlCommand cmd = new SqlCommand("Insert into CarTab(Brand,Model,Plate,Color,Service) Values (@brand,@model,@plate,@color,@year,@service)",conn))
- {
- cmd.CommandType = CommandType.Text;
- cmd.Parameters.AddWithValue("@brand", Label1.Text);
- cmd.Parameters.AddWithValue("@model", Label2.Text);
- cmd.Parameters.AddWithValue("@plate", Label3.Text);
- cmd.Parameters.AddWithValue("@color", Label4.Text);
- cmd.Parameters.AddWithValue("@year", Label5.Text);
- cmd.Parameters.AddWithValue("@service", Label6.Text);
- conn.Open();
- try
- {
- cmd.ExecuteNonQuery();
- }
- catch(Exception e)
- {
- string errorMessage = e.Message;
- //Set a label on the client to see the error message, or pause in the debugger and examine the property here.
- //throw;
- }
- }
- }
- Button2.Click += Button2_Click;
- INSERT INTO CarTab(Brand, Model, Plate, Color, Year, Service)