How to insert record in two different tables on a button clk

Question

I have two tables product(productId,productName,productRate) and ID(id)...i want to insert record on a button click.I want to insert record into product table and want productId to insert in ID table on same button click.

Satyapriya Nayak · Answer

Create two stored procedure as below ALTER PROCEDURE insert1 (@custid varchar(50),@custname varchar(50),@custaddress varchar(50),@prodid varchar (50)) AS insert Customer(custid,custname,custaddress,prodid) values (@custid,@custname,@custaddress,@prodid) ALTER PROCEDURE insert2 (@prodid varchar(50),@prodname varchar(50),@price int) AS insert Product(prodid,prodname,price) values (@prodid,@prodname,@price) Default.aspx code Untitled Page onclick="btn_insert_Click" /> Default.aspx.cs code using System; using System.Configuration; using System.Data; using System.Linq; using System.Web; using System.Web.Security; using System.Web.UI; using System.Web.UI.HtmlControls; using System.Web.UI.WebControls; using System.Web.UI.WebControls.WebParts; using System.Xml.Linq; using System.Data.SqlClient; public partial class _Default : System.Web.UI.Page { string strConnString = ConfigurationManager.ConnectionStrings["ConnectionString"].ConnectionString; SqlCommand com; protected void btn_insert_Click(object sender, EventArgs e) { SqlConnection con = new SqlConnection(strConnString); com = new SqlCommand("insert1", con); com.CommandType = CommandType.StoredProcedure; com.Parameters.Add("@custid", SqlDbType.VarChar).Value = TextBox1.Text; com.Parameters.Add("@custname", SqlDbType.VarChar).Value = TextBox2.Text; com.Parameters.Add("@custaddress", SqlDbType.VarChar).Value = TextBox3.Text; com.Parameters.Add("@prodid", SqlDbType.VarChar).Value = TextBox4.Text; con.Open(); com.ExecuteNonQuery(); con.Close(); com = new SqlCommand("insert2", con); com.CommandType = CommandType.StoredProcedure; com.Parameters.AddWithValue("@prodid", SqlDbType.VarChar).Value = TextBox4.Text; com.Parameters.Add("@prodname", SqlDbType.VarChar).Value = TextBox5.Text; com.Parameters.Add("@price", SqlDbType.Int).Value = TextBox6.Text; con.Open(); com.ExecuteNonQuery(); con.Close(); Label7.Text = "Record Successfully Inserted"; TextBox1.Text=""; TextBox2.Text=""; TextBox3.Text=""; TextBox4.Text=""; TextBox5.Text=""; TextBox6.Text=""; } }