How to dynamic increment with character for example E001

Question

How to dynamic increment with charactor using C# with Sql Server. For example: Customer ID E0001E0002 like that.. Please any one know this answer plz help me soon as possible.

Sam Hobbs · Answer

The easiest and most reliable way is to create an auto-incrementing database field and then when you insert a new customer get the number of the new customer record. The number will be only the number part; it would not include the E part. The E could be added whenever it is needed for formatting. Will the E part ever be something other than E? If so then what is the meaning of it?

magesh manavalan · Answer

Hi dear thanx for ur co operations and help ur codings. Thanks&Regs Magesh.A

Satyapriya Nayak · Answer

Hi Magesh Create table employee(empid varchar(50)empname varchar(50)sal int) Default.aspx code Untitled Page Width=100px> Text=Insert Data /> 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; string str; SqlCommand com; int count; protected void Page_Load(object sender EventArgs e) { SqlConnection con = new SqlConnection(strConnString); str = select count(*) from employee; com = new SqlCommand(str con); con.Open(); count = Convert.ToInt16(com.ExecuteScalar()) + 1; txt_empid.Text = E000 + count; con.Close(); } protected void btn_insert_Click(object sender EventArgs e) { SqlConnection con = new SqlConnection(strConnString); con.Open(); str = insert into employee values(' + txt_empid.Text.Trim() + '' + txt_empname.Text.Trim() + ' + txt_sal.Text.Trim() + ); com = new SqlCommand(str con); com.ExecuteNonQuery(); con.Close(); Label4.Text = Records successfully Inserted; } } Thanks If this post helps you mark it as answer