Re: problem in inserting record in ms access.

From:

=?ISO-8859-1?Q?Arne_Vajh=F8j?= <arne@vajhoej.dk>

Newsgroups:

comp.lang.java.programmer

Date:

Thu, 04 Oct 2012 17:35:39 -0400

Message-ID:

<506e0132$0$289$14726298@news.sunsite.dk>

On 10/4/2012 12:47 AM, Navnath Gadakh wrote:

package javaapplication3;
import java.sql.*;

public class JavaApplication3 {
     Connection con;
     Statement st;
     ResultSet rs;

Make them private.

     public JavaApplication3()
     {
         connect();
     }

     public void connect()
     {
         try
         {
             String driver = "sun.jdbc.odbc.JdbcOdbcDriver";

I will not recommend the JDBC ODBC bridge unless you are absolutely
forced to use it.

There are plenty of alternatives.

             Class.forName(driver);

             String db = "jdbc:odbc:db1";
             con = DriverManager.getConnection(db);
             st = con.createStatement();

As Martin has explained then this is very likely the cause of
your specific problem.

             String sql = "select * from Table1";
             rs = st.executeQuery(sql);

             while(rs.next())
             {
                 String fname = rs.getString("fname");
                 String lname = rs.getString("lname");
                 String address = rs.getString("address");
                 String email = rs.getString("email");
                 String mobile = rs.getString("mobile");

                 System.out.println(fname+lname+address+email+mobile);
             }

         }catch(Exception ex)
         {

Always print the exception.

         }

         try
     {
             rs.moveToInsertRow();
             rs.updateString("fname","abc");
             rs.updateString("lname","xyz");
             rs.updateString("address","mubmai");
             rs.updateString("email","abc@gmail.com");
             rs.updateString("mobile","99854874154");
             rs.insertRow();

I would suggest using plain INSERT instead of this.

             st.close();
             rs.close();

     }
     catch(Exception err)
     {
         System.out.println("Error!!!");

Always ...

     }

     }
     public static void main(String[] args) {
         // TODO code application logic here
         new JavaApplication3();

Don't do such heavy work in the constructor.

}
}

Arne