import java.awt.*;
import java.sql.*;
import java.awt.event.*;
import javax.swing.*;

class Search{
    public static void main(String[] args) {
        JLabel lab=new JLabel("Enter Name:");
        final JTextField t=new JTextField(20);
        JButton b = new JButton("Search");
        JPanel p = new JPanel(new GridLayout(2,2));
        p.add(lab);
        p.add(t);
        p.add(b);
        JFrame f = new JFrame();
        f.getContentPane().add(p);
        f.pack();
        f.setVisible(true);

        b.addActionListener(new ActionListener(){
            public void actionPerformed(ActionEvent e){
            String name=t.getText();
            try{
           Class.forName("com.mysql.jdbc.Driver");
           Connection con = DriverManager.getConnection("jdbc:mysql://localhost:3306/test", "root", "root");
           Statement st=con.createStatement();
           ResultSet rs=st.executeQuery("select * from data where name='"+name+"'");
           String n="",add="",contact="",email="";

           if(rs.next()){
               n=rs.getString("name");
                add=rs.getString("address");
                contact=rs.getString("contactNo");
                email=rs.getString("email");
           }
           String data[][]=new String[1][4];
           data[0][0]=n;
           data[0][1]=add;
           data[0][2]=contact;
           data[0][3]=email;
           JFrame frame=new JFrame();
           String labels[]={"Name","Address","Contact No","Email"};
           JTable table=new JTable(data,labels);
           JScrollPane pane=new JScrollPane(table);

            frame.add(pane);
            frame.setVisible(true);
            frame.pack();

            }
            catch(Exception ex){
            System.out.println(e);
            }
            }
        });
    }
}

thanks for you answer, now I can see by the code in my project ...

Thanks I managed to enter the code in my project so give me this error when I call my button, for example in another package a class called desktopapplication i have an interface that runs on my table this in mysql, so I do not interface desktopapplication I have a button called search, but the button is called I put the code:

search demo = new search (); demo.actionPerformed (evt);

but he did not recognize that the action perfomed. because it will be???

tou asking for your help once again .................