Help with a Color Calculator in Java

I want to know how to paint the background based on the color of the hex value inputed by the user. I have this:

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

public class SimpleColorCalc extends JFrame implements Runnable 
{
  ColorPanel cp;
  JButton show;
  JTextField hexCode;

  public SimpleColorCalc() 
  { 
    super("Simple Color Calculator"); 
    cp = new ColorPanel(); 
    show = new JButton("Show the Color"); 
    hexCode = new JTextField("ffffff");
    show.addActionListener(new ActionListener(){ 
      public void actionPerformed(ActionEvent e) 
      { 
        String text = hexCode.getText(); 
        try 
        { 
          int colorCode = Integer.parseInt(text, 16); 
          Color enteredColor = new Color(colorCode); 
          cp.setColor(enteredColor); 
        } 
        catch(NumberFormatException ex) 
        { 
          hexCode.setText("ffffff"); 
          cp.setColor(Color.white); 
        } 
        finally 
        { 
          repaint(); 
        }
      }
    });
  } 
  public static void main(String[] args) 
  {
    SimpleColorCalc scc = new SimpleColorCalc(); 
    javax.开发者_如何学运维swing.SwingUtilities.invokeLater(scc); 
  } 
  public void run() 
  { 
    setSize(400,300); 
    Container c = getContentPane();
    JPanel top = new JPanel(); 
    c.add(BorderLayout.NORTH, top); 
    top.setLayout(new GridLayout(1,2)); 
    top.add(show); 
    top.add(hexCode);
    setVisible(true); 
  }
}

But I want to know how to fix it such that incase the user decides to put a 0x in front of the hexadecimal code or not it will work. I also want to know how to convert the hex code into a color in java. I am having trouble with this.

This JUnit test may help you understand:

@Test
public void test1() {
    Integer hexInt = Integer.parseInt("FF0000", 16);
    Color createdColor = new Color(hexInt);
    assertEquals(Color.RED, createdColor);
}

You can use Integer.parseInt to turn the hexadecimal string into a number of base 16. Note this will throw an exception if the string is invalid (contains characters other than digits or a-f).

Color instances can then be created using the Integer.

I have included an assertion to show that the created is what we expect it to be.

Well, converting a hex-string into a Color:

Color myColor = Color.decode("0xFF0000");

Or you could do:

Color myColor2 = new Color(0xFF0000);

But then the input can't be a String, or else you'll get a NumberFormatException.

Figuring out if the input starts with an 0x should be easy