所以我正在用gui进行研究任务,我在左侧面板上有3个框"输入","处理"one_answers"显示",如果使用CardLayout选择"输入"则第一个面板显示在右侧,如果选择"显示"则显示最后一个面板,但我无法获得"处理"显示在这部分:
public void actionPerformed( ActionEvent event ) {
// show first card
if ( event.getSource() == controls[ 0 ] )
cardManager.first( deck );
else if ( event.getSource() == controls[ 1 ] )
cardManager.show( card2Panel(), "c2");
// show previous card
else if ( event.getSource() == controls[ 2 ] )
cardManager.last( deck );
完整形式的代码:
import java.awt.*;
进口java.awt.event。;进口javax.swing。,
公共类BookCentre扩展JFrame实现ActionListener {
private CardLayout cardManager;
private JPanel deck;
private JButton controls[];
private String names[] = { "Input", "Processing", "Display"};
public BookCentre(){
super( "CardLayout" );
Container container = getContentPane();
deck = new JPanel();
cardManager = new CardLayout();
deck.setLayout( cardManager );
deck.add( card1Panel(), "c1" );
deck.add( card2Panel(), "c2" );
deck.add( card3Panel(), "c3" );
JPanel buttons = new JPanel();
buttons.setLayout( new GridLayout( 2, 2 ) );
controls = new JButton[ names.length ];
for ( int count = 0; count < controls.length; count++ ) {
controls[ count ] = new JButton( names[ count ] );
controls[ count ].addActionListener( this );
buttons.add( controls[ count ] );
container.add( buttons, BorderLayout.WEST );
container.add( deck, BorderLayout.EAST );
setSize( 450, 200 );
setVisible( true );}
}
public JPanel card1Panel(){
JLabel label1 = new JLabel( "card one", SwingConstants.CENTER );
JPanel card1 = new JPanel();
card1.add( label1 );
return card1;
}
public JPanel card2Panel(){
JLabel label2 = new JLabel( "card two", SwingConstants.CENTER );
JPanel card2 = new JPanel();
card2.setBackground( Color.yellow );
card2.add( label2 );
return card2;
}
public JPanel card3Panel(){
JLabel label3 = new JLabel( "card three" );
JPanel card3 = new JPanel();
card3.setLayout( new BorderLayout() );
card3.add( new JButton( "North" ), BorderLayout.NORTH);
card3.add( new JButton( "West" ), BorderLayout.WEST );
card3.add( new JButton( "East" ), BorderLayout.EAST );
card3.add( new JButton( "South" ), BorderLayout.SOUTH);
card3.add( label3, BorderLayout.CENTER );
return card3;
}
public void actionPerformed( ActionEvent event ) {
// show first card
if ( event.getSource() == controls[ 0 ] )
cardManager.first( deck );
else if ( event.getSource() == controls[ 1 ] )
cardManager.show( card2Panel(), "c2");
// show previous card
else if ( event.getSource() == controls[ 2 ] )
cardManager.last( deck );
}
public static void main( String args[] ) {
BookCentre cardDeckDemo = new BookCentre();
cardDeckDemo.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE );
}}
我可以使用cardManager。第一,下一个,前一个,最后一个。我只希望显示"Processing"面板(所以card2),而不是用户首先看到,最后看到或循环遍历所有面板。查看CardLayout的Java Trail。ActionListener中的cardManager.show()调用应该是
cardManager.show( deck, "c2" );
作为第一个参数是带有CardLayout的父容器,而不是您想要显示的组件。