我正在尝试使用JTree的内容使用CardLayout更改Panel。它只适用于最后的选择。我应该对我的代码做什么样的听众或更改?
我也在控制台中编写文本,似乎得到了正确的值。这就是为什么让我感到沮丧的原因。
单击第一级节点时,我的代码应显示第一级,单击第二级节点时,我的代码应显示第二级。我用了
DefaultMutableTreeNode selectedNode = (DefaultMutableTreeNode) tree
.getLastSelectedPathComponent();
它只影响最后的组件。我该如何解决?谢谢!
源代码:
import java.awt.BorderLayout;
import java.awt.CardLayout;
import java.awt.Color;
import java.awt.Dimension;
import javax.swing.BorderFactory;
import javax.swing.JFrame;
import javax.swing.JLabel;
import javax.swing.JPanel;
import javax.swing.JScrollPane;
import javax.swing.JTree;
import javax.swing.event.TreeSelectionEvent;
import javax.swing.event.TreeSelectionListener;
import javax.swing.tree.DefaultMutableTreeNode;
import javax.swing.tree.DefaultTreeModel;
public class ProblemTree2 extends JFrame {
private DefaultMutableTreeNode root = new DefaultMutableTreeNode("Root");
private DefaultTreeModel model = new DefaultTreeModel(root);
private JTree tree = new JTree(model);
private JPanel card;
public ProblemTree2() {
card = new JPanel(new CardLayout());
card.setBorder(BorderFactory.createLineBorder(Color.BLACK));
JLabel label1 = new JLabel("1st level");
JLabel label2 = new JLabel("2nd level");
DefaultMutableTreeNode n1 = new DefaultMutableTreeNode(
"1st level: Child 1");
n1.add(new DefaultMutableTreeNode("2nd level: Child l"));
DefaultMutableTreeNode n2 = new DefaultMutableTreeNode(
"1st level: Child 2");
n2.add(new DefaultMutableTreeNode("2nd level: Child 2"));
DefaultMutableTreeNode n3 = new DefaultMutableTreeNode(
"1st level: Child 3");
n3.add(new DefaultMutableTreeNode("2nd level: Child 3"));
card.add(label1,"1st level: Child 1");
card.add(label1,"1st level: Child 2");
card.add(label1,"1st level: Child 3");
card.add(label2,"2nd level: Child l");
card.add(label2,"2nd level: Child 2");
card.add(label2,"2nd level: Child 3");
root.add(n1);
root.add(n2);
root.add(n3);
tree.setEditable(true);
tree.setSelectionRow(0);
tree.setRootVisible(true);
tree.setShowsRootHandles(true);
tree.getSelectionModel().addTreeSelectionListener(
new TreeSelectionListener() {
@Override
public void valueChanged(TreeSelectionEvent e) {
final CardLayout cards = (CardLayout) card
.getLayout();
DefaultMutableTreeNode selectedNode = (DefaultMutableTreeNode) tree
.getLastSelectedPathComponent();
System.out.println(selectedNode.toString());
cards.show(card,selectedNode.toString());
}
});
JScrollPane scrollPane = new JScrollPane(tree);
scrollPane.setPreferredSize(new Dimension(500,500));
getContentPane().add(scrollPane, BorderLayout.WEST);
getContentPane().add(card, BorderLayout.CENTER);
setSize(1000, 600);
setVisible(true);
}
public static void main(String[] arg) {
ProblemTree2 pt = new ProblemTree2();
}
}
答案 0 :(得分:4)
此...
card.add(label1, "1st level: Child 1");
card.add(label1, "1st level: Child 2");
card.add(label1, "1st level: Child 3");
card.add(label2, "2nd level: Child l");
card.add(label2, "2nd level: Child 2");
card.add(label2, "2nd level: Child 3");
导致您的问题,您无法为组件分配多个密钥的相同实例。
当您尝试再次添加该组件时,它首先会从其父容器中删除,这会导致CardLayout
删除" name"同样,只意味着......
card.add(label1, "1st level: Child 3");
// and...
card.add(label2, "2nd level: Child 3");
实际上正在运行(已添加到用户界面/ CardLayout
)
您必须为每个名称提供组件的新实例。
或者,您可以向TreeNode
提供更多详细信息,以便您确定该级别并显示该级别的正确组件,这意味着您CardLayout
中只有两个组件,每个级别一个。
DefaultMutableTreeNode
允许您提供"用户" Object
到它。默认情况下,它使用此对象的toString
方法作为节点的文本,但您可以使用TreeCellRenderer
自定义此方法或为您的toString
方法提供值#34;自定义级别"对象