我正在尝试使用JTree的内容使用CardLayout更改面板。它仅适用于最后一个选择。我应该对我的代码执行哪些侦听器或更改?
我也在控制台中编写了文本,它似乎获得了正确的值。这就是为什么它让我感到沮丧。
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();
并且只影响最后的组成部分。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();
}
}
这个...
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,您不能为一个组件的同一实例分配多个键。
当您尝试第二次添加组件时,首先将其从其父容器CardLayout
中删除,这也导致删除了“名称”,仅意味着行...
card.add(label1, "1st level: Child 3");
// and...
card.add(label2, "2nd level: Child 3");
正在实际运行(已添加到UI /中CardLayout
)
您将必须为每个名称提供该组件的新实例。
另外,您可以向您提供更多详细信息,TreeNode
以便确定它是哪个级别并显示该级别的正确组件,这意味着您只有两个组件,CardLayout
每个级别一个。
将DefaultMutableTreeNode
允许您在“用户”提供Object
给它。默认情况下,它使用toString
该对象的方法作为节点的文本,但是您可以使用TreeCellRenderer
来自定义此对象或为toString
“自定义级别”对象的方法提供值
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