使用JTree,假设根节点为0级并且根目录下可能有多达5个级别,我如何轻松扩展所有1级节点,以便所有1级和1级节点都可以。可以看到2个分支和叶子,但3级和3级以下不是?
答案 0 :(得分:21)
感谢快速回复的人。但是我现在找到了我正在寻找的简单解决方案。出于某种原因,我在JavaDocs中看不到DefaultMutableTreeNode.getLevel()!仅供参考我现在正在做的事情是:
DefaultMutableTreeNode currentNode = treeTop.getNextNode();
do {
if (currentNode.getLevel()==1)
myTree.expandPath(new TreePath(currentNode.getPath()));
currentNode = currentNode.getNextNode();
}
while (currentNode != null);
答案 1 :(得分:3)
你有一些Tree实用程序类正是这样做的:
赞this one:
public class SimpleNavigatorTreeUtil {
/**
* Expands/Collapse specified tree to a certain level.
*
* @param tree jtree to expand to a certain level
* @param level the level of expansion
*/
public static void expandOrCollapsToLevel(JTree tree, TreePath treePath,int level,boolean expand) {
try {
expandOrCollapsePath(tree,treePath,level,0,expand);
}catch(Exception e) {
e.printStackTrace();
//do nothing
}
}
public static void expandOrCollapsePath (JTree tree,TreePath treePath,int level,int currentLevel,boolean expand) {
// System.err.println("Exp level "+currentLevel+", exp="+expand);
if (expand && level<=currentLevel && level>0) return;
TreeNode treeNode = ( TreeNode ) treePath.getLastPathComponent();
TreeModel treeModel=tree.getModel();
if ( treeModel.getChildCount(treeNode) >= 0 ) {
for ( int i = 0; i < treeModel.getChildCount(treeNode); i++ ) {
TreeNode n = ( TreeNode )treeModel.getChild(treeNode, i);
TreePath path = treePath.pathByAddingChild( n );
expandOrCollapsePath(tree,path,level,currentLevel+1,expand);
}
if (!expand && currentLevel<level) return;
}
if (expand) {
tree.expandPath( treePath );
// System.err.println("Path expanded at level "+currentLevel+"-"+treePath);
} else {
tree.collapsePath(treePath);
// System.err.println("Path collapsed at level "+currentLevel+"-"+treePath);
}
}
}
基本上,您需要探索子节点,直到满足您的条件(此处为深度级别),然后展开所有节点直到该点。
答案 2 :(得分:1)
由于您的1级节点都是根节点的子节点,并且假设您正在使用DefaultMutableTreeNode
(调用getPath()
所需),您可以像根节点的子节点一样迭代这样:
Enumeration<?> topLevelNodes
= ((TreeNode)tree.getModel().getRoot()).children();
while(topLevelNodes.hasMoreElements()) {
DefaultMutableTreeNode node
= (DefaultMutableTreeNode)topLevelNodes.nextElement();
tree.expandPath(new TreePath(node.getPath()));
}
答案 3 :(得分:0)
对所有2级节点使用expand(TreePath)。
答案 4 :(得分:0)
这应该有效 -
import javax.swing.*;
import javax.swing.tree.*;
import java.awt.BorderLayout;
import java.awt.event.*;
import java.util.*;
public class Tree {
public static void main(String[] args) {
JPanel panel = new JPanel(new BorderLayout());
final JTree tree = new JTree();
panel.add(new JScrollPane(tree));
JButton btn = new JButton("Press Me");
btn.addActionListener(new ActionListener() {
public void actionPerformed(ActionEvent ae) {
for (Enumeration e = ((TreeNode)tree.getModel().getRoot()).children();e.hasMoreElements();) {
TreeNode tn = (TreeNode)e.nextElement();
tree.expandPath(new TreePath(((DefaultTreeModel)tree.getModel()).getPathToRoot(tn)));
}
}
});
panel.add(btn, BorderLayout.SOUTH);
JFrame frame = new JFrame("");
frame.getContentPane().add(panel);
frame.setSize(300, 300);
frame.setLocation(100, 100);
frame.pack();
frame.show();
}
}