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社区首页 >问答首页 >如何将list<String>转换为DefaultMutableTreeNode?

如何将list<String>转换为DefaultMutableTreeNode?
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Stack Overflow用户
提问于 2011-02-17 20:49:30
回答 1查看 2K关注 0票数 1

如何将list转换为DefaultMutableTreeNode?

我正在从列表中的数据库中获取数据。现在,我想将该列表转换为JTree。我该怎么做呢?

代码语言:javascript
复制
public class ExpressionTree extends JFrame implements ActionListener {

private JComboBox group;    
private JLabel groupLabel;
Container content;
JTree tree;
DefaultMutableTreeNode root, child;

public static void main(String[] args) {
    new ExpressionTree();
}


public ExpressionTree() {

    super("Creating a Simple JTree");
    WindowUtilities.setNativeLookAndFeel();
    addWindowListener(new ExitListener());
    content = getContentPane();

    group = new JComboBox();
    groupLabel = new JLabel("GROUP");

    ExpressionBuilder expressionBuilder = new ExpressionBuilder(); 
    Set<Integer> set = expressionBuilder.getGroup().keySet();
    Iterator<Integer> iterator = set.iterator();
    group.addItem("select");
    while(iterator.hasNext()){
        group.addItem(expressionBuilder.getGroup().get(iterator.next()));       
    }


    content.add(groupLabel);
    content.add(group);
    group.addActionListener(this);
    content.setLayout(new FlowLayout());
    root= new DefaultMutableTreeNode("root");
    tree = new JTree(root);
    content.add(new JScrollPane(tree), BorderLayout.CENTER);
    setSize(300, 475);
    setVisible(true);

}
public String groupString = null;
public int groupId;

public void actionPerformed(ActionEvent e){

    groupString = group.getSelectedItem().toString();
    ExpressionBuilder builder=new ExpressionBuilder();
    builder.getGroupId(groupString);
    //tree.setModel(new NodeTree());.........//what to write over here
    //root.add(new NodeTree());
}
}


public class NodeTree  extends DefaultMutableTreeNode{

    private JTree tree;
    DefaultMutableTreeNode top; 
    public NodeTree(){
            ExpressionTree expressionTree= new ExpressionTree();
            String groupstring = expressionTree.groupString;
            int groupId = expressionTree.groupId;
            top = new DefaultMutableTreeNode(groupstring);
            tree = new JTree(top);
            tree.setEditable(true);
            JScrollPane treeView = new JScrollPane(tree);
            createNode(top,groupId);

    }

    public void createNode(DefaultMutableTreeNode top,int group){
        DefaultMutableTreeNode expression[] = null;
        ExpressionBuilder builder=new ExpressionBuilder();
        Object object = (Object)builder.getExpression(group,0);
        expression[1] = new DefaultMutableTreeNode(object);  
        top.add(expression[1]);
        expression[1].add(new DefaultMutableTreeNode((Object)builder.getExpression(group,1)));

    }

}
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回答 1

Stack Overflow用户

发布于 2011-02-17 23:36:42

假设您正在从数据库中正确获取节点名称,并且节点是平面的(无层次结构),您可以使用以下代码创建一棵树

代码语言:javascript
复制
private JTree createTree(List<String> nodeNames) {

    DefaultMutableTreeNode root = new DefaultMutableTreeNode(new Object());
    TreeModel model = new DefaultTreeModel(root);
    for (int i = 0; i < nodeNames.size(); i++) {
      root.add(new DefaultMutableTreeNode(nodeNames.get(i)));
    }
    JTree tree = new JTree(model);
    tree.setRootVisible(false);
    return tree;
  }

注意:我创建了虚拟节点作为根,并使用了setRootVisible(false)。如果您有一个有效的根节点,您可以使用它并删除隐藏根节点的行。

票数 0
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页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接:

https://stackoverflow.com/questions/5029124

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