獲取選定的節點jtree java

Question

這是我的JTree

javax.swing.tree.DefaultMutableTreeNode treeNode1 = new javax.swing.tree.DefaultMutableTreeNode(".Net Framework");
javax.swing.tree.DefaultMutableTreeNode treeNode2 = new javax.swing.tree.DefaultMutableTreeNode("Install");
javax.swing.tree.DefaultMutableTreeNode treeNode3 = new javax.swing.tree.DefaultMutableTreeNode("V1");
treeNode2.add(treeNode3);

treeNode3 = new javax.swing.tree.DefaultMutableTreeNode("V2");
treeNode2.add(treeNode3);

treeNode3 = new javax.swing.tree.DefaultMutableTreeNode("V3");
treeNode2.add(treeNode3);

treeNode3 = new javax.swing.tree.DefaultMutableTreeNode("V4");
treeNode2.add(treeNode3);
treeNode1.add(treeNode2);

treeNode2 = new javax.swing.tree.DefaultMutableTreeNode("Repair");
treeNode3 = new javax.swing.tree.DefaultMutableTreeNode("V1");
treeNode2.add(treeNode3);

treeNode3 = new javax.swing.tree.DefaultMutableTreeNode("V2");
treeNode2.add(treeNode3);

treeNode3 = new javax.swing.tree.DefaultMutableTreeNode("V3");
treeNode2.add(treeNode3);

treeNode3 = new javax.swing.tree.DefaultMutableTreeNode("V4");
treeNode2.add(treeNode3);
treeNode1.add(treeNode2);

treeNode2 = new javax.swing.tree.DefaultMutableTreeNode("Removal");
treeNode3 = new javax.swing.tree.DefaultMutableTreeNode("Remove All");
treeNode2.add(treeNode3);
treeNode1.add(treeNode2);

jTree2.setModel(new javax.swing.tree.DefaultTreeModel(treeNode1));
jScrollPane5.setViewportView(jTree2);

這是幾乎可行的代碼

DefaultMutableTreeNode node = (DefaultMutableTreeNode) jTree2.getLastSelectedPathComponent();

if (node == null) {
      //Nothing is selected.     
    return;
}

if (node.isLeaf()) {
    System.out.println(node);
} 

我希望能夠做出if語句，說明安裝節點中的葉節點V1，然后執行此操作。 只是不確定使用if(node.equals(thisleaf)))該thisleaf對象應該等於什么。

Answer 1

聲明一個領域

private DefaultMutableTreeNode v1Node;

將第三行更改為

this.v1Node = new DefaultMutableTreeNode("V1");

和使用

if (node.equals(this.v1Node))

旁注：您應該導入javax.swing.tree.DefaultMutableTreeNode並通過簡單地使用DefaultMutableTreeNode而不是標准名稱來使代碼更具可讀性。 您還應該選擇有意義的變量名稱，而不是treeNode1，jTree2等。這樣的不良名稱使代碼很難理解。