如何在JavaFX中获取父节点中的所有节点？

Question

In C# I found a method that was pretty sweet that allowed you to get all the descendants and all of THEIR descendants from a specified control.

I'm looking for a similar method for JavaFX.

I saw that the Parent class is what I want to work with since it is the class from which all Node classes that bear children are derived.

This is what I have so far (and I haven't really found anything on google with searches like "JavaFX get all nodes from a scene"):

public static ArrayList<Node> GetAllNodes(Parent root){
    ArrayList<Node> Descendents = new ArrayList<>();
    root.getChildrenUnmodifiable().stream().forEach(N -> {
        if (!Descendents.contains(N)) Descendents.add(N);
        if (N.getClass() == Parent.class) Descendents.addAll(
            GetAllNodes((Parent)N)
        );
    });
}

So how do I tell if N is a parent (or extended from a parent)? Am I doing that right? It doesn't seem to be working... It's grabbing all the nodes from the root (parent) node but not from the nodes with children in them. I feel like this is something that's probably got an answer to it but I'm just asking the question... wrong. How do I go about doing this?

Answer 1

public static ArrayList<Node> getAllNodes(Parent root) {
    ArrayList<Node> nodes = new ArrayList<Node>();
    addAllDescendents(root, nodes);
    return nodes;
}

private static void addAllDescendents(Parent parent, ArrayList<Node> nodes) {
    for (Node node : parent.getChildrenUnmodifiable()) {
        nodes.add(node);
        if (node instanceof Parent)
            addAllDescendents((Parent)node, nodes);
    }
}

Answer 2

Unfortunately this won't get subnodes for most container components. If you try a TabPane as parent, you'll find no children, but you can find tabs in it with getTabs() . The same is with SplitPane and other. So every container will require a specific approach.

You could use node.lookupAll("*") , but it also doesn't look inside.

The solution could be a "Prototype" pattern - creating a meta class with common interface of getChildren() method, which is realized in subclasses - one for each type.

Approach example is given here .

Answer 3

I use this,

public class NodeUtils {

    public static <T extends Pane> List<Node> paneNodes(T parent) {
        return paneNodes(parent, new ArrayList<Node>());
    }

    private static <T extends Pane> List<Node> paneNodes(T parent, List<Node> nodes) {
        for (Node node : parent.getChildren()) {
            if (node instanceof Pane) {
                paneNodes((Pane) node, nodes);
            } else {
                nodes.add(node);
            }
        }

        return nodes;
    }
}

Usage,

List<Node> nodes = NodeUtils.paneNodes(aVBoxOrAnotherContainer);

This source code uses the references of the existing nodes. It does not clone them.

Answer 4

I'd like to add to Hans' answer, that you have to check if parent is a SplitPane . Because SplitPane s have an empty list using getUnmodifiableChildren() , you'll have to use getItems() instead. (I do not know if there are other parents that do not provide their children via getUnmodifiableChildren() . SplitPane was the first I found...)

Answer 5

This seems to get ALL nodes. (In Kotlin)

fun getAllNodes(root: Parent): ArrayList<Node> {
    var nodes = ArrayList<Node>()
    fun recurseNodes(node: Node) {
        nodes.add(node)
        if(node is Parent)
            for(child in node.childrenUnmodifiable) {
                recurseNodes(child)
            }
    }
    recurseNodes(root)
    return nodes
}

Answer 6

This works for me:

public class FXUtil {

    public static final List<Node> getAllChildren(final Parent parent) {
        final List<Node> result = new LinkedList<>();

        if (parent != null) {

            final List<Node> childrenLvl1 = parent.getChildrenUnmodifiable();
            result.addAll(childrenLvl1);

            final List<Node> childrenLvl2 =
                    childrenLvl1.stream()
                                .filter(c -> c instanceof Parent)
                                .map(c -> (Parent) c)
                                .map(FXUtil::getAllChildren)
                                .flatMap(List::stream)
                                .collect(Collectors.toList());
            result.addAll(childrenLvl2);
        }

        return result;
    }

}