我用一条线连接两个节点。我可以通过拖放CubicCurve将一条线从一个节点拖到另一个节点的圆圈。
我的节点看起来像这样:
我的问题是,在我放下CubicCurve并设置起点和终点后,'Anchor'点是我的DragNode的高度/ 2和宽度/ 2。但我希望它们位于我的圈子的中心(我的节点的左侧或右侧)。
我当前的bindEnds() - 函数,我将曲线链接到我的DragNode中心(AnchorPane):
public void bindEnds (DragNode source, DragNode target) {
cubicCurve.startXProperty().bind(
Bindings.add(source.layoutXProperty(), (source.getWidth() / 2.0)));
cubicCurve.startYProperty().bind(
Bindings.add(source.layoutYProperty(), (source.getWidth() / 2.0)));
cubicCurve.endXProperty().bind(
Bindings.add(target.layoutXProperty(), (target.getWidth() / 2.0)));
cubicCurve.endYProperty().bind(
Bindings.add(target.layoutYProperty(), (target.getWidth() / 2.0)));
source.registerLink (getId());
target.registerLink (getId());
}
我正在考虑将我的bindEnds() - 函数改为这样的东西,我有节点,还有他们的子圈和它们的中心,我想绑定我的链接曲线:
public void bindEnds (DragNode source, DragNode target, Circle c1, Circle c2) {
source.getChildren().add(c1);
target.getChildren().add(c2);
cubicCurve.startXProperty().bind(
Bindings.add(source.layoutXProperty(), (c1.getLayoutX())));
cubicCurve.startYProperty().bind(
Bindings.add(source.layoutYProperty(), (c1.getLayoutY())));
cubicCurve.endXProperty().bind(
Bindings.add(target.layoutXProperty(), (c2.getLayoutX())));
cubicCurve.endYProperty().bind(
Bindings.add(target.layoutYProperty(), (c2.getLayoutY())));
source.registerLink (getId());
target.registerLink (getId());
}
并在我的Window ControllerClass中:
private void buildDragHandlers() {
this.setOnDragDone (new EventHandler <DragEvent> (){
@Override
public void handle (DragEvent event) {
DragContainer container = (DragContainer) event.getDragboard().getContent(DragContainer.AddNode);
container = (DragContainer) event.getDragboard().getContent(DragContainer.AddLink);
if (container != null) {
String sourceId = container.getValue("source");
String targetId = container.getValue("target");
if (sourceId != null && targetId != null) {
NodeLink link = new NodeLink();
rightAnchor.getChildren().add(0,link);
DragNode source = null;
DragNode target = null;
for (Node n: rightAnchor.getChildren()) {
if (n.getId() == null)
continue;
if (n.getId().equals(sourceId)){
source = (DragNode) n;
}
if (n.getId().equals(targetId)){
target = (DragNode) n;
}
}
if (source != null && target != null){
source.link(target);
link.bindEnds(source, target, c1, c2);
}
}
}
}
});
在我的DragNode控制器类中:
private ArrayList<Circle> circles = new ArrayList<Circle>();
private Circle getNearestCircle(DragNode source) {
Circle nearestCircle = null;
for (Circle circle : circles) {
if (nearestCircle == null) {
nearestCircle = circle;
} else {
}
}
return nearestCircle;
}
public void link (DragNode source) {
getNearestCircle(source).centerXProperty().bindBidirectional(source.getNearestCircle(this).centerXProperty());
getNearestCircle(source).centerYProperty().bindBidirectional(source.getNearestCircle(this).centerYProperty());
}
我必须让我可以访问我使用的圈子,并将它们放在link.indEnds(source, target);
任何人都可以帮助我吗?
答案 0 :(得分:1)
只需让您的圈子成为DragNode的成员。然后你可以创建一个方法,例如target.getNearestCircle()获取最接近拖动节点的圆。下面的代码或多或少是元代码,但我希望你能得到这个想法:
DragNode:
class Foo {
constructor () {
this.name = 'foo';
}
}
function Bar () {
this.name = 'bar';
}
function doSmth (anyArg) {
if (typeof anyArg === 'function') {
var obj = { someProp: 'qux' };
try {
anyArg.call(obj);
} catch(e) {
var x = Reflect.construct(anyArg, []);
Object.assign(obj, x);
}
return obj;
}
}
doSmth(Bar);
doSmth(Foo);