是否有可能在JavaFX中创建动态Bindings.OR？

Question

让我们考虑以下情况。有一个Pane parentPane，并有Pane firstChildPane, secondChildPane, thirdChildPane ...。子窗格添加到父窗格。我如何才能让parentPane在只有在这种情况下才可见，如果任何子窗格是可见的，考虑到可以动态添加和删除子窗格而没有任何限制和以任何顺序。当然childPane可见状态也可以随时更改。是否有可能创建动态Bindings.OR，以便我可以动态地向/从它添加/删除子可视属性？如果是，那么如何？如果不是，那么在这种情况下使用什么解决方案？

Answer 1

您可以尝试大意如下的内容：

// list that fires updates if any members change visibility: 
ObservableList<Node> children = 
    FXCollections.observableArrayList(n -> new Observable[] {n.visibleProperty()}); 
// make the new list always contain the same elements as the pane's child list: 
Bindings.bindContent(children, parentPane.getChildren()); 
// filter for visible nodes: 
ObservableList<Node> visibleChildren = children.filter(Node::isVisible); 
// and now see if it's empty: 
BooleanBinding someVisibleChildren = Bindings.isNotEmpty(visibleChildren); 
// finally: 
parentPane.visibleProperty().bind(someVisibleChildren); 

另一种方法是创建自己的BooleanBinding直接：

Pane parentPane = ... ; 

BooleanBinding someVisibleChildren = new BooleanBinding() { 


    { 
     parentPane.getChildren().forEach(n -> bind(n.visibleProperty())); 

     parentPane.getChildren().addListener((Change<? extends Node> c) -> { 
      while (c.next()) { 
       c.getAddedSubList().forEach(n -> bind(n.visibleProperty())); 
       c.getRemoved().forEach(n -> unbind(n.visibleProperty())) ; 
      } 
     }); 

     bind(parentPane.getChildren()); 
    } 

    @Override 
    public boolean computeValue() { 
     return parentPane.getChildren().stream() 
      .filter(Node::isVisible) 
      .findAny() 
      .isPresent(); 
    } 
} 

parentPane.visibleProperty().bind(someVisibleChildren);