从 heightProperty() 和 widthProperty() 中获取最大值和最小值

Question

是否有任何代码，例如 Math.max(num1, num2) 但用于比较两个 DoubleProperty s？ 我目前正在尝试显示一个可以根据窗口大小自动调整自身大小的圆圈（从窗格延伸）。我想尝试在两者之间取较小的值来设置圆的半径。

import javafx.application.Application;
import javafx.scene.Scene;
import javafx.scene.shape.Circle;
import javafx.scene.layout.Pane;
import javafx.stage.Stage;

public class DisplayCircle extends Application {
   @Override
   public void start(Stage primaryStage) {
      Scene scene = new Scene(ResizableCircle(), 400, 300);
      primaryStage.setTitle("DisplayCircle");
      primaryStage.setScene(scene);
      primaryStage.show();
   }

   public class ResizableCircle extends Pane {
      public ResizableCircle() {
         Circle c = new Circle(getWidth()/2, getHeight()/2);
         c.centerXProperty().bind(widthProperty().subtract(10));
         c.centerYProperty().bind(heightProperty().subtract(10));


         // Need help setting the radius and binding it


         getChildren().add(c);
   }
}

Answer 1

您可以使用Bindings.max(ObservableNumberValue op1, ObservableNumberValue op2) 和Bindings.min(ObservableNumberValue op1, ObservableNumberValue op2) 将ObservableNumberValue 绑定到其他两个可观察值的最小值或最大值：

DoubleProperty min = new SimpleDoubleProperty();
DoubleProperty max = new SimpleDoubleProperty();

min.bind(Bindings.min(widthProperty(), heightProperty()));
max.bind(Bindings.max(widthProperty(), heightProperty()));