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JavaFX : StackPane Sequential Transition

转载 作者:塔克拉玛干 更新时间:2023-11-02 07:59:50 28 4
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我正在尝试使用 FadeTransition 在 3 个不同的 AnchorPane 之间切换(点击 Button),下面是我的代码,

public class TestSlide extends Application {
private ObjectBinding<Node> frontNode;

@Override
public void start(Stage primaryStage) {

StackPane root = new StackPane();
AnchorPane pane1 = new AnchorPane(new Button("1"));
AnchorPane pane2 = new AnchorPane(new Button("2"));
AnchorPane pane3 = new AnchorPane(new Button("3"));
root.getChildren().addAll(pane1, pane2, pane3);

handleAnimation(root);

BorderPane border= new BorderPane(root);

HBox bottom = new HBox(10);
Button front1 = new Button("Pane 1");
Button front2 = new Button("Pane 2");
Button front3 = new Button("Pane 3");
front1.setOnAction((ActionEvent event) -> {
pane1.toFront();
});
front2.setOnAction((ActionEvent event) -> {
pane2.toFront();
});
front3.setOnAction((ActionEvent event) -> {
pane3.toFront();
});
bottom.getChildren().addAll(front1, front2, front3);
border.setBottom(bottom);

Scene scene = new Scene(border,400,400);
primaryStage.setScene(scene);
primaryStage.show();

}

private void handleAnimation(StackPane root) {
frontNode = Bindings.valueAt(root.getChildren(),
Bindings.size(root.getChildren()).subtract(1));
frontNode.addListener((obs, oldNode, newNode) -> {
SequentialTransition fadeOutIn = new SequentialTransition();
if (oldNode != null) {
FadeTransition fadeOut = new FadeTransition(Duration.millis(500), oldNode);
fadeOut.setToValue(0);
fadeOutIn.getChildren().add(fadeOut);
}
if (newNode != null) {
FadeTransition fadeIn = new FadeTransition(Duration.millis(500), newNode);
fadeIn.setFromValue(0);
fadeIn.setToValue(1);
fadeOutIn.getChildren().add(fadeIn);
}
fadeOutIn.play();
});
}

public static void main(String[] args) {
launch(args);
}
}

handleAnimation 方法引用自另一个 SO 帖子。

问题是,

  1. 启动应用程序后,单击 Pane 1 按钮。 -> Transition 将首先显示 pane2,然后是 pane1

  2. 现在点击 Pane 3 Button -> Transition 将首先显示 pane2 然后是 pane3 .

  3. 现在点击Pane 2 Button -> Transition会显示pane2,上面2点提到的问题不会出现再次出现。

为什么在第 1 点和第 2 点显示实际 Pane 之前转换显示 pane2?是由于不透明度设置吗?

为什么在第 3 点之后问题得到解决?

如何在不显示第三个 Pane 的情况下使过渡工作到 FadeInFadeOut 各自的 Pane

最佳答案

这里的问题是 StackPane 的子级的初始状态是错误的:所有节点的不透明度为 1。当没有动画运行时,您想要的状态具有所有节点,但最后一个节点完全透明(不透明度 = 0)和最后一个完全不透明(不透明度 = 1)。您应该能够通过正确初始化不透明度来解决此问题:

root.getChildren().addAll(pane1, pane2, pane3);

// set opacity for all but the last child to 0
List<Node> children = root.getChildren();
for (int i = children.size()-2; i >= 0; i--) {
children.get(i).setOpacity(0);
}

否则会发生以下情况:

就在 pane1.toFront() 之后。请注意,(SequentialTransition 确保建立动画开始的状态。

最上面的节点是列表中的最后一个子节点,----... 放在可见“层”的旁边。

Pane 1: opacity = 0
Pane 3: opacity = 1 ------------------------------
Pane 2: opacity = 1

现在 SequentialTransition 的前半部分完成后,看起来如下:

Pane 1: opacity = 0
Pane 3: opacity = 0
Pane 2: opacity = 1 ------------------------------

动画完成后:

Pane 1: opacity = 1 ------------------------------
Pane 3: opacity = 0
Pane 2: opacity = 1

使用 pane3.toFront() 产生类似的结果:

Pane 3: opacity = 0
Pane 1: opacity = 1 ------------------------------
Pane 2: opacity = 1
Pane 3: opacity = 0
Pane 1: opacity = 0
Pane 2: opacity = 1 ------------------------------
Pane 3: opacity = 1 ------------------------------
Pane 1: opacity = 0
Pane 2: opacity = 1

关于JavaFX : StackPane Sequential Transition,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/57471869/

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