java - 无法从 JavaFX 操作标签

Question

每次用户转到该页面时，我都试图更改 JavaFX 软件 GUI 标签的文本。为此，我必须能够更改该标签的值，例如将其解析为变量并执行这些操作。

问题是，由于标签都是用FXML写的，我不知道如何在Java上正确使用它。我试过:

@FXML public Label WordID;

为了使标签(以“WordID”作为 ID)成为对象，但我失败了。当我这样做时，程序将不再转到该帧。我该怎么办？

代码:

public class Screen3Controller implements Initializable, ControlledScreen {

ScreensController myController;

  @Override
public void initialize(URL url, ResourceBundle rb) {
    // TODO
}

public void setScreenParent(ScreensController screenParent){
    myController = screenParent;
}

@FXML
private void goToScreen1(ActionEvent event){
   myController.setScreen(ScreensFramework.screen1ID);
}

@FXML
private void goToScreen2(ActionEvent event){
   myController.setScreen(ScreensFramework.screen2ID);
}

}

FXML:

<Label fx:id="WordID" layoutX="393.0" layoutY="148.0" prefWidth="177.0" text="">
  <font>
    <Font name="Avenir Roman" size="40.0" />
  </font>
</Label>

来自 Screen2 的按钮监听器:

@FXML
private void goToScreen3(ActionEvent event){
   myController.setScreen(ScreensFramework.screen3ID);
   WordID.setText("LOl");
}

错误信息:

Glass detected outstanding Java exception at -[GlassViewDelegate sendJavaMouseEvent:]:src/com/sun/mat/ui/GlassViewDelegate.m:543 Exception in thread "JavaFX Application Thread" java.lang.RuntimeException: ... com.sun.javafx.event.CompositeEventHandler.dispatchBubblingEvent(CompositeEventHandler.java:64) at com.sun.javafx.event.EventHandlerManager.dispatchBubblingEvent(EventHandlerManager.java:217) at com.sun.javafx.event.EventHandlerManager.dispatchBubblingEvent(EventHandlerManager.java:170) at com.sun.javafx.event.CompositeEventDispatcher.dispatchBubblingEvent(CompositeEventDispatcher.java:38)com.sun.javafx.tk.quantum.GlassViewEventHandler$MouseEventNotification.run(GlassViewEventHandler.java:250) at com.sun.javafx.tk.quantum.GlassViewEventHandler$MouseEventNotification.run(GlassViewEventHandler.java:173) at java.security.AccessController.doPrivileged(Native Method) at javafx.fxml.FXMLLoader$ControllerMethodEventHandler.handle(FXMLLoader.java:1444) ... 40 more Caused by: java.lang.NullPointerException at screensframework.Screen3Controller.mostrarPalavra(Screen3Controller.java:89) ... 50 more

Answer 1

要以编程方式使用 Label，您需要做的就是在 Controller 文件中声明一个变量，其 id 与在您的 中声明的相同.fxml 文件:

你的FXML.fxml

<AnchorPane xmlns:fx="http://javafx.com/fxml/1" xmlns="http://javafx.com/javafx/2.2" fx:controller="com.sample.Screen3Controller ">
    ...
    <Label fx:id="WordID" layoutX="393.0" layoutY="148.0" prefWidth="177.0" text="">
      <font>
        <Font name="Avenir Roman" size="40.0" />
      </font>
    </Label>
    ...
</AnchorPane>

com.sample.Screen3Controller 必须引用 com.sample 包中的现有 .java 文件。

Screen3Controller.java

import javafx.scene.control.Label;  //make sure you're importing the JavaFX Label

public class Screen3Controller implements Initializable, ControlledScreen {

    ScreensController myController;
    @FXML private Label WordID;

    ...

    WordID.setText("some text...");

    ...
}