Question

这并没有改变我的标签。只有第一个运行而其他人不会改变。在Java中它起作用但在Javafx中并没有。

<script>
    jQuery(document).on("click",".consult-header .consult-primary-menu li a",function(event){
      event.preventDefault();
      var thishref =jQuery(this).attr("href");
      var url = thishref.substr(thishref.indexOf("#"));
      jQuery('html, body').animate({
                  scrollTop: $(url).offset().top
      }, 2000);
    });
</script>

Answer 1

当您调用Thread.sleep(40);时，您实际上是在停止JavaFX应用程序线程，并且不会有GUI更新。正如评论中指出的那样，您不应该暂停申请线程。

您可以从其他主题进行更新：

Task<Void> loadTask = new Task<Void>() {
    @Override
    protected Void call() throws Exception {
        for (int i = 0; i <= 200; i++) {
            Thread.sleep(40);
            String text = "";
            if (i <= 30)
                text = "Initilizing Components...";
            else if (i <= 50)
                text = "Initializing Database Connection...";
            else if (i <= 80)
                text = "Initializing User Interface...";
            else
                text = "Please wait...";

            String finalText = text;
            Platform.runLater(() -> lblLoad.setText(finalText));
        }
        return null;
    }
};


new Thread(loadTask).start();

Answer 2

您可以将javafx.animation.Timeline用于定期活动：

int counter=0;

Timeline doEvery40sec = new Timeline(new KeyFrame(Duration.seconds(40), new EventHandler<ActionEvent>() {

    @Override
    public void handle(ActionEvent event) {
if(counter==0) 
        lblLoad.setText("Initilizing Components...");
    }else if(counter==1){
        lblLoad.setText("Initializing Database Connection...");
    }else if(counter==2){
        lblLoad.setText("Initializing User Interface...");                           
    }else if(counter==3)
        lblLoad.setText("Please wait...");
counter++;  
    }
}));
fiveSecondsWonder.setCycleCount(4);
fiveSecondsWonder.play();

将JavaFX中的Label更改为不同的值

2 个答案: