Question

我知道如何在控制台中显示线程中的文本，但是如何将它放在TextArea中呢？此外，由于我的文本是字符和数字，我怎么能将它们转换为字符串，因为如果我是对的，TextArea只接受字符串？

我需要在TextArea中添加PrintNum，LetterSmall和LetterBig。有人知道吗？任何帮助将不胜感激！

public class pracice extends Application {


@Override
public void start(Stage primaryStage) {

    TextArea ta = new TextArea();

    Button btn = new Button();
    btn.setText("Show");
    btn.setOnAction(new EventHandler<ActionEvent>() {


        Runnable run = new PrintNum(25);
        Thread th1 = new Thread(run);

        char lett;
        char lettUp;

        Runnable let = new LetterSmall(lett);
        Thread th2 = new Thread(let);

        Runnable lUp = new LetterBig(lettUp);
        Thread th3 = new Thread(lUp);

        @Override
        public void handle(ActionEvent event) {
            System.out.append("\nBegin\n");

            th1.start();
            try{
                th1.join(2000);
            } catch (InterruptedException e){
                e.printStackTrace();
            }

            th2.start();
            try{
                th2.join();
            } catch (InterruptedException e){
                e.printStackTrace();
            }

            th3.start();
        }
    });


    BorderPane root = new BorderPane();
    root.setTop(btn);
    root.setCenter(ta);

    Scene scene = new Scene(root, 300, 250);

    primaryStage.setTitle("Practice");
    primaryStage.setScene(scene);
    primaryStage.show();
}

public static void main(String[] args) {

    launch(args);
}

}

class PrintNum implements Runnable {

private int lastNum;
Random r = new Random();

public PrintNum(int n){
    lastNum = n;
}

public void run(){
    System.out.append("\n");
    for(int i = 1; i <= lastNum; i++){
        int rN = r.nextInt(25) + 1;
        System.out.append((lastNum - rN) + " ");
    }
}

}

class LetterSmall implements Runnable {

Random r = new Random();

private char lett;

public LetterSmall(char s){
    lett = s;
}

public void run(){
   System.out.append("\n");
    for(int i = 1; i <= 25; i++){
        char c = (char) (r.nextInt(26) + 'a');
        lett = c;
        System.out.append(lett + " ");
    }
}

}

class LetterBig implements Runnable {

Random r = new Random();

private char lettUp;

public LetterBig(char up){
    lettUp = up;
}

public void run(){
    System.out.append("\n");
    for(int i = 1; i <= 25; i++){
        char c = (char) (r.nextInt(26) + 'A');
        lettUp = c;
        System.out.append(lettUp + " ");
    }
}

}

Answer 1

所以我找到了TextArea输入的解决方案。我没有创建三个类线程，而是在主JavaFX类中移动它们。谢谢大家的帮助。

看起来像这样：

new Thread(new Runnable() {
        private int lastNum = 25;
        private char lett;
        private char lettUp;

        Random r = new Random();


        @Override
        public void run() {

            Thread th1 = new Thread();
            ta.appendText("\n");
            for(int i = 1; i <= lastNum; i++){
                int rN = r.nextInt(25) + 1;
                ta.appendText((lastNum - rN) + " ");
            }

            Thread th2 = new Thread();
            ta.appendText("\n\n");
            for(int i = 1; i <= 25; i++){
                char c = (char) (r.nextInt(26) + 'a');
                lett = c;
                ta.appendText(lett + " ");
            }

            Thread th3 = new Thread();
            ta.appendText("\n\n");
            for(int i = 1; i <= 25; i++){
                char c = (char) (r.nextInt(26) + 'A');
                lettUp = c;
                ta.appendText(lettUp + " ");
            }

            th1.start();
            try{
                th1.join(2000);
            } catch (InterruptedException e){
                e.printStackTrace();
            }
            th2.start();
            try{
                th1.join();
            } catch (InterruptedException e){
                e.printStackTrace();
            }
            th3.start();
            try{
                th1.join();
                th2.join();
                th3.join();
            } catch (InterruptedException e){
                e.printStackTrace();
            }

        }
    }).start();

线程放在TextArea JavaFX中

1 个答案: