Question

我正在为我的一个朋友写一个小笑话应用程序，我无法弄清楚为什么我的标签没有升级并且GUI冻结了。我知道我必须使用线程进行后台操作才能继续使用GUI，但是在进入线程之前它甚至都会冻结。

Ui课程：

private class CheckButtonListener implements ActionListener {

    @Override
    public void actionPerformed(ActionEvent e) {
        check();
    }
}

public void check() {
    this.jbCheck.setEnabled(false);

    this.rand = (int) (Math.random() * 4) + 1;

            // Offending lines
    this.jlMessage1.setText("");
    this.jlMessage2.setText(this.messages.getString("wait_" + this.rand));
    this.jlMessage2.setForeground(Color.BLACK);
            // End of the offending lines

    ExecutorService service = null;
    try {
        service = Executors.newFixedThreadPool(1);
        Future<Integer> task = service.submit(new BackgroundJob());

        ping = task.get();
    } catch (final InterruptedException ex) {
        ex.printStackTrace();
    } catch (final ExecutionException ex) {
        ex.printStackTrace();
    }

    this.jlMessage1.setText("Tienes " + ping + " ms de ping.");
    if (ping <= 125) {
        this.jlMessage2.setText(this.messages.getString("bien_" + this.rand));
        this.jlMessage2.setForeground(Color.GREEN);
    } else if (ping > 126 && ping <= 200) {
        this.jlMessage2.setText(this.messages.getString("medio_" + this.rand));
        this.jlMessage2.setForeground(Color.YELLOW);
    } else {
        this.jlMessage2.setText(this.messages.getString("mal_" + this.rand));
        this.jlMessage2.setForeground(Color.RED);
    }

    service.shutdownNow();
    this.jbCheck.setEnabled(true);
}

这是后台工作：

public class BackgroundJob implements Callable<Integer> {

private int ping;

@Override
public Integer call() throws Exception {
    ProcessBuilder pb = new ProcessBuilder("cmd.exe", "/c", "ping " + Constants.IP + " | find \"Media\"");
    pb.redirectErrorStream(true);

    try {
        Process p = pb.start();

        BufferedReader br = new BufferedReader(new InputStreamReader(p.getInputStream()));
        String line = br.readLine().trim();
        ping = Integer.parseInt(line.split(" ")[8].split("ms")[0]);

    } catch (IOException e) {
        // TODO Auto-generated catch block
        e.printStackTrace();
    }
    return ping;
}

}

我不知道为什么在调用Future<Integer> task = service.submit(new BackgroundJob());

之前没有更改标签

Answer 1

GUI不会更改标签，因为您位于＆＃34; main＆＃34;线程，负责重新绘制GUI。你打电话

ExecutorService service = null;
try {
    service = Executors.newFixedThreadPool(1);
    Future<Integer> task = service.submit(new BackgroundJob());

    ping = task.get();

以异步方式生成线程。但是呼叫task.get();再次成为阻止操作。因此，您生成后台线程，但等待阻塞其结果。

即使我使用线程，摇摆jLabel也不会更新

1 个答案: