Question

What I m trying to create is a simple progress bar, that would load for ~10 sec. So what I want is a for loop like this:

for(int i = 1; i <= 100; i++) {
                    progressDialog.setProgress(i);
                    //100ms delay
                }

Thanks

Answer 1

下面的法典可能对你有用。

public void startProgress(View view) {
    // Do something long
    Runnable runnable = new Runnable() {
        @Override
        public void run() {
            for (int i = 0; i <= 10; i++) {
                final int value = i;
                try {
                    Thread.sleep(1000);
                } catch (InterruptedException e) {
                    e.printStackTrace();
                }
                handler.post(new Runnable() {
                    @Override
                    public void run() {
                        progressDialog.setProgress(value);
                    }
                });
            }
        }
    };
    new Thread(runnable).start();
}

Answer 2

你们可以为此目的利用阿森特任务

(a) 预支指数为0;

在背景过程中睡觉的10秒钟,然后用最新方法发送进展。

在紧急更新后,进展很难在参数上取得进展。

Answer 3

如果不明确建立新通道,那么解决办法就会越短:

final Handler handler = new Handler(Looper.getMainLooper());
final int sleepMs = 100;
for (int i = 1; i <= 100; ++i) {
    handler.postDelayed(() -> progressDialog.setProgress(value), i * sleepMs);
}

友情链接