android - "Launch timeout has expired, giving up wake lock"上的监听器或通知

Question

在我的一个测试设备中，我在 LogCat 中收到了这个可怕的警告:

07-20 09:57:02.093: W/ActivityManager(1159): Launch timeout has expired, giving up wake lock!
07-20 09:57:02.218: W/ActivityManager(1159): Activity idle timeout for HistoryRecord{4072b5e8 com.rero.myapp/.MyActivity}

我的初步研究表明这是一种常见的挑眉行为:

1. “应用程序可能有 exceeded the VM budget 并且内存不足。”
2. “这个 can be ignored 。”
3. > Network problem ？
4. Activity 是 taking to long to start (在 UI 线程上处理太多？)
5. 涉及 HTTP 的服务和广播.
6. 继续 ...

我的问题是:

1. “启动超时已过期”的含义是什么？系统有什么作用？ (例如，一个消息来源声称它终止了我的应用程序，但实际上我看到我的应用程序在 1-2 分钟的感知卡住后正常进行)
2. 有没有办法在我的应用程序中收到有关此警告的通知？

Answer 1

这是我将要尝试的一般想法，但是您如何处理 AsyncTask 取决于您在获得状态后正在做什么。我会做这样的事情:

private class GetHttpStatus extends AsyncTask<String, Void, Boolean> {
    @Override
    protected Boolean doInBackground(String[] params) {
        private boolean status;

        //this will be the string you pass in execute()
        String urlString = params[0];

        HttpURLConnection httpConnection;
        try {
            URL gurl = new URL(urlString);
            URLConnection connection = gurl.openConnection();
            connection.setConnectTimeout(5 * 1000);
            httpConnection = (HttpURLConnection) connection;
            int responseCode = httpConnection.getResponseCode();
            if(responseCode == HttpURLConnection.HTTP_OK) {
                status = true;
            }
        } catch (Exception e) {
            status = false;
        } finally { 
            if(httpConnection != null) httpConnection.disconnect();
        }
        return status;
    }

    @Override
    protected Void onPostExecute(Boolean result) {
        //Here you'll do whatever you need to do once the connection
        //status has been established
        MyActivity.notifyHttpStatus(result);
    }
}

//in your Activity somewhere, you would call...

new GetHttpStatus.execute("http://www.amazon.com");