java - 将应用程序从服务返回到前台

Question

Android 中相当新的内容。

这里的问题很简单。我的应用程序需要在一段时间内被唤醒。问题是我无法正确调用它。

在评论中//这里是问题代码正在破坏，给我错误，无法找到上下文并声明它。

公共(public)类BackgroundService扩展服务{

Integer background_connect = null;
String SERVER_IP;
String APP_ID;
private Context context;

@Override
public IBinder onBind(Intent arg0) {
    return null;
}

@Override
public void onCreate() {
    super.onCreate();

    DB_Queries db = new DB_Queries(getApplicationContext());
    Cursor c = db.getSettings();
    StringBuilder app_id = new StringBuilder();
    StringBuilder server_ip = new StringBuilder();
    if (c.moveToFirst()) {
        do {
            server_ip.append(c.getString(c.getColumnIndex("server_ip")));
            app_id.append(c.getString(c.getColumnIndex("app_id")));
        } while (c.moveToNext());
    }
    db.close();

    SERVER_IP = server_ip.toString();
    APP_ID = app_id.toString();
}

@Override
public void onDestroy() {
    super.onDestroy();
}

@Override
public int onStartCommand(Intent intent, int flags, int startId) {

    requestServerDelay delay = new requestServerDelay();
    delay.execute("http://" + SERVER_IP + "/index.php/app/requestAppRecall");

    return super.onStartCommand(intent, flags, startId);
}

private class requestServerDelay extends AsyncTask<String, Void, Void> {

    @Override
    protected Void doInBackground(String... params) {
        HttpClient client = new DefaultHttpClient();
        HttpPost post = new HttpPost(params[0]);
        post.addHeader("Content-type", "application/x-www-form-urlencoded");
        List<NameValuePair> pair = new ArrayList<NameValuePair>();
        pair.add(new BasicNameValuePair("app_id", APP_ID));

        try {
            post.setEntity(new UrlEncodedFormEntity(pair));
            HttpResponse response = client.execute(post);
            InputStream is = response.getEntity().getContent();
            InputStreamReader isr = new InputStreamReader(is);
            BufferedReader br = new BufferedReader(isr);
            StringBuilder str = new StringBuilder();
            String chunk = null;

            while ((chunk = br.readLine()) != null) {
                str.append(chunk);
            }

            background_connect = Integer.parseInt(str.toString());

        } catch (Exception e) {
            e.printStackTrace();
        }

        return null;
    }

    @Override
    protected void onPostExecute(Void result) {
        Handler handler = new Handler();
        handler.postDelayed(new Runnable() {
            public void run() {
                //HERE IS THE PROBLEM
                 Intent intent = new Intent(getApplication().getApplicationContext(), getApplication().getApplicationContext().getClass());
                    intent.addFlags(Intent.FLAG_ACTIVITY_REORDER_TO_FRONT);
                    intent.addFlags(Intent.FLAG_ACTIVITY_RESET_TASK_IF_NEEDED);
                    intent.addFlags(Intent.FLAG_ACTIVITY_PREVIOUS_IS_TOP);
                    intent.addFlags(Intent.FLAG_ACTIVITY_NEW_TASK);
                    getApplication().getApplicationContext().startActivity(intent);
            }
        }, background_connect * 60 * 1000);
    }

}

}

Answer 1

如果requestServerDelay任务在服务中，您可以使用getApplicationContext().startActivity(...)来启动 Activity 。

之前不需要getApplication()。

或者，您可以在 Service 类 (mContext) 中保留 Context 成员并使用它。