如何使天文钟表永远不会在颤动中散漫时间?

时间:2019-05-25 17:08:13

标签: flutter chronometer

我成功创建了一个计时器,但是当我关闭应用程序时,我浪费了时间。我不知道在处理和计算最后日期的差值并从开放时间重新计算计数器之前不调用sharedpref怎么办...也许是不同的方法,通过比较开始时间和当前时间并增加每一秒,但是不知道该怎么做

Stopwatch stopwatch = new Stopwatch();

  void rightButtonPressed() {
    setState(() {
      if (stopwatch.isRunning) {
        stopwatch.reset();
      } else {
        stopwatch.reset();
        stopwatch.start();
      }
    });
  }
 @override
  Widget build(BuildContext context)
  {

     new Container(height: 80.0,
            child: new Center(
              child: new TimerText(stopwatch: stopwatch),
            )),

class TimerText extends StatefulWidget {
  TimerText({this.stopwatch});
  final Stopwatch stopwatch;

  TimerTextState createState() => new TimerTextState(stopwatch: stopwatch);
}

class TimerTextState extends State<TimerText> {

  Timer timer;
  final Stopwatch stopwatch;

  TimerTextState({this.stopwatch}) {
    timer = new Timer.periodic(new Duration(milliseconds: 30), callback);
  }

  void callback(Timer timer) {
    if (stopwatch.isRunning) {
      setState(() {

      });
    }
  }

  @override
  Widget build(BuildContext context) {
    final TextStyle timerTextStyle = const TextStyle(fontSize: 50.0, fontFamily: "Open Sans");
    String formattedTime = TimerTextFormatter.format(stopwatch.elapsedMilliseconds);
    return new Text(formattedTime, style: timerTextStyle);
  }
}

class TimerTextFormatter {
  static String format(int milliseconds) {

    int seconds = (milliseconds / 1000).truncate();
    int minutes = (seconds / 60).truncate();
    int hours = (minutes / 60).truncate();
    int days = (hours / 24).truncate();
    String minutesStr = (minutes % 60).toString().padLeft(2, '0');
    String secondsStr = (seconds % 60).toString().padLeft(2, '0');
    String hoursStr = (hours % 60).toString().padLeft(2, '0');
    String daysStr = (days % 24).toString().padLeft(2, '0');
    return "$daysStr:$hoursStr:$minutesStr:$secondsStr";
  }
}

0 个答案:

没有答案