Question

How to set the onStart() method of window service so that after installing it first execute at 12 am,time interval is working fine,and service is executing after mentioned time interval, but not start at given time.

            public static System.Timers.Timer Timer;
            Double _timeinterval = 300 * 1000;// 6 mins
            protected override void OnStart(string[] args)
            {
            Timer = new System.Timers.Timer();

            Timer.Elapsed += new ElapsedEventHandler(OnTimedEvent);
            Timer.Interval = _timeinterval;
            Timer.Enabled = true;
            //method call to do operation
           }
            private void OnTimedEvent(object source, ElapsedEventArgs e)
              {
                //method call to do operation
              }

Answer 1

  protected override void OnStart(string[] args)
    {

        aTimer = new System.Timers.Timer();

        string starttime = "01.25";
        //start time is 01.25 means 01:15 AM

        double mins = Convert.ToDouble(starttime);
        DateTime t = DateTime.Now.Date.AddHours(mins);
        TimeSpan ts = new TimeSpan();
        // ts = t - System.DateTime.Now;
        ts = t.AddDays(1) - System.DateTime.Now;
        if (ts.TotalMilliseconds < 0)
        {
            ts = t.AddDays(1) - System.DateTime.Now;
            // ts = t - System.DateTime.Now;
        }
        _timeinterval = ts.TotalMilliseconds;
       // _timeinterval now set to 1:15 am (time from now to 1:15AM)
        aTimer.Elapsed += new ElapsedEventHandler(OnTimedEvent);
        aTimer.Interval = _timeinterval;
        aTimer.Enabled = true;
       }
        private static void OnTimedEvent(object source, ElapsedEventArgs e)
         {

           // operation to perform
           aTimer.Interval = 86400000; // now interval sets to 24 hrs

    }