Is it possible to automatically check time then execute certain codes?
timer = os.date('%H:%M:%S', os.time() - 13 * 60 * 60 )
if timer == "18:04:40" then
print("hello")
end
I am trying to print hello on "18:04:40" everyday (os.date's time) without setting up a timer (which counts how much time past since the program's initiation) as I can't run the program 24 hours non-stop...
Thanks for reading.
This may not be the best solution but, when using a library like love2d for example you could run something like this:
function love.update(dt)
timer = os.date('%H:%M:%S', os.time() - 13 * 60 * 60 )
if timer >= value then
--stuff here
end
end
Or if you wanna make it so you have a whole number something like
tick = 0
function love.update(dt)
tick = tick + dt
if tick > 1 then
timer = os.date('%H:%M:%S', os.time() - 13 * 60 * 60 )
if timer >= value then
--stuff here
end
end
end
Lua has to check the time in some way.
Without a loop that can be realized with debug.sethook().
Example with Lua 5.1 typed in an interactive Lua (lua -i)...
> print(_VERSION)
Lua 5.1
> debug.sethook() -- This clears a defined hook
> -- Next set up a hook function that fires on 'line' events
> debug.sethook(function() local hour, min, sec = 23, 59, 59 print(os.date('%H:%M:%S', os.time({year = 2021, month = 12, day = 11, hour = hour, min = min, sec = sec}))) end, 'l')
-- just hit return/enter or do other things
23:59:59
5.9 - The Debug Library
https://www.lua.org/manual/5.1/manual.html#5.9
Hello fellow Stack Overflowers,
I have a situation, where I need some help choosing the best way to make an algorithm work, the objective is to manage the occupation of a resource (Lets consider the resource A) to have multiple tasks, and where each task takes a specified amount of time to complete. At this first stage I don't want to involve multiple variables, so lets keep it the simple way, lets consider he only has a schedule of the working days.
For example:
1 - We have 1 resource, resource A
2 - Resource A works from 8 am to 4 pm, monday to friday, to keep it simple by now, he doesn't have lunch for now, so, 8 hours of work a day.
3 - Resource A has 5 tasks to complete, to avoid complexity at this level, lets supose each one will take exactly 10 hours to complete.
4 - Resource A will start working on this tasks at 2018-05-16 exactly at 2 pm.
Problem:
Now, all I need to know is the correct finish date for all the 5 tasks, but considering all the previous limitations.
In this case, he has 6 working days and additionaly 2 hours of the 7th day.
The expected result that I want would be: 2018-05-24 (at 4 pm).
Implementation:
I thought about 2 options, and would like to have feedback on this options, or other options that I might not be considering.
Algorithm 1
1 - Create a list of "slots", where each "slot" would represent 1 hour, for x days.
2 - Cross this list of slots with the hour schedule of the resource, to remove all the slots where the resource isn't here. This would return a list with the slots that he can actually work.
3 - Occupy the remaining slots with the tasks that I have for him.
4 - Finnaly, check the date/hour of the last occupied slot.
Disadvantage: I think this might be an overkill solution, considering that I don't want to consider his occupation for the future, all I want is to know when will the tasks be completed.
Algorithm 2
1 - Add the task hours (50 hours) to the starting date, getting the expectedFinishDate. (Would get expectedFinishDate = 2018-05-18 (at 4 pm))
2 - Cross the hours, between starting date and expectedFinishDate with the schedule, to get the quantity of hours that he won't work. (would basically get the unavailable hours, 16 hours a day, would result in remainingHoursForCalc = 32 hours).
3 - calculate new expectedFinishDate with the unavailable hours, would add this 32 hours to the previous 2018-05-18 (at 4 pm).
4 - Repeat point 2 and 3 with new expectedFinishDate untill remainingHoursForCalc = 0.
Disadvantage: This would result in a recursive method or in a very weird while loop, again, I think this might be overkill for calculation of a simple date.
What would you suggest? Is there any other option that I might not be considering that would make this simpler? Or you think there is a way to improve any of this 2 algorithms to make it work?
Improved version:
import java.util.Calendar;
import java.util.Date;
public class Main {
public static void main(String args[]) throws Exception
{
Date d=new Date();
System.out.println(d);
d.setMinutes(0);
d.setSeconds(0);
d.setHours(13);
Calendar c=Calendar.getInstance();
c.setTime(d);
c.set(Calendar.YEAR, 2018);
c.set(Calendar.MONTH, Calendar.MAY);
c.set(Calendar.DAY_OF_MONTH, 17);
//c.add(Calendar.HOUR, -24-5);
d=c.getTime();
//int workHours=11;
int hoursArray[] = {1,2,3,4,5, 10,11,12, 19,20, 40};
for(int workHours : hoursArray)
{
try
{
Date end=getEndOfTask(d, workHours);
System.out.println("a task starting at "+d+" and lasting "+workHours
+ " hours will end at " +end);
}
catch(Exception e)
{
System.out.println(e.getMessage());
}
}
}
public static Date getEndOfTask(Date startOfTask, int workingHours) throws Exception
{
int totalHours=0;//including non-working hours
//startOfTask +totalHours =endOfTask
int startHour=startOfTask.getHours();
if(startHour<8 || startHour>16)
throw new Exception("a task cannot start outside the working hours interval");
System.out.println("startHour="+startHour);
int startDayOfWeek=startOfTask.getDay();//start date's day of week; Wednesday=3
System.out.println("startDayOfWeek="+startDayOfWeek);
if(startDayOfWeek==6 || startDayOfWeek==0)
throw new Exception("a task cannot start on Saturdays on Sundays");
int remainingHoursUntilDayEnd=16-startHour;
System.out.println("remainingHoursUntilDayEnd="+remainingHoursUntilDayEnd);
/*some discussion here: if task starts at 12:30, we have 3h30min
* until the end of the program; however, getHours() will return 12, which
* substracted from 16 will give 4h. It will work fine if task starts at 12:00,
* or, generally, at the begining of the hour; let's assume a task will start at HH:00*/
int remainingDaysUntilWeekEnd=5-startDayOfWeek;
System.out.println("remainingDaysUntilWeekEnd="+remainingDaysUntilWeekEnd);
int completeWorkDays = (workingHours-remainingHoursUntilDayEnd)/8;
System.out.println("completeWorkDays="+completeWorkDays);
//excluding both the start day, and the end day, if they are not fully occupied by the task
int workingHoursLastDay=(workingHours-remainingHoursUntilDayEnd)%8;
System.out.println("workingHoursLastDay="+workingHoursLastDay);
/* workingHours=remainingHoursUntilDayEnd+(8*completeWorkDays)+workingHoursLastDay */
int numberOfWeekends=(int)Math.ceil( (completeWorkDays-remainingDaysUntilWeekEnd)/5.0 );
if((completeWorkDays-remainingDaysUntilWeekEnd)%5==0)
{
if(workingHoursLastDay>0)
{
numberOfWeekends++;
}
}
System.out.println("numberOfWeekends="+numberOfWeekends);
totalHours+=(int)Math.min(remainingHoursUntilDayEnd, workingHours);//covers the case
//when task lasts 1 or 2 hours, and we have maybe 4h until end of day; that's why i use Math.min
if(completeWorkDays>0 || workingHoursLastDay>0)
{
totalHours+=8;//the hours of the current day between 16:00 and 24:00
//it might be the case that completeWorkDays is 0, yet the task spans up to tommorrow
//so we still have to add these 8h
}
if(completeWorkDays>0)//redundant if, because 24*0=0
{
totalHours+=24*completeWorkDays;//for every 8 working h, we have a total of 24 h that have
//to be added to the date
}
if(workingHoursLastDay>0)
{
totalHours+=8;//the hours between 00.00 AM and 8 AM
totalHours+=workingHoursLastDay;
}
if(numberOfWeekends>0)
{
totalHours+=48*numberOfWeekends;//every weekend between start and end dates means two days
}
System.out.println("totalHours="+totalHours);
Calendar calendar=Calendar.getInstance();
calendar.setTime(startOfTask);
calendar.add(Calendar.HOUR, totalHours);
return calendar.getTime();
}
}
You may adjust the hoursArray[], or d.setHours along with c.set(Calendar.DAY_OF_MONTH, to test various start dates along with various task lengths.
There is still a bug , due to the addition of the 8 hours between 16:00 and 24:00:
a task starting at Thu May 17 13:00:00 EEST 2018 and lasting 11 hours will end at Sat May 19 00:00:00 EEST 2018.
I've kept a lot of print statements, they are useful for debugging purposes.
Here is the terminology explained:
I agree that algorithm 1 is overkill.
I think I would make sure I had the conditions right: hours per day (8), working days (Mon, Tue, Wed, Thu, Fri). Would then divide the hours required (5 * 10 = 50) by the hours per day so I know a minimum of how many working days are needed (50 / 8 = 6). Slightly more advanced, divide by hours per week first (50 / 40 = 1 week). Count working days from the start date to get a first shot at the end date. There was probably a remainder from the division, so use this to determine whether the tasks can end on this day or run into the next working day.
I have to write a spring scheduler which runs once in a week , say every Monday at 1 AM .What will be the cron expression for that , can we archive this using fixedDely.
You can use cron for that:
#Scheduled(cron = "0 0 1 * * MON")
In order from left to right of a cron expression:
second, minute, hour, day of month, month, day(s) of week
A fixed delay is a delay, not a recurring event. With fixedDelay you could setup an event that runs after a fixed period in milliseconds between the end of the last invocation and the start of the next. That's not what you want here.
To have a job run every monday at 1 AM you can setup a cron expression
#Scheduled(cron = "0 0 1 * * MON")
I have a sequence where values are generated at random times (real time stock market prices). I have a requirement to find the highest and lowest value of the sequence between a one minute period. I know you can use something like Buffer for this.
But the minute window should start with 00 seconds and finish at 59 seconds. e.g. the minute should start from 8:00:00 and finish at 8:00:59 the second minute should start from 8:01:00 to 8:01:59. Can we do this with Rx? Thanks. Vipter
I believe this would work:
var query =
source
.Publish(ss =>
ss
.GroupByUntil(
x => x.Timestamp.ToUnixTimeSeconds() / 60,
x => x.Value,
x => ss.Where(s => x.Key != s.Timestamp.ToUnixTimeSeconds() / 60))
.Select(gxs => gxs.ToArray().Select(xs => new
{
min = xs.Min(), max = xs.Max()
})));
I want to trigger a notification for all my users at a specific time in their time zone. I want to compute the delay the server should wait before firing the notification. I can compute the time at the users Time Zone using Time.now.in_time_zone(person.time_zone)
I can strip out the hours, minutes and seconds from that time and find out the seconds remaining to the specific time. However, I was wondering if there's a more elegant method where I could set 9:00 AM on today and tomorrow in a timezone and compare it with Time.now.in_time_zone(person.time_zone) and just find out the number of seconds using arithmetic operations in the ruby Time Class.
Or in short my question is: (was: before the downvote!)
How do I compute the number of seconds to the next 9:00 AM in New York?
What about this
next9am = Time.now.in_time_zone(person.time_zone).seconds_until_end_of_day + 3600 * 9
next9am -= 24 * 60 * 60 if Time.now.in_time_zone(person.time_zone).hour < 9
NOTIFICATION_HOUR = 9
local_time = Time.now.in_time_zone(person.time_zone)
desired_time = local_time.hour >= NOTIFICATION_HOUR ? local_time + 1.day : local_time
desired_time = Time.new(desired_time.year, desired_time.month, desired_time.day, NOTIFICATION_HOUR, 0, 0, desired_time.utc_offset)
return desired_time - local_time