What is the best way to calculate the numbered day of the week over an 8 week period?
For example, my app tracks a users progress over an 8 week course.
I'm stuck on how to determine the current numbered day of the week according to the weeks progress, for example, Week 6 - Day 5
I have the users course start date as DateTime from Firebase, and can obviously get DateTime.now() to calculate the difference between course start date and now. How do I go from this so that I can display for example:
Week 2 - Day 3 or Week 6 - Day 4
Let's say the course is started 15 days ago, you can get what you want by something like this:
DateTime now = DateTime.now();
DateTime start = now.subtract(Duration(days:15));
int allDays = now.difference(start).inDays;
int w = allDays ~/ 7 + 1;
int d = allDays % 7;
print('Week $w - day $d');
~/ is the Truncating Division Operator which gives you the Integer result of a division
Related
I use Jiffy for calculate the difference in between 2 date in month.
But I don't the good result.
For end = 2/4/2023 and start = 1/4/2022 I have 12 months.
For end = 1/4/2023 and start = 1/4/2022 I have 11 months (error: expected 12).
Thanks,
num month = Jiffy(end).diff(Jiffy(start), Units.MONTH);
Right now it is checking 12 at midnight to 12 at midnight of the end date which is 11 months 30 days. Now if the end date is one second greater than the correct date it should work. So a hackey solution is to add one day duration to the end date and check
end.add(duration(day:1))
How can I in a smart way get the first business day of a month after the weekend?
To get the first business day of the month (given dateInput), we can do:
firstBusdayMonth = fbusdate(year(dateInput),month(dateInput));
As an example, for November, using the above function will return Thursday November 1 as the first business day. However, the first business day of the month after the first weekend is Monday November 5. How can I get this latter date?
Notes:
The weekend does not have to be in the same month.
If the Monday is not a working day, then I would like it to return the next business day
This function will do the trick. Here is the logic:
Create a datetime array for all days in the given month.
Get the day numbers.
Create a logical array, true from the first Monday onwards (so after the first weekend, accounting for the last day of the previous month being a Sunday).
Create another logical array using isbusday to exclude Mondays which aren't working days.
Finding the first day number where these two logical arrays are true, therefore the first business day after the weekend.
Code:
function d = fbusdateAferWE( y, m )
% Inputs: y = year, m = month
% Outputs: day of the month, first business day after weekend
% Create array of days for the given month
dates = datetime( y, m, 1 ):days(1):datetime( y, m, eomday( y, m ) );
% Get the weekday numbers to find first Monday, 1 = Sunday
dayNum = weekday( dates );
% Create the logical array to determine days from first Monday
afterFirstWeekend = ( cumsum(dayNum==2) > 0 ).';
% Get first day which is afterFirstWeekend and a business day.
d = find( afterFirstWeekend & isbusday( dates ), 1 );
end
You could probably speed this up (although it will be pretty rapid already) by not looking at the whole month, but say just 2 weeks. I used eomday to get the last day of the month, which means I don't have to make assumptions about a low number of holiday days in the first week or anything.
Edit: Working with datenum speeds it up by half (C/O JohnAndrews):
function d = fbusdateAferWE( y, m )
% Inputs: y = year, m = month
% Outputs: day of the month, first business day after weekend
% Create array of days for (first 2 weeks of) the given month
dates = datenum(datetime(y,m,1)):datenum(datetime(y,m,eomday(y,m)))-14;
% Get the weekday numbers to find first Monday, 1 = Sunday
dayNum = weekday( dates );
% Create the logical array to determine days from first Monday
afterFirstWeekend = ( cumsum(dayNum==2) > 0 ).';
% Get first day which is afterFirstWeekend and a business day.
d = find( afterFirstWeekend & isbusday( dates ), 1 );
end
I would add something like this after your statement:
[DayNumber, DayName] = weekday(firstBusdayMonth);
if DayNumber > 2
day = 10 - DayNumber;
else
This works because 'weekday' will return a number between 1 (Sunday) and 7 (Saturday).
The fbusdate() function will never return a 1 or 7, so we can ignore those cases.
If weekday(fbusdate()) == 2, the first is on a Monday and the firstBusdayMonth variable doesn't need to be changed.
If weekday(firstBusdayMonth) returns between 2 and 6, we need to skip to the next week, so we subtract the weekday value from 10 to find the next Monday.
It might not be the most elegant solution, but should work.
I am building a calendar application that needs to calculate the remaining days of (1) the current week and (2) the current month.
What I have tried was using the java.util.Calendar API, but it seems not to be fully supported.
int days = cal.getActualMaximum(Calendar.DAY_OF_MONTH);
How would I do this in CN1?
Many thanks in advance.
We would love to add support for JSR310 in the future which would probably solve this in a more elegant way. This is something that can be done as a cn1lib without modifying Codename One but would require some work...
A bit of a hack to do this is:
int month = cal.get(Calendar.MONTH);
cal.set(Calendar.DAY_OF_MONTH, 1);
if(month == cal.DECEMBER) {
cal.set(Calendar.YEAR, cal.get(Calendar.YEAR) + 1);
cal.set(Calendar.MONTH, Calendar.JANUARY);
} else {
cal.set(Calendar.MONTH, month + 1);
}
int day = 24 * 60 * 60000;
cal.setTime(new Date(cal.getTime().getTime() - day));
I don't know CN1, but you can calculate the values using a subtraction between the max and the current value:
(1) cal.getActualMaximum(Calendar.DAY_OF_WEEK) - cal.get(Calendar.DAY_OF_WEEK)
(2) cal.getActualMaximum(Calendar.DAY_OF_MONTH) - cal.get(Calendar.DAY_OF_MONTH)
For your notification: The first day of the week is sunday in the api of calendar. When your week should start with monday you have to reduce the offset by 1.
This question already has answers here:
Closed 11 years ago.
Possible Duplicate:
Calculate date from week number
Given Year, Month,Day and the Week number, is it possible to get the Date?
e.g. Year = 2010
Month =Jan
Day = Sun
WeekNumber = 3
output : 2010-01-10
I am trying it in c#
Thanks
I would make it like this:
int Year = 2010;
int Month = 1; //Jan=1, Feb=2, ...
int Day = 0; //Sun=0, Mon=1, ...
int WeekNumber = 3; // greater than 0
DateTime dateValue = new DateTime(Year, Month, 1);
int firstDay = (int)dateValue.DayOfWeek;
dateValue = dateValue.AddDays(Day - firstDay + (WeekNumber - 1) * 7);
I don't think there's something for such date calculations in plain .NET BCL. But there are libraries that can help you, see i.e. Fluent DateTime. Using this library, you can try something like that:
var firstWeekInYearBeginning = new DateTime(2010, 1, 2).Previous(DayOfWeek.Monday); // note 2 to not miss a week if the year begins on Monday
var expectedDate = 3.Weeks().From(firstWeekInYearBeginning);
Based on the APIs here, don't this its possible to initialize a DateTime Object from the information given. You would need to develop an algorithm to get the exact date of the year. A simple strategy would be to get the first Day of the month and based on that, find first Monday of the month. This is the start of WeekNumber 1 for that month, and you can locate your required Week by simpl loop and locate the exact date. You would then know the calendar date you are interested in.
BTW: I am assuming WeekNumber of a year/month starts from the first Monday of that Year/Month. Someone please correct me if I am wrong.
Maybe you should check out System.Globalization.Calendar class. It might be useful.
If I've got a time_t value from gettimeofday() or compatible in a Unix environment (e.g., Linux, BSD), is there a compact algorithm available that would be able to tell me the corresponding week number within the month?
Ideally the return value would work in similar to the way %W behaves in strftime() , except giving the week within the month rather than the week within the year.
I think Java has a W formatting token that does something more or less like what I'm asking.
[Everything below written after answers were posted by David Nehme, Branan, and Sparr.]
I realized that to return this result in a similar way to %W, we want to count the number of Mondays that have occurred in the month so far. If that number is zero, then 0 should be returned.
Thanks to David Nehme and Branan in particular for their solutions which started things on the right track. The bit of code returning [using Branan's variable names] ((ts->mday - 1) / 7) tells the number of complete weeks that have occurred before the current day.
However, if we're counting the number of Mondays that have occurred so far, then we want to count the number of integral weeks, including today, then consider if the fractional week left over also contains any Mondays.
To figure out whether the fractional week left after taking out the whole weeks contains a Monday, we need to consider ts->mday % 7 and compare it to the day of the week, ts->wday. This is easy to see if you write out the combinations, but if we insure the day is not Sunday (wday > 0), then anytime ts->wday <= (ts->mday % 7) we need to increment the count of Mondays by 1. This comes from considering the number of days since the start of the month, and whether, based on the current day of the week within the the first fractional week, the fractional week contains a Monday.
So I would rewrite Branan's return statement as follows:
return (ts->tm_mday / 7) + ((ts->tm_wday > 0) && (ts->tm_wday <= (ts->tm_mday % 7)));
If you define the first week to be days 1-7 of the month, the second week days 8-14, ... then the following code will work.
int week_of_month( const time_t *my_time)
{
struct tm *timeinfo;
timeinfo =localtime(my_time);
return 1 + (timeinfo->tm_mday-1) / 7;
}
Assuming your first week is week 1:
int getWeekOfMonth()
{
time_t my_time;
struct tm *ts;
my_time = time(NULL);
ts = localtime(&my_time);
return ((ts->tm_mday -1) / 7) + 1;
}
For 0-index, drop the +1 in the return statement.
Consider this pseudo-code, since I am writing it in mostly C syntax but pretending I can borrow functionality from other languages (string->int assignment, string->time conversion). Adapt or expand for your language of choice.
int week_num_in_month(time_t timestamp) {
int first_weekday_of_month, day_of_month;
day_of_month = strftime(timestamp,"%d");
first_weekday_of_month = strftime(timefstr(strftime(timestamp,"%d/%m/01")),"%w");
return (day_of_month + first_weekday_of_month - 1 ) / 7 + 1;
}
Obviously I am assuming that you want to handle weeks of the month the way the standard time functions handle weeks of the year, as opposed to just days 1-7, 8-13, etc.