Day of the year is a number between 1 and 365 (in 2023), January 1 is day 1.
After today 140 days are remaining in this year.
This page uses the ISO-8601 ordinal date format.
There is also another less-used format: the 'ISO day of year' numbers, this is a number between 1 and 371, day 1 of the year is Monday of the first ISO week (where the first Thursday of the new year is in week 1).
1 2 3=TODAY()-DATE(YEAR(TODAY()),1,0) // or, for any date entered in cell A1, calculate the corresponding day-number in that date’s year: =A1-DATE(YEAR(A1),1,0)Calculate today's day-number, starting from the day before Jan 1, so that Jan 1 is day 1.
1=DATEDIF(CONCAT("1-1-";year(now()));today();"D")+1Calculates the difference between Jan 1 and today (=days past) then add 1 for today's daynumber. (Your date format (1-1-year) may be different)
1=ROUNDDOWN(DAYS(NOW(),DATE(YEAR(NOW()),1,1))) + 11 2$dayNumber = date("z") + 1; date("z", epoch) + 1You can use an epoch to find other day numbers
date("z") starts counting from 0 (0 through 365)!
1 2from datetime import datetime day_of_year = datetime.now().timetuple().tm_yday1 2 3 4 5 6 7 8 9use Time::Piece; my $day_of_year = localtime->yday + 1; # ... or ... my $day_of_year = (localtime)[7] +1; # ... or (if you really want to use POSIX) ... use POSIX; my $day_of_year = POSIX::strftime("%j", time);Replace time with other epochs for other days.
1 2 3SELECT DAYOFYEAR(NOW()) // or SELECT DAYOFYEAR('2022-02-20');Day number between 1 and 366. Replace now() with other dates.
1 2select to_char(sysdate, 'DDD') from dual select to_char(to_date('2022-02-20','YYYY-MM-DD'), 'DDD') from dual1 2using DateUtils, SysUtils; DayOfTheYear(Date);1DatePart("y", Now())1Dim dayOfYear As Integer = DateTime.Now.DayOfYear1 2 3 4 5 6 7 8var today = new Date(); Math.ceil((today - new Date(today.getFullYear(),0,1)) / 86400000); Date.prototype.getDOY = function() { var onejan = new Date(this.getFullYear(),0,1); return Math.ceil((this - onejan) / 86400000); } var today = new Date(); var daynum = today.getDOY();1LocalDate.now().getDayOfYear();1date +%j1#dayofyear(now())#1 2 3 4 5int currentDay; dateFormatter = [[NSDateFormatter alloc] init]; [dateFormatter setDateFormat:@"D"]; date = [NSDate date]; currentDay = [[dateFormatter stringFromDate:date] intValue];1int iDayOfYear = System.DateTime.UtcNow.DayOfYear;1format(Sys.Date(), "%j")1 2time = Time.new puts time.yday1 2$DayOfYear = (Get-Date).DayofYear Write-Host $DayOfYear1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33on mouseUp put "January 1," && the last word of the long date into firstDayofYear --append current year convert firstDayofYear to seconds -- from Jan 1, 1970 to first day of this year put the long date into currentDay convert currentDay to seconds -- from Jan 1, 1970 GMT to today put currentDay - firstDayofYear into totalSeconds answer the round of (totalSeconds / (60*60*24)) + 1 --display total days in dialog box end mouseUp // Or on mouseUp answer DayOfYear() end mouseUp function DayOfYear put the short date into currentDate convert currentDate to dateItems --list of date elements separated by commas put item 1 of currentDate into year put item 2 of currentDate into month put item 3 of currentDate into day put floor (275 * month / 9) into N1 put floor ( (month + 9) / 12) into N2 put (1 + floor ( (year - 4 * floor (year / 4) + 2) / 3) ) into N3 put N1 - (N2 * N3) + day - 30 into N return N end DayOfYear function floor pNumber -- LiveCode has no built-in floor() function put Round (pNumber) into theInteger if theInteger > pNumber then put theInteger - 1 into theInteger end if return theInteger end floor1 2 3 4SELECT DATEPART(DAYOFYEAR, SYSDATETIME()) // Or SELECT datediff(day,CAST(datepart(year,getdate()) AS CHAR(4)) + '-01-01',getdate()+1) AS number_of_today1day := time.Now().YearDay()1dayNumber = today-datenum(['1-Jan-' year(today)])+1