calculate diferent between years, month, days

[Dré]

Hi,

I try to make a script to calculate years, month days between 2 dates.
I will check like 2 September 2006 and today (16 April 2022)

output must be: 15 years, 7 months, 14 days

i try to make this, but it is really hard for,
i can calculate years 2022 - 2006 - 1 = 15

but if i m with months like this, April = 4, September is 9
4 - 9 = crashing for me
Because it is negative, and it has to take an extra year of the previous calculation.

And the same with days.

Did someone already think how to do this, or is there already an example for doing this?

Before I was thinking to do it with seconds, but that's even harder I think, because there are months like 28 / 29(once a 4 years) / 30 or 31

[Erwin van der Zwart]

Not fully tested but something like this could work (however always tricky with leap years):

Code:

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startdate = {   day = 2,   month = 9,   year = 2006 } enddate = {   day = 16,   month = 4,   year = 2022 } unixstart = os.time(startdate) unixend = os.time(enddate) if unixend < unixstart then   log('Start date is higer then the end date, exiting script')   return end seconds = os.difftime(unixend, unixstart) timediff = os.date('!*t', seconds) years = timediff.year - 1970 months = timediff.month - 1 days = timediff.day - 1 log(years,months,days)

[Dré]

Hi Erwin,

Mostly looks great, sometimes is has a miscalculation of 1 day, but i couldn't figure it out why.

Another thing, what i will try to fix, is when the first date is before 1970.
i changed

Code:

1

unixend = os.time(enddate)

with

Code:

1

unixend = os.time()

[Erwin van der Zwart]

Hi, I the day difference is probably due to a leap year in your start or end date, you could try to add something like this in your code:

Code:

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function isLeapYear(year)    if ((((year % 4 ==0) and (year % 100~=0)) or (year % 400==0)))then       return true    else       return false   end end if isLeapYear(startdate.year) then   -- add or remove a day   log('startdate is a leap year') end if isLeapYear(enddate.year) then   -- add or remove a day   log('enddate is a leap year') end

[admin]

Here's my solution

Code:

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sday, smonth, syear = 1, 2, 1999 eday, emonth, eyear = 2, 3, 2022 monthdiff = emonth - smonth yeardiff = eyear - syear daydiff = eday - sday if daydiff < 0 then   leap = ((eyear % 4 == 0) and (eyear % 100 ~= 0)) or (eyear % 400 == 0)   mdays = { 31, (leap and 29 or 28), 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 }   if emonth == 1 then     emonth = 12   else     emonth = emonth - 1   end   daydiff = daydiff + mdays[ emonth ]   monthdiff = monthdiff - 1 end if monthdiff < 0 then   monthdiff = monthdiff + 12   yeardiff = yeardiff - 1 end log(daydiff, monthdiff, yeardiff)

[Dré]

Thanks, I choose to use the one of admin and changed it a little, so he calculates the different between today and the set day.
But both thanks for helping me.

Code:

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now = os.date('*t') eday = now.day emonth = now.month eyear = now.year -- bevrijdingsdag sday, smonth, syear = 5, 5, 1945 monthdiff = emonth - smonth yeardiff = eyear - syear daydiff = eday - sday if daydiff < 0 then   leap = ((eyear % 4 == 0) and (eyear % 100 ~= 0)) or (eyear % 400 == 0)   mdays = { 31, (leap and 29 or 28), 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 }   if emonth == 1 then     emonth = 12   else     emonth = emonth - 1   end   daydiff = daydiff + mdays[ emonth ]   monthdiff = monthdiff - 1 end if monthdiff < 0 then   monthdiff = monthdiff + 12   yeardiff = yeardiff - 1 end grp.checkupdate('62/7/51', yeardiff ..' jaar ' .. monthdiff ..' maanden ' .. daydiff ..' dagen ')