Day of Project (subtracting calendar days

[bbalmerTotalFluency]

Hi:

Imagine you started a project today at 00:00:00. We'll call that Day 1. At 1 second past midnight you'd be on day 2. You could calculate this easily using the difference in seconds. But what if you started your project today at 23:59:59. In calendar days you'd be on day 2 two seconds later. That's what I need. I need to know the day of a project in CALENDAR days given today and a start date.

Put differently let's suppose you start a project on a Monday. Any time during Monday should be day 1 and ANY time during Tuesday should be day two. It seems to me this must be a common problem and thus there is likely an elegant solution. Time zone are a thing for me since I have clients in many different time zones.

UPDATE: I have a solution. It's ugly. I'd be very happy if someone gave me the good looking / elegant / sophisticated version.

convert tProjectStarted to the internet date -- internet date format = Fri, 21 Jun 2024 09:51:36 +0700
put "00:00:00" into word 5 of tProjectStarted
convert tProjectStarted to seconds -- I believe this gets me the start of my day local time in seconds
put the seconds into tNow
return ceiling((tNow - tProjectStarted) / 86400) -- give me the correct day 0.0001 => Day 1 1.0001 => day 2

Bruce

[bn]

Hi Bruce,

here I post a library made by Malte Brill (derBrill) which he has put it into the public domain in 2014.

It has a very complete variety of handlers for I guess for most calendar related problems.

for days between dates you would have to add 1 to the result since you want to include the first date

Code: Select all

-- from Malte Brill's libDate
function libDate_DaysBetween pYear1,pMonth1,pDay1,pYear2,pMonth2,pDay2
   return libDate_DayNumber(pYear2,pMonth2,pDay2) - libDate_DayNumber(pYear1,pMonth1,pDay1)
end libDate_DaysBetween

-- taken from http://alcor.concordia.ca/~gpkatch/gdate-algorithm.html

function libDate_DayNumber pYear,pMonth,pDay
put  (pMonth + 9) mod 12 into pMonth
put pYear - pMonth div 10 into pYear
return 365*pYear + pYear div 4 - pYear div 100 + pYear div 400 + (pMonth*306 + 5) div 10 + ( pDay - 1 )
end libDate_DayNumber

The complete library is attached here.

Kind regards
Bernd

[bbalmerTotalFluency]

Thank you very much Bernd.

I feel like I should know about such things. How / where does one find out about these libraries that exist?

Bruce

[bn]

Hi Bruce,

I feel like I should know about such things. How / where does one find out about these libraries that exist?

I don't know. It was announced and referenced here in the forum a couple of times.

Here is Malte's announcement.
viewtopic.php?f=108&t=23885

When I saw your post I remembered the library.

When going the "seconds" way you have to watch out for "daylight saving time changes". Malte calculates the days directly.

Kind regards
Bernd