OpenEuphoria

Re: [DOS]: Help with a date() problem

• Posted by CChris <christian.cuvier at agricult?re.?ouv.fr> Nov 08, 2007
• 507 views

Alex Caracatsanis wrote:
> 
> I'd appreciate some help coding for the following problem:
> 
> I have a client who commenced a "self-improvement activity" at 00:01am
> on 1 July 2007. In order to give him "encouragement" for persisting with the
> activity, he would like to have a program that he can run each day, that
> will display: "It's now day <x> of your activiy! Keep going!", where <x>
> increments with each passing day. We'll assume that he persists every day
> without fail, and that he can run the program any number of times on any one
> day (so that if he ran the program at 6am, midday, and 10pm on 2 July 2007,
> he would see "It's now day 2 of your activity! Keep going!").
> 
> I thought of using date() like this...
> 
> }}}
<eucode>
> constant DAY_ZERO = 181 -- represents 30 June 07, the 181st day of this year
> constant DAY_NUM  = 8
> 
> sequence today
> integer dayNumber, daysOnActivity
> 
> today = date()
> dayNumber = today[DAY_NUM]
> daysOnActivity = dayNumber - DAY_ZERO
> 
> printf( 1, "It's now day %d of your activity!", daysOnActivity )
> </eucode>
{{{

> 
> ... but I'll have a problem from 1 Jan 08, which is day 1!
> 
> Thanks for any suggestions.
> 
> Alex Caracatsanis

function days_in_year(integer year)
-- year = actual year - 1900
    if and_bits(year,3) then return 365
    elsif remainder(year,100) then return 366 -- extra day
    else  -- secular years usually don't have an extra day, 
          -- but quadrisecular years have 
        return 365 + (remainder(year,400)=100)
    end if
end function
-- this is valid as long as dates are using the gregorian calendar

function num_of_days
(integer start_day,integer start_year,integer end_day,integer end_year)
-- number of elapsed days, including both ends
    integer result
    -- start from Jan 1st of start_year 
    result = end_day
    for i=start_year to end_year-1 do
         result += days_in_year(i)
    end for
-- and then remove start_day
    return result - start_day + 1
end function

constant DAY_ONE = 182,
         YEAR_ONE = 2007-- represent 01 July 07, the 182nd day of that year

constant DATE_DAY_NUM  = 8, DATE_YEAR = 1

sequence today
integer dayNumber, daysOnActivity, this_year

today = date()
dayNumber = today[DATE_DAY_NUM]
thisYear = today[DATE_YEAR]
daysOnActivity = num_of_days(DAY_ONE,YEAR_ONE,dayNumber,thisYear)

printf( 1, "It's now day %d of your activity!", daysOnActivity )

CChris