calculate age down to years, mths, days?

[Mike Doty]

Pardon? ABS(1-2) + 1 = 2

[Donald Darden]

Interest payments are more complicated than just intervals between birthdays because you have to take into consideration two things: The rate that the interest is compounted (if not a simple interest loan), and the point at which payment is posted. A lot of loan institutions assume that the payment is always made according to schedule, so that as a consequence, even if your actual playment varies somewhat from the schedule, the payment terms remain in force (though you may have to pay a late payment fee if you are excessively late on making the payment). Others use the fact that the payments are late to allow the compound interest to accumulate - this is the reason that compound interest loans were created in the first place. Another benefit of compound interest rate loans is that they are mistakenly believed to infer a lower interest rate than similar simple rate loans.

However, there is a point missed about using two dates to determine the period between by just subtracting one from the other. The better way is to use ABS(date1 - date2) +1. The ABS() ensures that daycount is always positive, regardless of which order the two dates are entered. And the +1 is necessary because the ending date is always inclusive. That is, instead of the day count between March 1 and March 2 being zero (the result of just subtracting the two), you would add one so that the day count between them is 1 day.

[Paul Purvis]

i am sure many of you deal with the calculation of interest.
my head is tried of thinking of detailed calculation right now, but if i am not mistaken, the above code can produce an excellent way of calculating interest to pay if interest is paid based on a monthly basis.

when calculation is based on days that have already passed this is not problem for interest to be calculated.
but when your schedule to calculate interest is not regular this can be a problem and this program could be used for such means.

here is an example.

let's say i owe you money of 1000 and interest is 12 percent a year.
and i pay you monthly but i also need a way to pay you for some period that is shorter than a month, an example would be if i paid you off before the end of the next month. also lets say i do not have a way to keep track of compounding periods other than one year.
by using and comparing the both months and days returned in the two routines i can choose the right month and days for calculating interest.

if my logic is correct, i would compare the first months results with the second month results, if the first figure is equal to the second, use the first results for calculating interest, if the first figure is less than the second figure, use the second results for calculating interest.


paul

[Paul Purvis]

the only changes in the future will be in an alternative for date string to julian date numbers and julian date numbers to date strings.

this listing will work with dates in the year 0000 and the next will not.
both routines will give the same julian date routines from the year 0001 forward.

the date routines listed here will allow the year 0000 and the julian date numbers will match up to the iso 8601 date standards that are now the most current ISO date standard.

but once again this program was written mostly for the today's useage and the program as tested will handle two different aging methods, one based on where end of months are the same in length and the second method, anniversary style dating for such things as how old a person is in months and days.

paul

'compymmdd.bas
'compute years month and days differences
'program considers end of months dates to be the same no matter the days in the month.
'
'this program is somewhat different in that if the first date is the last day of a month
'a whole month will not increase until any month has passed
'if you wanted to be calculate birthdays, it would be best to subtract 30 from the days variable and add one month to the totalmonths

#COMPILE EXE
#DIM ALL
DECLARE FUNCTION date2julian(mm_dd_yyyy AS STRING) AS LONG
DECLARE FUNCTION julian2date(jd AS LONG) AS STRING
DECLARE FUNCTION computeymdd(lowdate AS STRING, highdate AS STRING,usebirthdaytypemonth AS LONG) AS STRING


FUNCTION datevalidate(mm_dd_yyyy AS STRING) AS LONG
LOCAL temp1&,temp2&,temp3&,temp4&,templeapyearflag&
LOCAL tbuffer AS BYTE PTR
FUNCTION=0&
temp4&=LEN(mm_dd_yyyy)
IF temp4<10& THEN
     IF temp4<9& THEN EXIT FUNCTION
     mm_dd_yyyy=RIGHT$("0"+mm_dd_yyyy,10&)
END IF
tbuffer=STRPTR(mm_dd_yyyy)
temp1&=(PEEK(BYTE,tbuffer)-48)*10+PEEK(BYTE,tbuffer+1?)-48
temp2&=(PEEK(BYTE,tbuffer+3?)-48)*10+PEEK(BYTE,tbuffer+4?)-48
temp3&=(PEEK(BYTE,tbuffer+6?)-48)*1000+(PEEK(BYTE,tbuffer+7?)-48)*100+(PEEK(BYTE,tbuffer+8?)-48)*10+(PEEK(BYTE,tbuffer+9?)-48)

IF temp3&<1 THEN
    IF temp3& THEN EXIT FUNCTION
    IF temp2&=0& AND temp1&=0& THEN FUNCTION=-1&
    'Remove or remark out the next line if you want to include the year 0000
    EXIT FUNCTION
END IF

IF temp1&<1 OR TEMP2&<1 THEN EXIT FUNCTION
IF temp2&<29 THEN FUNCTION=1&:EXIT FUNCTION
temp4&=CHOOSE&(temp1&,32,30,32,31,32,31,32,32,31,32,31,32)
IF temp4& THEN
      IF temp1&<>2 THEN                  ' here if the month is february pass over test
         IF temp2&<temp4& THEN FUNCTION=1& ' if true the #of days are equal to or less than the total days in a month
         EXIT FUNCTION  'here the day is greater than the # of days in a month
      ELSE
         IF temp2&>29& THEN EXIT FUNCTION
        'here is where we know the day is the 29th of february
        IF temp3& MOD 4& = 0& THEN templeapyearflag&=1&
        IF temp3& MOD 100& =0& THEN templeapyearflag&=0&
        IF temp3& MOD 400& =0& THEN templeapyearflag&=1&
        IF templeapyearflag& THEN FUNCTION=1&
      END IF
END IF
'this function will return a 0 if the program gets here
END FUNCTION




FUNCTION PBMAIN AS LONG
' ex computeymdd(lowdate,highdate,0) 'mm/dd/yyyy required
' ex for counting birthdays computeymdd(lowdate,highdate,1) 'mm/dd/yyyy required


STDOUT "the same dates with two methods"
STDOUT "1st method could be called a whole month method because"
STDOUT "  it treats days at the end of each month being the same"
STDOUT "  ex. one month from 06/30/0001 equals 07/30/0001 and 07/31/0001"
STDOUT "2nd method could be called a calendar method because it counts"
STDOUT "  the months and days since a certain date."
STDOUT "  ex. one month from 06/30/0001 equals 07/30/0001"
STDOUT "   07/31/001 is one month and one day, closer to counting a birthday"
STDOUT
STDOUT "09/22/1918  04/04/1994"
STDOUT computeymdd("09/22/1918","04/04/1994",0)
STDOUT computeymdd("09/22/1918","04/04/1994",1)
STDOUT "-----------------------------------------------------"
STDOUT "01/01/0004  02/03/0004"
STDOUT computeymdd("01/01/0004","02/03/0004",0)
STDOUT computeymdd("01/01/0004","02/03/0004",1)
STDOUT "-----------------------------------------------------"
STDOUT "01/01/0004  02/29/0004"
STDOUT computeymdd("01/01/0004","02/29/0004",0)
STDOUT computeymdd("01/01/0004","02/29/0004",1)
STDOUT "-----------------------------------------------------"
STDOUT "01/27/0001  02/28/0001"
STDOUT computeymdd("01/27/0001","02/28/0001",0)
STDOUT computeymdd("01/27/0001","02/28/0001",1)
STDOUT "-----------------------------------------------------"
STDOUT "01/27/1989  02/28/1989"
STDOUT computeymdd("01/27/1989","02/28/1989",0)
STDOUT computeymdd("01/27/1989","02/28/1989",1)
STDOUT "-----------------------------------------------------"
STDOUT "01/31/0004  02/28/0004"
STDOUT computeymdd("01/31/0004","02/28/0004",0)
STDOUT computeymdd("01/31/0004","02/28/0004",1)
STDOUT "-----------------------------------------------------"
STDOUT "01/31/0004  02/29/0004"
STDOUT computeymdd("01/31/0004","02/29/0004",0)
STDOUT computeymdd("01/31/0004","02/29/0004",1)
STDOUT "-----------------------------------------------------"
STDOUT "01/31/0001  02/28/0001"
STDOUT computeymdd("01/31/0001","02/28/0001",0)
STDOUT computeymdd("01/31/0001","02/28/0001",1)
STDOUT "-----------------------------------------------------"
STDOUT "01/31/0001  03/30/0001"
STDOUT computeymdd("01/31/0001","03/30/0001",0)
STDOUT computeymdd("01/31/0001","03/30/0001",1)
STDOUT "-----------------------------------------------------"
STDOUT "01/31/0001  03/31/0001"
STDOUT computeymdd("01/31/0001","03/31/0001",0)
STDOUT computeymdd("01/31/0001","03/31/0001",1)
STDOUT "-----------------------------------------------------"
STDOUT "02/28/0001  03/30/0001"
STDOUT computeymdd("02/28/0001","03/30/0001",0)
STDOUT computeymdd("02/28/0001","03/30/0001",1)
STDOUT "-----------------------------------------------------"
STDOUT "02/28/0001  03/31/0001"
STDOUT computeymdd("02/28/0001","03/31/0001",0)
STDOUT computeymdd("02/28/0001","03/31/0001",1)
STDOUT "-----------------------------------------------------"
STDOUT "02/28/0003  03/30/0004"
STDOUT computeymdd("02/28/0003","03/30/0004",0)
STDOUT computeymdd("02/28/0003","03/30/0004",1)
STDOUT "-----------------------------------------------------"
STDOUT "02/28/0003  03/31/0004"
STDOUT computeymdd("02/28/0003","03/31/0004",0)
STDOUT computeymdd("02/28/0003","03/31/0004",1)
STDOUT "-----------------------------------------------------"
STDOUT "02/29/0004  03/30/0004"
STDOUT computeymdd("02/29/0004","03/30/0004",0)
STDOUT computeymdd("02/29/0004","03/30/0004",1)
STDOUT "-----------------------------------------------------"
STDOUT "02/29/0004  03/31/0004"
STDOUT computeymdd("02/29/0004","03/31/0004",0)
STDOUT computeymdd("02/29/0004","03/31/0004",1)
STDOUT "-----------------------------------------------------"
STDOUT "02/29/1988  03/31/1989"
STDOUT computeymdd("02/29/1988","03/31/1989",0)
STDOUT computeymdd("02/29/1988","03/31/1989",1)
STDOUT "-----------------------------------------------------"
STDOUT "02/29/1988  02/28/1989"
STDOUT computeymdd("02/29/1988","02/28/1989",0)
STDOUT computeymdd("02/29/1988","02/28/1989",1)
STDOUT "-----------------------------------------------------"
STDOUT "02/29/1988  06/30/1988"
STDOUT computeymdd("02/29/1988","06/30/1988",0)
STDOUT computeymdd("02/29/1988","06/30/1988",1)
STDOUT "-----------------------------------------------------"
STDOUT "06/30/0004  07/31/0004"
STDOUT computeymdd("06/30/0004","07/31/0004",0)
STDOUT computeymdd("06/30/0004","07/31/0004",1)
STDOUT "-----------------------------------------------------"
STDOUT "12/31/0004  01/31/0005"
STDOUT computeymdd("12/31/0004","01/31/0005",0)
STDOUT computeymdd("12/31/0004","01/31/0005",1)
STDOUT "-----------------------------------------------------"
STDOUT "check the validatiy of date routines "
STDOUT "date 1875-05-20 should be equal to julian date 2406029"
STDOUT "calculated julian date of 05/20/1875 ="+STR$(date2julian("05/20/1875"))
STDOUT "calculated date from julian  2406029 = "+julian2date(2406029)

WAITKEY$
END FUNCTION



FUNCTION computeymdd(lowdate AS STRING, highdate AS STRING,usebirthdaytypemonth AS LONG) AS STRING
'mm/dd/yyyy required for lowdate and highdate
'usebirthdaytypemonth will add a month if the days are over 30 days and subtract 30 from the days
'usebirthdaytypemonth could have an difference in years,months and days
LOCAL years, months, days, totaldays,totalmonths AS LONG
LOCAL lowdateateom, I AS LONG
LOCAL lowdatemth,lowdateday,lowdateyear,highdatemth,highdateday,highdateyear AS LONG
LOCAL tempdate AS STRING
'LOCAL  usebirthdaytypemonth as long
'usebirthdaytypemonth=0&

IF datevalidate(lowdate)<1& OR datevalidate(highdate)<1& THEN
   YEARS=-1&
   MONTHS=-1&
   DAYS=-1&
   TOTALDAYS=-1&
   totalmonths=-1&
  ' PRINT "error in dates"
   GOTO EXITTHEFUNCTION
END IF

totaldays=date2julian(highdate)-date2julian(lowdate)
IF totaldays<0& THEN
   YEARS=-1&
   MONTHS=-1&
   DAYS=-1&
   TOTALDAYS=-1&
   totalmonths=-1&
   'PRINT "error 2nd date prior to first date"
   GOTO exitthefunction
END IF

REM days are valid and computation can be done

lowdatemth=VAL(MID$(lowdate,1&,2&))
lowdateday=VAL(MID$(lowdate,4&,2&))
lowdateyear=VAL(MID$(lowdate,7&,4&))
highdatemth=VAL(MID$(highdate,1&,2&))
highdateday=VAL(MID$(highdate,4&,2&))
highdateyear=VAL(MID$(highdate,7&,4&))

IF lowdateyear=highdateyear THEN
   IF lowdatemth=highdatemth THEN
   YEARS=0&
   MONTHS=0&
   DAYS=totaldays
   totalmonths=0&
   GOTO EXITTHEFUNCTION
   END IF
END IF

'compute to see if the low date is a date at the end of a month and if so check the high date
'  if the high date is also at the end of the month do date math
IF usebirthdaytypemonth&=0 THEN
IF VAL(MID$(julian2date(date2julian(lowdate)+1&),1&,2&))<>lowdatemth THEN
        IF VAL(MID$(julian2date(date2julian(highdate)+1&),1&,2&))<>highdatemth THEN
       'both dates are end of the month, compute months"
        totalmonths=(highdatemth+(highdateyear*12&))-(lowdatemth+(lowdateyear*12&))
        years=totalmonths\12&
        months=totalmonths MOD 12&
        days=0&
        GOTO exitthefunction
        END IF
      lowdateateom=1&
    ELSE
      lowdateateom=0&
END IF
END IF

'set a temporary date
tempdate=highdate
FOR I=0& TO 3&
MID$(tempdate,4,2)=RIGHT$("0"+TRIM$(STR$(lowdateday-I)),2)
IF datevalidate(tempdate)=1 THEN EXIT FOR
NEXT I
IF date2julian(tempdate)=date2julian(highdate) THEN
        totalmonths=(highdatemth+(highdateyear*12&))-(lowdatemth+(lowdateyear*12&))
        years=totalmonths\12&
        months=totalmonths MOD 12&
        days=0&
        GOTO exitthefunction
    END IF

tempdate=highdate
MID$(tempdate,4,2)=RIGHT$("0"+TRIM$(STR$(lowdateday)),2)
IF datevalidate(tempdate)=1 THEN
    IF date2julian(tempdate)<=date2julian(highdate) THEN
      totalmonths=(VAL(MID$(tempdate,1,2))+(VAL(MID$(tempdate,7,4))*12&))-(lowdatemth+(lowdateyear*12&))
      years=totalmonths\12&
      months=totalmonths MOD 12&
      days=date2julian(highdate)-date2julian(tempdate)
      GOTO EXITTHEFUNCTION
    END IF
END IF
'find the last day of the month previous to the highdate
IF lowdateateom THEN
    tempdate=highdate
    MID$(tempdate,4,2)="01
    tempdate=julian2date(date2julian(tempdate)-1)
END IF
IF date2julian(tempdate)<date2julian(highdate) THEN
    totalmonths=(VAL(MID$(tempdate,1,2))+(VAL(MID$(tempdate,7,4))*12&))-(lowdatemth+(lowdateyear*12&))
    years=totalmonths\12&
    months=totalmonths MOD 12&
    days=date2julian(highdate)-date2julian(tempdate)
    GOTO EXITTHEFUNCTION
END IF

tempdate=highdate
MID$(tempdate,4,2)="01
tempdate=julian2date(date2julian(tempdate)-1)
FOR I=0& TO 3&
MID$(tempdate,4,2)=RIGHT$("0"+TRIM$(STR$(lowdateday-I)),2)
IF datevalidate(tempdate)=1 THEN EXIT FOR
NEXT I
totalmonths=(VAL(MID$(tempdate,1,2))+(VAL(MID$(tempdate,7,4))*12&)) - (lowdatemth+(lowdateyear*12&))
years=totalmonths\12&
months=totalmonths MOD 12&
days=date2julian(highdate)-date2julian(tempdate)


EXITTHEFUNCTION:
IF usebirthdaytypemonth THEN
       computeymdd=_lowdate+" to "+highdate+ "  using birthdate date math"+$CRLF+_
         STR$(years)+" yrs"+STR$(months)+" mths"+STR$(days)+" days    "+_
         STR$(totalmonths)+" total months actual days "+STR$(totaldays)
       ELSE
     computeymdd=_lowdate+" to "+highdate+ "  using simple calendar date math"+$CRLF+_
         STR$(years)+" yrs"+STR$(months)+" mths"+STR$(days)+" days    "+_
         STR$(totalmonths)+" total months actual days "+STR$(totaldays)
      END IF

END FUNCTION


'=============================================================================================================================================================


FUNCTION date2julian(mm_dd_yyyy AS STRING) AS LONG
'this function does not check the validity of a date, use the datevalidate function above for that purpose
LOCAL month,day,year,k,julian AS LONG
LOCAL tbuffer AS BYTE PTR
tbuffer=STRPTR(mm_dd_yyyy)
month=(PEEK(BYTE,tbuffer)-48)*10+PEEK(BYTE,tbuffer+1?)-48
day=(PEEK(BYTE,tbuffer+3?)-48)*10+PEEK(BYTE,tbuffer+4?)-48
year=(PEEK(BYTE,tbuffer+6?)-48)*1000+(PEEK(BYTE,tbuffer+7?)-48)*100+(PEEK(BYTE,tbuffer+8?)-48)*10+(PEEK(BYTE,tbuffer+9?)-48)
k = INT((14 - month) / 12)
IF month THEN
    date2julian = (day + INT(367 * (month + (k * 12) - 2) / 12)+ INT(1461 * (year + 4800 - k) / 4) - 32113)-_
               (INT(3& * INT((year + 100& - k) / 100&) / 4&) - 2&) 'convert to gregorian
    ELSE
    date2julian=0
END IF

END FUNCTION

FUNCTION julian2date(jd AS LONG) AS STRING
'any julian date number below 1721060 will return a date of "00/00/0000"
LOCAL ss AS STRING
ss="00/00/0000"
IF jd<1721060 THEN GOTO EXITTHEFUNCTION
LOCAL  m,d,y, q, r, s, t, u, v AS LONG
q = INT((jd / 36524.25) - 51.12264)
r = jd + q - INT(q / 4&) + 1&
s = r + 1524&
t = INT((s / 365.25) - 0.3343)
u = INT(t * 365.25)
v = INT((s - u) / 30.61)
d = s - u - INT(v * 30.61)
m = (v - 1&) + 12& * (v > 13.5)
y = t - (m < 2.5) - 4716&
LOCAL remains AS LONG
LOCAL digit AS LONG
LOCAL nextdigit AS LONG
LOCAL nextcharacter AS LONG

 ASC(ss,1)  = m \10  +48
  ASC(ss,2)  = m MOD 10 + 48
  ASC(ss,4)  = d\10 + 48
  ASC(ss,5)  = d MOD 10 + 48
  Remains&=y
  FOR digit& = 4 TO 1 STEP -1
    NextDigit&=Remains& MOD 10
    Remains&=Remains& \ 10
    NextCharacter&=NextDigit& + 48
    ASC(ss,digit+6) = NextCharacter
  NEXT
EXITTHEFUNCTION:
FUNCTION=ss
END FUNCTION

[Paul Purvis]

please go to this thread at this time

http://www.powerbasic.com/support/pbforums/showthread.php?t=37153

new code listed on internet is pointed to in the above thread
i found an error in the code and fix it prior to asking for test from members at the thread above

once testing is felt to be completed, i will repost in this section.
follow above thread for testing

in response to Mike's note about a validation error on a month being larger that 12, i could not see an error or i could not see what he was seeing.
non valid dates return dates of -1(negative) on calculation returns.

paul

[Michael Mattias]

The problem with 'accuracy' is there are multiple possible correct answers:

Is the difference between April 18 2007 and April 18 2008 Exactly one year? Or it is it "one year, one day" because there was a Feb 29th in 2008 which had to be lived, a day which never happened between 4/18/06 and 4/18/07?

IIRC there was a fairly lengthy thread on this exact same subject about six months ago, and the answer came out the same: "Correct = Whatever the User Thinks It Should Be"

When "what the user thinks" is involved, written confirmation is mandatory.

[Mike Doty]

If someone is performing date math they would format the string after obtaining the result.
This would also remove the string handling. The current version is for ease of use in displaying the result.

DECLARE SUB DateDif(FromYear AS LONG, FromMonth AS LONG, FromDay AS LONG, _
                 ToYear   AS LONG, ToMonth   AS LONG, ToDay   AS LONG, _
                 OPTIONAL BYVAL Birthday AS LONG)

[Mike Doty]

STDOUT computeymdd("13/18/2008","04/18/2008",0)
This incorrectly passes the date validation routine.
Just about there!!!!

MM: The responses are from the authors of internet date sites, not casual users.
A lot of research has gone into this and this pssting is the best I've seen.
We are looking for accuracy as #1 and speed as #2. Many routines handle a single
day into the next month as an month. This routine takes that into consideration.
This would not be used for making headstones because birthdays are handled differently.
The author was asked for for permission to post the email response.
This is the best date routine I've seen for returning the years, months and days on the internet.

Another idea would be to use 0 as failing the date verification instead of -1 and zero.

Paul, great work.

[Michael Mattias]

>What have I missed?

The discussion with the user and the associated written confirmation re how they want the calculations performed. Correct=whatever they agreed to.

MCM

[Paul D. Elliott]

It may be that I've missed something or just a problem of my
misunderstanding the wording. Say my birthday is tomorrow and if I plug in
tomorrow's date ( but prior year ) & today's date and use Birthdate math then I am 1 year older today.
If I use simple calendar math then I get so many years 11 months and
30 days ( which I've always considered right ).

What have I missed?

Yes, it is a PAIN. I remember having a lot of problems with this when I did a program
for hospital data that needed to calculate the age for patients. All sorts of discussions
dealing with newborns and those up to 1 month of age ( had to calculate days or weeks
or months or years ).

[Paul Purvis]

i had that thought too,
but went the other way because somebody might not want to calculate dates, just validate a date, maybe for key entry, or for importing purposes, also the routine can accept a date of 00/00/0000.
which your routine does not if i am not mistaken, nothing wrong about that.

i could have made changes in your date2julian code to validate the date there, but if the dates are valid, much like what would be inside of database, your date2julian would be much faster and also code in my date functions where it validates the lowdate and highdate could be removed, adding more speed to the routine of calculating date differences.

i am not suggesting any change to your code to allow for 00/00/0000 dates
and a julian(really georgian) date number to be 0, but i do believe many times a programmer will place code just before your code anyways to test entry or convert strings to match strings in the routine.

the code before your date2julian could simply be

on datevalidate("xx/xx/xxxx")+1& GOTO LINE1, LINE2,LINE3
LINE1:
some kind of an error line
exit function or goto somewhere
LINE2:
juliandate=0
exit function or goto somewhere
LINE3:
date2julian("xx/xx/xxxx")
goto somewhere

OR YOU CAN DO THE CASE OR IF THEN BLOCKS

i do not know what is the fastest code here
if you are going to use case block or if block then probably the fastest would be
line3 should be done first
line 2 done second
line 3 done last

the validate routine is now written to be streamlined and as fast as i could make it for getting into the routine, maybe a sub routine would be faster than a function routine in this program, but left it as a function for more portability. also, to test for dates of 00/00/0000, even if you moved he code to the date2julian function, you only save code from the ON GOTO statement to the end of the function. i have also streamlined the code from the ON GOTO statement to the end of the function.

by streamlined, get out of the function fast if certain test are meet and planning what variables equaling certain values maybe occur more often than others and also trying to keep variables inside the cpu cache.

paul

[Mike Doty]

http://www.timeanddate.com/date/durationresult.html?m1=02&d1=29&y1=1988&m2=02&d2=28&y2=1989

timeanddate.com computes 02/29/1988 to 02/28/1989 as
11 months and 30 days.

I sent to the author at timeanddate.com and received a very fast response and was given permission to post here.

Hello Mike,

Unfortunately, there are several ways to count days/months in some
cases, in the future we hope to be able to show all alternative ways a
duration can be counted as, to pinpoint these differences.

So yes, you might also say that it is one year, given the assumption
that you
"round" February 29, 1989 down to February 28 (and not roll over to
March 1).

Best regards,
Steffen

I like your answer better of 1 year, 0 months and 0 days.
If 1 is added to their computation it still shows 1 year and 1 day.
They never show just 1 year with this date range.
-----------------------------------------------------------------------------------------------
This one comes up with 0 years, 11 months and 28 days.

http://calendarhome.com/cgi-bin/date2.pl?month1=2&date1=29&year1=1988&wd=Monday&month2=2&date2=28&year2=1989&wd2=Tuesday

[Mike Doty]

'Validate date
jd = Date2Julian(LowDate)
IF Julian2Date(Jd) <> LowDate THEN ? "Invalid date"

[Paul Purvis]

until code is reposted in this thread
see website below for newer code

http://pdptemp.dyndns.org/zipcode/dates/

[Mike Doty]

Is my solution above flawed?

[Mike Doty]

I missed that this was DOS. Hopefully, an easy conversion.

[Mel Bishop]

Keeping in mind that Mike's code was written for and in PB/CC. Needs to be converted to PB/DOS.

[Mike Doty]

How about this?

#COMPILE EXE
#DIM ALL
'4/13/08 pbDate.Bas by Mike Doty 4/13/08
'Tested against: [URL]http://www.timeanddate.com/date/duration.html[/URL]
'
'Can you break this code?  
'
DECLARE SUB pbDate(BYVAL LowDate AS STRING, BYVAL HighDate AS STRING, Years AS LONG, Months AS LONG, Days AS LONG, TotalDays AS LONG)
DECLARE FUNCTION date2julian(mm_dd_yyyy AS STRING) AS LONG
DECLARE FUNCTION julian2date(jd AS LONG) AS STRING
FUNCTION PBMAIN AS LONG
LOCAL Years, Months, Days, MonthCorrection, TotalDays AS LONG, LowDate, HighDate AS STRING
LowDate$ = "09/22/1918" 'mm/dd/yyyy required
HighDate$ = "04/04/1994" 'mm/dd/yyyy required
pbDate LowDate$,HighDate$,Years&,Months&,Days&,TotalDays&
'
'Display results: 75 years, 6 months and 13 days. Total days 27588
? LowDate$ + " to "+ HighDate$ + _
STR$(Years) + " years, " + FORMAT$(Months) + " months, " + FORMAT$(Days) + " days. Total days" + STR$(TotalDays)+ "."
END FUNCTION
'
SUB pbDate(BYVAL LowDate AS STRING, BYVAL HighDate AS STRING, Years AS LONG, Months AS LONG, Days AS LONG, TotalDays AS LONG)
'Pass LowDate (mm/dd/yyyy)
'Pass HighDate (mm/dd/yyyy)
'Years&,Months&,Days&,TotalDays& returned '4/13/08 Mike Doty
LOCAL A AS LONG, B AS LONG, offset1 AS LONG, Offset2 AS LONG, MonthCorrection AS LONG
A = Date2Julian&(LowDate$) '(A) = low date
B= Date2Julian&(HighDate$) '(B) = high date
TotalDays = B-A 'Total days difference
offset1 = B - Date2Julian&("01/01/" + RIGHT$(HighDate,4)) '01/01/ of (B)
B = B - Offset1 'move high date (B) back to the beginning of the year
A = A - Offset1 'move low date (A) back by same number of days as (B)
offset2 = Date2Julian&("12/31/" + RIGHT$(LowDate,4)) - A '12/31/ of (A) the low date
A = A + Offset2 'move low date (A) forward to end of year
B = B + Offset2 'move high date (B) forward same number of days
Years = (B - A)\365 'questionable, open to suggestions (can you break it?)
IF LEFT$(HighDate,5) < LEFT$(LowDate,5) THEN MonthCorrection =1 'correct month (if needed)
Months = VAL(LEFT$(Julian2Date$(B),2)) - MonthCorrection 'get number of months from julian date
Days = VAL(MID$(Julian2Date$(B),4,2)) 'get number of days from julian date
END SUB
'
FUNCTION date2julian(mm_dd_yyyy AS STRING) AS LONG
DIM month&,day&,year&
month = VAL(LEFT$(mm_dd_yyyy$,2))
day = VAL(MID$(mm_dd_yyyy$,4,2))
year = VAL (RIGHT$(mm_dd_yyyy$,4))
LOCAL k AS LONG, julian AS LONG
k = INT((14 - month) / 12)
julian = day + INT(367 * (month + (k * 12) - 2) / 12) _
+ INT(1461 * (year + 4800 - k) / 4) - 32113
julian = julian - (INT(3 * INT((year + 100 - k) / 100) / 4) - 2) 'convert to gregorian
date2julian = julian
END FUNCTION
'
FUNCTION julian2date(jd AS LONG) AS STRING
DIM yr$,mo$,da$
DIM m&,d&,y&, q&, r&, s&, t&, u&, v&
q = INT((jd / 36524.25) - 51.12264)
r = jd + q - INT(q / 4) + 1
s = r + 1524
t = INT((s / 365.25) - 0.3343)
u = INT(t * 365.25)
v = INT((s - u) / 30.61)
d = s - u - INT(v * 30.61)
m = (v - 1) + 12 * (v > 13.5)
y = t - (m < 2.5) - 4716
FUNCTION = FORMAT$(M,"00") + "-" + _ ' Return the date string, thank you Larry Cooke
FORMAT$(D,"00") + "-" + _ ' using the new values...
FORMAT$(Y,"0000") ' " " " "
END FUNCTION

[Mike Doty]

I'm thinking that an offset from the beginning/ending of the julian dates
could be used and push the highest julian date back to the beginning of
the year. Then push the date forward to obtain the reminder represented as the month and days.

[Don Schullian]

http://www.dasoftvss.com/basic/index.html

Robert,

Here's some code on my page that can get you the # of days between the dates. Figuring from the number of days you'll use this formula to get the number of elapsed full years:

Years = ( 4 * (ToDay& - Bdate&) ) \ 1461

Computing the # of elapsed months and days is going to be brute work however.