* User Supplied ISO Date
UserDate = '20040203'
UserDateYear = UserDate[1,4] + 0
UserDateMonth = UserDate[5,2] + 0
UserDateDay = UserDate[7,2] + 0
* First Day Of The Next Month
NextMonthYear = UserDateYear + (IF (UserDateMonth + 1) > 12 THEN 1 ELSE 0)
NextMonthMonth = (IF (UserDateMonth + 1) > 12 THEN 1 ELSE (UserDateMonth+1))
NextMonthDay = 1
NextMonthIDate = ICONV(NextMonthMonth:"/":NextMonthDay:"/":NextMonthYear,"D4/")
*Subtract 1 day from the first day on the next month to get the last day of the
month supplied
UserDateIEOM = NextMonthIDate -1
UserDateEOM =
OCONV(UserDateIEOM,'DY'):OCONV(OCONV(UserDateIEOM,"DM"),"MR(%2)"):OCONV(OCONV(UserDateIEOM,"DD"),"MR(%2)")
*
PRINT 'NextMonthYear = ':NextMonthYear
PRINT 'NextMonthMonth = ':NextMonthMonth
PRINT 'NextMonthDay = ':NextMonthDay
PRINT 'EOM Internal = ':UserDateIEOM
PRINT 'EOM ISO Date = ':UserDateEOM
*
--
Donald Verhagen
Application Development Manager
[EMAIL PROTECTED]
Tandem Staffing Solutions, Inc.
1690 S Congress Avenue, Suite 210
Delray Beach, FL 33445 USA
Voice Phone: 561.454.3592 Fax Phone: 561.454.3640
>>> [EMAIL PROTECTED] 1:31:17 PM 06/02/2005 >>>
Hi,
Given a date like 20040203, I want to return the last valid date for that month
and year (20040229 in this case). What is the shortest code fragment to achieve
this?
At the moment I'm replacing the day with 01, then iconv, add 35 days to the
internal date and then oconv and replace the day again with 01. I'm then on the
first day of the next month. I then iconv, subtract 1 day and oconv.
Thanks for any help.
---------------------------------
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