Re: [U2] Finding last day of month
The original request was for a shortest code fragment. This likely meant as line count, but could have been execution time. I am using UniVerse 10.1.3 on a 1.7 GHz pentium. Here are some current timings. Except for test 0001 which is an empty FOR NEXT loop, to the left of the | is a hint of the initialization, and to the right is a hint of the code under test. An interesting result is that it is faster to assign SPACE(2) than to a variable than assigning a 1 or 10 character string in a variable to another variable. The basic takeaway is that use of ICONV and OCONV should be minimized if you optimize for time. Test Loop Core.. (uSec) per Iteration. 0001 FOR/NEXT | (EMPTY) 0.081 0005 A=0 0.037 0012 |A=1+1 0.034 0050 IF (B) THEN-WIDE.IF 0.025 0051 IF (B) THEN WIDE.IF 0.031 0072 A=2;B=2|C=A*B0.033 0201 A= 10x|B=A[1,1]0.303 0221 A=SPACE(2)|A=SPACE(2)0.283 0231 A:='12345':@VM...|B=A[char(253) 0.457 ,10,1] 0338 A='1'|B=OCONV(A,D4/) 2.503 0352 A='1'|B=OCONV(A,D4-YMD[4,2 2.821 ,2]) 0353 A='1'|B=OCONV(A,D4/);[7,4 3.986 ]-[1,2]-[4,2] 0365 A=2;B=2|C=A/B0.042 0369 A=2|B=INT(A) 0.293 0370 A=2.2|B=INT(A) 0.294 0371 A=2000|B=MOD(A,4)0.236 0372 A=2001|B=MOD(A,4)0.238 0374 A=SPACE(1)|B=A 0.321 0375 A=SPACE( 10)|B=A 0.323 0376 A=SPACE( 100)|B=A 0.369 0377 A=SPACE( 1000)|B=A 0.559 0378 A=SPACE(1)|B=A 1.727 0385 A='20040825'|B=ICONV(A,D4) 0.879 0386 A='13387'|B=OCONV(A,D4)2.796 0387 A=13387|B=OCONV(A,D4) 3.186 Mark A. Baldridge Principal Consultant North American Lab Services DB2 and U2 Information Management, IBM Software Group (508) 524-5666 --- u2-users mailing list u2-users@listserver.u2ug.org To unsubscribe please visit http://listserver.u2ug.org/
Re: [U2] Finding last day of month
In a message dated 6/2/2005 9:40:08 PM Pacific Daylight Time, [EMAIL PROTECTED] writes: [EMAIL PROTECTED] wrote on 06/02/2005 10:34:09 PM: input oDate ; *format mmdd iDate = iconv(oDate,'dymd') loop until oconv(iDate+1,'dd) = 1 do iDate += 1 repeat lastDay = oconv(iDate,'dymd') Not sure if it's worth bragging about - but mine's smaller (146 v. 161 bytes). But your's takes more CPU cycles so there :) Smaller in opcode? or smaller in milliseconds? Ah Will --- u2-users mailing list u2-users@listserver.u2ug.org To unsubscribe please visit http://listserver.u2ug.org/
Re: [U2] Finding last day of month
In a message dated 6/2/2005 8:17:25 PM Pacific Daylight Time, [EMAIL PROTECTED] writes: That's an interesting approach, but it comes at the price of an extra ICONV, which is more expensive than a simple IF statement. Granted, the difference isn't significant for a small number of invocations, but after a while it adds up. I just hate special logic like If its the 12th month If its a leap year If its year 1999 or I just find it more pure to use the internal system information to calculate all that. Will Johnson --- u2-users mailing list u2-users@listserver.u2ug.org To unsubscribe please visit http://listserver.u2ug.org/
Re: [U2] Finding last day of month
But your's takes more CPU cycles so there :) Smaller in opcode? or smaller in milliseconds? Ah Yeah, but now that you've challenged me, this one's smaller in size, op codes and presumably cpu. input oDate ; *format mmdd lastDayOfMonth = oconv(iconv(oconv(iconv(oDate,'ml#6':@vm:'dym')+31,'dym'),'dym')-1,'dymd[4,2]') S ** This email message and any files transmitted with it are confidential and intended solely for the use of addressed recipient(s). If you have received this email in error please notify the Spotless IS Support Centre (61 3 9269 7555) immediately who will advise further action. This footnote also confirms that this email message has been scanned for the presence of computer viruses. ** --- u2-users mailing list u2-users@listserver.u2ug.org To unsubscribe please visit http://listserver.u2ug.org/
Re: [U2] Finding last day of month
Will, This was exactly my method, except I was adding 35 days (it could have been 30 as well) to the first day of the month to get to next month. I also dislike the lookup table idea i.e. 31,28,30,31,etc, the leap year or 12th month logic. Thanks for all the responses guys. [EMAIL PROTECTED] wrote: Ok smartass you say let's see you do it First of all, you can always get from the month you are in, to the next month by simply add 30 days to the middle of the month. And you never need to worry about leap years nor about 12th month logic Input Date ; * format mmdd Middleofthismonth = date[1,6]:'15' I.Middle = Iconv(Middleofthismonth,'D') I.Next = I.Middle + 30 NextMonth = Oconv(I.NEXT,'D2-') FirstDayNextMonth = NextMonth[1,2]:'01':NextMonth[5,2] I.FirstDay = Iconv(FirstDayNextMonth,'D') I.LastDayPreviousMonth = I.FirstDay - 1 Now can we scrunch this down? Input Date ; *format mmdd NextMonth = Oconv(Iconv(date[1,6]:'15','D') + 30,'D2') I.LastDayPreviousMonth = Iconv(NextMonth[1,2]:'01':NextMonth[5,2],'D') + 1 Do I win the big stuffed pickle? Will Johnson Fast Forward Technologies --- u2-users mailing list u2-users@listserver.u2ug.org To unsubscribe please visit http://listserver.u2ug.org/ - Yahoo! Messenger NEW - crystal clear PC to PCcalling worldwide with voicemail --- u2-users mailing list u2-users@listserver.u2ug.org To unsubscribe please visit http://listserver.u2ug.org/
Re: [U2] Finding last day of month
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RE: [U2] Finding last day of month
i've not read this entire thread so someone may have said this but i get the first day of the next month and then -1 works a treat -Original Message- From: [EMAIL PROTECTED] on behalf of Marco Manyevere Sent: Fri 03/06/2005 09:40 To: u2-users@listserver.u2ug.org Cc: Subject: Re: [U2] Finding last day of month Will, This was exactly my method, except I was adding 35 days (it could have been 30 as well) to the first day of the month to get to next month. I also dislike the lookup table idea i.e. 31,28,30,31,etc, the leap year or 12th month logic. Thanks for all the responses guys. [EMAIL PROTECTED] wrote: Ok smartass you say let's see you do it First of all, you can always get from the month you are in, to the next month by simply add 30 days to the middle of the month. And you never need to worry about leap years nor about 12th month logic Input Date ; * format mmdd Middleofthismonth = date[1,6]:'15' I.Middle = Iconv(Middleofthismonth,'D') I.Next = I.Middle + 30 NextMonth = Oconv(I.NEXT,'D2-') FirstDayNextMonth = NextMonth[1,2]:'01':NextMonth[5,2] I.FirstDay = Iconv(FirstDayNextMonth,'D') I.LastDayPreviousMonth = I.FirstDay - 1 Now can we scrunch this down? Input Date ; *format mmdd NextMonth = Oconv(Iconv(date[1,6]:'15','D') + 30,'D2') I.LastDayPreviousMonth = Iconv(NextMonth[1,2]:'01':NextMonth[5,2],'D') + 1 Do I win the big stuffed pickle? Will Johnson Fast Forward Technologies --- u2-users mailing list u2-users@listserver.u2ug.org To unsubscribe please visit http://listserver.u2ug.org/ - Yahoo! Messenger NEW - crystal clear PC to PCcalling worldwide with voicemail --- u2-users mailing list u2-users@listserver.u2ug.org To unsubscribe please visit http://listserver.u2ug.org/ __ This email has been scanned by the MessageLabs Email Security System. For more information please visit http://www.messagelabs.com/email __ __ This email has been scanned by the MessageLabs Email Security System. For more information please visit http://www.messagelabs.com/email __ [demime 1.01d removed an attachment of type application/ms-tnef which had a name of winmail.dat] --- u2-users mailing list u2-users@listserver.u2ug.org To unsubscribe please visit http://listserver.u2ug.org/
[U2] Finding last day of month
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. - Yahoo! Messenger NEW - crystal clear PC to PCcalling worldwide with voicemail --- u2-users mailing list u2-users@listserver.u2ug.org To unsubscribe please visit http://listserver.u2ug.org/
Re: [U2] Finding last day of month
From: Marco Manyevere [EMAIL PROTECTED] 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? Find the internal date of the first day of the next month, subtract 1, and convert it back to external format. ORIG. = MMDD[1,4] ORIG.MM = MMDD[5,2] ORIG.DD = MMDD[7,2] I.LAST.DAY = (ICONV(ORIG.:(ORIG.MM + 1): 01), D4/) - 1 O.LAST.DAY = OCONV(I.LAST.DAY, D4/ --- u2-users mailing list u2-users@listserver.u2ug.org To unsubscribe please visit http://listserver.u2ug.org/
Re: [U2] Finding last day of month
You can try ICONV(date,code) and check STATUS(). If status() returns other than 0 then date is valid Marco Manyevere wrote: 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. - Yahoo! Messenger NEW - crystal clear PC to PCcalling worldwide with voicemail --- u2-users mailing list u2-users@listserver.u2ug.org To unsubscribe please visit http://listserver.u2ug.org/ -- Lembit Pirn 7+7 Software Tondi 1 Tallinn 11313 tel. +372 6566232 [EMAIL PROTECTED] --- u2-users mailing list u2-users@listserver.u2ug.org To unsubscribe please visit http://listserver.u2ug.org/
RE: [U2] Finding last day of month
I might offer a small modification to your code snippet as noted below. BobW From: Marco Manyevere [EMAIL PROTECTED] 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? Find the internal date of the first day of the next month, subtract 1, and convert it back to external format. ORIG. = MMDD[1,4] ORIG.MM = MMDD[5,2] ORIG.DD = MMDD[7,2] * IN CASE IT'S DECEMBER IF ORIG.MM = 12 THEN ORIG. += 1 ORIG.MM = 0 END I.LAST.DAY = (ICONV(ORIG.:(ORIG.MM + 1): 01), D4/) - 1 O.LAST.DAY = OCONV(I.LAST.DAY, D4/) --- u2-users mailing list u2-users@listserver.u2ug.org To unsubscribe please visit http://listserver.u2ug.org/
Re: [U2] Finding last day of month
Marco Manyevere [EMAIL PROTECTED] wrote on 06/02/2005 01:31:17 PM: 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? Ooh! A good old-fashioned programming challenge. I think the following is simpler and, because it uses half as many date conversions, will burn fewer CPU cycles than your method. INPUT ORIG.DATE YEAR = ORIG.DATE[1,4] MONTH = ORIG.DATE[5,2] MONTH += 1 IF MONTH EQ 13 THEN MONTH = 1; YEAR += 1 NEW.DATE.I = ICONV(MONTH:'/01/':YEAR, 'D') - 1 NEW.DATE = CHANGE(OCONV(NEW.DATE.I, 'DYMD'), ' ', '') CRT NEW.DATE This may need to be modified if your default date convention is not MM/DD/YY. It also assumes that the input date is in the correct format. Tim Snyder Consulting I/T Specialist , U2 Professional Services North American Lab Services DB2 Information Management, IBM Software Group [EMAIL PROTECTED] --- u2-users mailing list u2-users@listserver.u2ug.org To unsubscribe please visit http://listserver.u2ug.org/
Re: [U2] Finding last day of month
Hi, Marco. I always ICONV the first day of the next month, then subtract a day. For example: ANY.DATE = '02-03-2004' MO = ANY.DATE[1,2] YR = ANY,DATE[7,4] MO += 1 IF MO 12 THEN MO = 1 YR += 1 END EOM.DATE = ICONV(MO:'-01-':YR, 'D') - 1 This lends itself to a subroutine or function as well. Regards, Cliff Marco Manyevere wrote: 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. - Yahoo! Messenger NEW - crystal clear PC to PCcalling worldwide with voicemail --- u2-users mailing list u2-users@listserver.u2ug.org To unsubscribe please visit http://listserver.u2ug.org/ --- u2-users mailing list u2-users@listserver.u2ug.org To unsubscribe please visit http://listserver.u2ug.org/
Re: [U2] Finding last day of month
That method can actually backfire, for example, if your starting date is 20040130, you'll end up on Feb 29 instead of Jan 31. What I would do is replace the day with 01 and add 1 to the month, then iconv and subtract 1 day. -Dianne Marco Manyevere wrote: 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. --- u2-users mailing list u2-users@listserver.u2ug.org To unsubscribe please visit http://listserver.u2ug.org/
RE: [U2] Finding last day of month
Here's how I do it... JJ = ICONV(THE.DATE, 'D') MO = OCONV(JJ, 'DM') YR= OCONV(JJ, 'DY') DAY=28 LOOP TRY = OCONV(ICONV(MO:'/':DAY+1:'/':YR, 'D'), 'DM') WHILE TRY = MO DO REPEAT LAST.DAY = MO:'/':DAY:'/':YR Perry Taylor -Original Message- From: Marco Manyevere [mailto:[EMAIL PROTECTED] Sent: Thu Jun 02 14:05:42 2005 To: u2-users@listserver.u2ug.org Subject:[U2] Finding last day of month 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. - Yahoo! Messenger NEW - crystal clear PC to PCcalling worldwide with voicemail --- u2-users mailing list u2-users@listserver.u2ug.org To unsubscribe please visit http://listserver.u2ug.org/ CONFIDENTIALITY NOTICE: This e-mail message, including any attachments, is for the sole use of the intended recipient(s) and may contain confidential and privileged information. Any unauthorized review, use, disclosure or distribution is prohibited. ZirMed, Inc. has strict policies regarding the content of e-mail communications, specifically Protected Health Information, any communications containing such material will be returned to the originating party with such advisement noted. If you are not the intended recipient, please contact the sender by reply e-mail and destroy all copies of the original message. --- u2-users mailing list u2-users@listserver.u2ug.org To unsubscribe please visit http://listserver.u2ug.org/
RE: [U2] Finding last day of month
same idea with a few less conversions cdate=20040203 yr=cdate[1,4] mn=cdate[5,2] if mn12 then mn+=1 else mn=1 ; yr+=1 edate=oconv(iconv(mn'R%2':'-01-':yr,D)-1,D) -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Behalf Of Marco Manyevere Sent: Thursday, June 02, 2005 01:31 PM To: u2-users@listserver.u2ug.org Subject: [U2] Finding last day of month 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. - Yahoo! Messenger NEW - crystal clear PC to PCcalling worldwide with voicemail --- u2-users mailing list u2-users@listserver.u2ug.org To unsubscribe please visit http://listserver.u2ug.org/ --- u2-users mailing list u2-users@listserver.u2ug.org To unsubscribe please visit http://listserver.u2ug.org/
RE: [U2] Finding last day of month
0001: INPUT DATE 0002: USE.DATE = ICONV(DATE,D) 0003: MTH = OCONV(USE.DATE,DM) 0004: YR = OCONV(USE.DATE,DY4) 0005: IF MTH LT 12 THEN 0006: MTH += 1 0007: END ELSE 0008: MTH = 1 0009: YR += 1 0010: END 0011: NEW.DATE = ICONV(MTH:/01/:YR,D) - 1 0012: CRT OCONV(NEW.DATE,D) -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Marco Manyevere Sent: Thursday, June 02, 2005 10:31 AM To: u2-users@listserver.u2ug.org Subject: [U2] Finding last day of month 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. - Yahoo! Messenger NEW - crystal clear PC to PCcalling worldwide with voicemail --- u2-users mailing list u2-users@listserver.u2ug.org To unsubscribe please visit http://listserver.u2ug.org/ --- u2-users mailing list u2-users@listserver.u2ug.org To unsubscribe please visit http://listserver.u2ug.org/
RE: [U2] Finding last day of month
I've always used the simple method. *Initialization section END.DATES = '31,28,31,30,31,30,31,31,30,31,30,31' END.DATES = CHANGE(END.DATES,',',@AM) *Main loop section LAST.DATE = END.DATESMONTH IF MONTH = 2 AND NOT(MOD(YEAR,4)) THEN LAST.DATE+=1 btw, there may be an error in this, my allergies are truly messing with my head today.and for the last weekGaaa Allen -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Behalf Of Dianne Ackerman Sent: Thursday, June 02, 2005 11:21 AM To: u2-users@listserver.u2ug.org Subject: Re: [U2] Finding last day of month That method can actually backfire, for example, if your starting date is 20040130, you'll end up on Feb 29 instead of Jan 31. What I would do is replace the day with 01 and add 1 to the month, then iconv and subtract 1 day. -Dianne Marco Manyevere wrote: 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. --- u2-users mailing list u2-users@listserver.u2ug.org To unsubscribe please visit http://listserver.u2ug.org/ --- u2-users mailing list u2-users@listserver.u2ug.org To unsubscribe please visit http://listserver.u2ug.org/
RE: [U2] Finding last day of month
Need to take month mod 12, when adding... D=20040203 Y=D[1,4] M=D[5,2] M+=1 IF M12 THEN M=1;Y+=1 NEXT.M=ICONV(M:/01/:Y,D4/)-1 PRINT OCONV(NEXT.M,D4Y):OCONV(NEXT.M,DM):OCONV(NEXT.M,DD) -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Marco Manyevere Sent: Thursday, June 02, 2005 10:31 AM To: u2-users@listserver.u2ug.org Subject: [U2] Finding last day of month 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. - Yahoo! Messenger NEW - crystal clear PC to PCcalling worldwide with voicemail --- u2-users mailing list u2-users@listserver.u2ug.org To unsubscribe please visit http://listserver.u2ug.org/ --- u2-users mailing list u2-users@listserver.u2ug.org To unsubscribe please visit http://listserver.u2ug.org/
RE: [U2] Finding last day of month
I believe you need to add a month check for December using that method: IN.DATE='20040203' NEW. = IN.DATE[1,4] NEW.MM = (IN.DATE[5,2]+1) 'R%2' IF NEW.MM=13 THEN NEW.MM='01' NEW.+=1 END I.LAST.DAY = ICONV(NEW.MM:-01-:NEW.),D4-)-1 -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Allen Egerton Sent: Thursday, June 02, 2005 10:57 AM To: u2-users@listserver.u2ug.org Subject: Re: [U2] Finding last day of month From: Marco Manyevere [EMAIL PROTECTED] 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? Find the internal date of the first day of the next month, subtract 1, and convert it back to external format. ORIG. = MMDD[1,4] ORIG.MM = MMDD[5,2] ORIG.DD = MMDD[7,2] I.LAST.DAY = (ICONV(ORIG.:(ORIG.MM + 1): 01), D4/) - 1 O.LAST.DAY = OCONV(I.LAST.DAY, D4/ --- u2-users mailing list u2-users@listserver.u2ug.org To unsubscribe please visit http://listserver.u2ug.org/ --- u2-users mailing list u2-users@listserver.u2ug.org To unsubscribe please visit http://listserver.u2ug.org/
RE: [U2] Finding last day of month
Won't this method choke if ORIG.MM is 12? === Norman Morgan [EMAIL PROTECTED] http://www.brake.com === Unscrewing an Oreo lets all the calories out. === -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Behalf Of Allen Egerton Sent: Thursday, June 02, 2005 12:57 PM To: u2-users@listserver.u2ug.org Subject: Re: [U2] Finding last day of month From: Marco Manyevere [EMAIL PROTECTED] 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? Find the internal date of the first day of the next month, subtract 1, and convert it back to external format. ORIG. = MMDD[1,4] ORIG.MM = MMDD[5,2] ORIG.DD = MMDD[7,2] I.LAST.DAY = (ICONV(ORIG.:(ORIG.MM + 1): 01), D4/) - 1 O.LAST.DAY = OCONV(I.LAST.DAY, D4/ --- u2-users mailing list u2-users@listserver.u2ug.org To unsubscribe please visit http://listserver.u2ug.org/ -- No virus found in this incoming message. Checked by AVG Anti-Virus. Version: 7.0.322 / Virus Database: 267.4.1 - Release Date: 6/2/2005 -- No virus found in this outgoing message. Checked by AVG Anti-Virus. Version: 7.0.322 / Virus Database: 267.4.1 - Release Date: 6/2/2005 --- u2-users mailing list u2-users@listserver.u2ug.org To unsubscribe please visit http://listserver.u2ug.org/
Re: [U2] Finding last day of month
* 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. - Yahoo! Messenger NEW - crystal clear PC to PCcalling worldwide with voicemail --- u2-users mailing list u2-users@listserver.u2ug.org To unsubscribe please visit http://listserver.u2ug.org/ --- u2-users mailing list u2-users@listserver.u2ug.org To unsubscribe please visit http://listserver.u2ug.org/
RE: [U2] Finding last day of month
If it's for a UniData dictionary you can use the function LAST_DAY(date) that will return the last day of the month. I'm not sure if UniVerse has this function as well. Eg: 001: I 002: LAST_DAY(INVOICE.DATE) 003: D4 004: Last Day of Month 005: 15R 006: S Colin Alfke Calgary, Canada -Original Message- From: Cliff Bennett Hi, Marco. I always ICONV the first day of the next month, then subtract a day. For example: ANY.DATE = '02-03-2004' MO = ANY.DATE[1,2] YR = ANY,DATE[7,4] MO += 1 IF MO 12 THEN MO = 1 YR += 1 END EOM.DATE = ICONV(MO:'-01-':YR, 'D') - 1 This lends itself to a subroutine or function as well. Regards, Cliff Marco Manyevere wrote: 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. --- u2-users mailing list u2-users@listserver.u2ug.org To unsubscribe please visit http://listserver.u2ug.org/
Re: [U2] Finding last day of month
you cant just use mod(x,4) logic the following from some website (dont remember where/when I found it) *** The rule for leap years is that all years divisable by 4 are leap years, except those years divisable by 100. The exception is that years divisible by 400 are leap years of course only us old foggies that used pick way back when remember this g Rich Allen E. Elwood wrote: I've always used the simple method. *Initialization section END.DATES = '31,28,31,30,31,30,31,31,30,31,30,31' END.DATES = CHANGE(END.DATES,',',@AM) *Main loop section LAST.DATE = END.DATESMONTH IF MONTH = 2 AND NOT(MOD(YEAR,4)) THEN LAST.DATE+=1 btw, there may be an error in this, my allergies are truly messing with my head today.and for the last weekGaaa Allen -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Behalf Of Dianne Ackerman Sent: Thursday, June 02, 2005 11:21 AM To: u2-users@listserver.u2ug.org Subject: Re: [U2] Finding last day of month That method can actually backfire, for example, if your starting date is 20040130, you'll end up on Feb 29 instead of Jan 31. What I would do is replace the day with 01 and add 1 to the month, then iconv and subtract 1 day. -Dianne Marco Manyevere wrote: 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. --- u2-users mailing list u2-users@listserver.u2ug.org To unsubscribe please visit http://listserver.u2ug.org/ --- u2-users mailing list u2-users@listserver.u2ug.org To unsubscribe please visit http://listserver.u2ug.org/ -- Richard A. Wilson Lakeside Systems Smithfield, RI, USA --- u2-users mailing list u2-users@listserver.u2ug.org To unsubscribe please visit http://listserver.u2ug.org/
RE: [U2] Finding last day of month
Wow, when did they sneak that in? And you forgot the {current weather
condition} in Calgary ;-)
Allen in the foggy San Fernando Valley
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] Behalf Of
[EMAIL PROTECTED]
Sent: Thursday, June 02, 2005 12:22 PM
To: u2-users@listserver.u2ug.org
Subject: RE: [U2] Finding last day of month
If it's for a UniData dictionary you can use the function LAST_DAY(date)
that will return the last day of the month. I'm not sure if UniVerse has
this function as well.
Eg:
001: I
002: LAST_DAY(INVOICE.DATE)
003: D4
004: Last Day of Month
005: 15R
006: S
Colin Alfke
Calgary, Canada
-Original Message-
From: Cliff Bennett
Hi, Marco. I always ICONV the first day of the next month,
then subtract a day. For example:
ANY.DATE = '02-03-2004'
MO = ANY.DATE[1,2]
YR = ANY,DATE[7,4]
MO += 1
IF MO 12 THEN
MO = 1
YR += 1
END
EOM.DATE = ICONV(MO:'-01-':YR, 'D') - 1
This lends itself to a subroutine or function as well.
Regards, Cliff
Marco Manyevere wrote:
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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RE: [U2] Finding last day of month
unidata has a LAST_DAY function? what version is that? -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Behalf Of [EMAIL PROTECTED] Sent: Thursday, June 02, 2005 2:22 PM To: u2-users@listserver.u2ug.org Subject: RE: [U2] Finding last day of month If it's for a UniData dictionary you can use the function LAST_DAY(date) that will return the last day of the month. I'm not sure if UniVerse has this function as well. Eg: 001: I 002: LAST_DAY(INVOICE.DATE) 003: D4 004: Last Day of Month 005: 15R 006: S Colin Alfke Calgary, Canada -Original Message- From: Cliff Bennett Hi, Marco. I always ICONV the first day of the next month, then subtract a day. For example: ANY.DATE = '02-03-2004' MO = ANY.DATE[1,2] YR = ANY,DATE[7,4] MO += 1 IF MO 12 THEN MO = 1 YR += 1 END EOM.DATE = ICONV(MO:'-01-':YR, 'D') - 1 This lends itself to a subroutine or function as well. Regards, Cliff Marco Manyevere wrote: 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. --- u2-users mailing list u2-users@listserver.u2ug.org To unsubscribe please visit http://listserver.u2ug.org/ --- u2-users mailing list u2-users@listserver.u2ug.org To unsubscribe please visit http://listserver.u2ug.org/
RE: [U2] Finding last day of month
Here's yet another way... END.OF.MONTH 0001 EDATES = 31:@FM:29:@FM:31:@FM:30:@FM:31:@FM:30:@FM:31:@FM:31:@FM: 30:@FM:31 :@FM:30:@FM:31 0002 PRINT INPUT DATE MMDD - : ; INPUT ODATE 0003 EYEAR = ODATE[1,4] 0004 EMONTH = ODATE[5,2] 0005 EDAY = EDATESEMONTH 0006 EDATE = EYEAR:EMONTH:EDAY 0007 IDATE = ICONV(EDATE,'D') 0008 IF STATUS() THEN 0009 EDATE -= 1 0010 END 0011 PRINT ODATE,EDATE Leon. --- u2-users mailing list u2-users@listserver.u2ug.org To unsubscribe please visit http://listserver.u2ug.org/
RE: [U2] Finding last day of month
Wow, there is a LAST_DAY function in unidate. I have 6.0 on HPUX and it works in virtual attributes but not in a basic program. Is that documented anywhere? What other cool functions are there undocumented? -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Behalf Of [EMAIL PROTECTED] Sent: Thursday, June 02, 2005 2:22 PM To: u2-users@listserver.u2ug.org Subject: RE: [U2] Finding last day of month If it's for a UniData dictionary you can use the function LAST_DAY(date) that will return the last day of the month. I'm not sure if UniVerse has this function as well. Eg: 001: I 002: LAST_DAY(INVOICE.DATE) 003: D4 004: Last Day of Month 005: 15R 006: S Colin Alfke Calgary, Canada -Original Message- From: Cliff Bennett Hi, Marco. I always ICONV the first day of the next month, then subtract a day. For example: ANY.DATE = '02-03-2004' MO = ANY.DATE[1,2] YR = ANY,DATE[7,4] MO += 1 IF MO 12 THEN MO = 1 YR += 1 END EOM.DATE = ICONV(MO:'-01-':YR, 'D') - 1 This lends itself to a subroutine or function as well. Regards, Cliff Marco Manyevere wrote: 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. --- u2-users mailing list u2-users@listserver.u2ug.org To unsubscribe please visit http://listserver.u2ug.org/ --- u2-users mailing list u2-users@listserver.u2ug.org To unsubscribe please visit http://listserver.u2ug.org/
RE: [U2] Finding last day of month
So...it won't be a problem until 2,400? I can live with that. And I'm gettin' up there in my years too. Just turned 49 last March. -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Behalf Of Richard A. Wilson Sent: Thursday, June 02, 2005 12:29 PM To: u2-users@listserver.u2ug.org Subject: Re: [U2] Finding last day of month you cant just use mod(x,4) logic the following from some website (dont remember where/when I found it) *** The rule for leap years is that all years divisable by 4 are leap years, except those years divisable by 100. The exception is that years divisible by 400 are leap years of course only us old foggies that used pick way back when remember this g Rich Allen E. Elwood wrote: I've always used the simple method. *Initialization section END.DATES = '31,28,31,30,31,30,31,31,30,31,30,31' END.DATES = CHANGE(END.DATES,',',@AM) *Main loop section LAST.DATE = END.DATESMONTH IF MONTH = 2 AND NOT(MOD(YEAR,4)) THEN LAST.DATE+=1 btw, there may be an error in this, my allergies are truly messing with my head today.and for the last weekGaaa Allen -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Behalf Of Dianne Ackerman Sent: Thursday, June 02, 2005 11:21 AM To: u2-users@listserver.u2ug.org Subject: Re: [U2] Finding last day of month That method can actually backfire, for example, if your starting date is 20040130, you'll end up on Feb 29 instead of Jan 31. What I would do is replace the day with 01 and add 1 to the month, then iconv and subtract 1 day. -Dianne Marco Manyevere wrote: 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. --- u2-users mailing list u2-users@listserver.u2ug.org To unsubscribe please visit http://listserver.u2ug.org/ --- u2-users mailing list u2-users@listserver.u2ug.org To unsubscribe please visit http://listserver.u2ug.org/ -- Richard A. Wilson Lakeside Systems Smithfield, RI, USA --- u2-users mailing list u2-users@listserver.u2ug.org To unsubscribe please visit http://listserver.u2ug.org/ --- u2-users mailing list u2-users@listserver.u2ug.org To unsubscribe please visit http://listserver.u2ug.org/
RE: [U2] Finding last day of month
It's in my version 5 documentation - not sure when it was put in. There
is also ADD_MONTHS, LAST_DAY, MONTHS_BETWEEN and NEXT_DAY. See Virtual
Attribute Functions in chapter 5 Creating Virtual Attributes.
Colin Alfke
Calgary, Canada
Where it's Sunny, no pouring rain, no wait it's just cloudy, oops now
it's raining again - OK I give up
-Original Message-
From: Allen E. Elwood
Wow, when did they sneak that in? And you forgot the
{current weather condition} in Calgary ;-)
Allen in the foggy San Fernando Valley
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RE: [U2] Finding last day of month
This is documented back to at least UniData 5.2 and probably earlier than that though I don't have easy access to the old manuals. It is in the Using UniData manual in the Creating Virtual Attributes section. You may wish to search for Virtual Attribute Functions. HTH. Regards, Jim [snip] Wow, there is a LAST_DAY function in unidate. I have 6.0 on HPUX and it works in virtual attributes but not in a basic program. Is that documented anywhere? What other cool functions are there undocumented? [snip] --- u2-users mailing list u2-users@listserver.u2ug.org To unsubscribe please visit http://listserver.u2ug.org/
Re: [U2] Finding last day of month
2100 2300 aren't leap years. I guess it depends on the application, antiques, historical dates etc. or future dates like when you will be fully vested in your retirement acct. g Rich Allen E. Elwood wrote: So...it won't be a problem until 2,400? I can live with that. And I'm gettin' up there in my years too. Just turned 49 last March. -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Behalf Of Richard A. Wilson Sent: Thursday, June 02, 2005 12:29 PM To: u2-users@listserver.u2ug.org Subject: Re: [U2] Finding last day of month you cant just use mod(x,4) logic the following from some website (dont remember where/when I found it) *** The rule for leap years is that all years divisable by 4 are leap years, except those years divisable by 100. The exception is that years divisible by 400 are leap years of course only us old foggies that used pick way back when remember this g Rich Allen E. Elwood wrote: I've always used the simple method. *Initialization section END.DATES = '31,28,31,30,31,30,31,31,30,31,30,31' END.DATES = CHANGE(END.DATES,',',@AM) *Main loop section LAST.DATE = END.DATESMONTH IF MONTH = 2 AND NOT(MOD(YEAR,4)) THEN LAST.DATE+=1 btw, there may be an error in this, my allergies are truly messing with my head today.and for the last weekGaaa Allen -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Behalf Of Dianne Ackerman Sent: Thursday, June 02, 2005 11:21 AM To: u2-users@listserver.u2ug.org Subject: Re: [U2] Finding last day of month That method can actually backfire, for example, if your starting date is 20040130, you'll end up on Feb 29 instead of Jan 31. What I would do is replace the day with 01 and add 1 to the month, then iconv and subtract 1 day. -Dianne Marco Manyevere wrote: 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. --- u2-users mailing list u2-users@listserver.u2ug.org To unsubscribe please visit http://listserver.u2ug.org/ --- u2-users mailing list u2-users@listserver.u2ug.org To unsubscribe please visit http://listserver.u2ug.org/ -- Richard A. Wilson Lakeside Systems Smithfield, RI, USA --- u2-users mailing list u2-users@listserver.u2ug.org To unsubscribe please visit http://listserver.u2ug.org/ --- u2-users mailing list u2-users@listserver.u2ug.org To unsubscribe please visit http://listserver.u2ug.org/ -- Richard A. Wilson Lakeside Systems Smithfield, RI, USA Voice 401-231-3959 Fax 206-202-2064 [EMAIL PROTECTED] www.lakeside-systems.com --- u2-users mailing list u2-users@listserver.u2ug.org To unsubscribe please visit http://listserver.u2ug.org/
RE: [U2] Finding last day of month
Mongo only small pawn in game of leap years I'm sure by then all of my code will be obsolete and that the technology that replaces computers will be beyond anything I can even imagine I read somewhere that some smart guy figured out how to transport a photon and one of the possible applications will be a zero wait state memory device. Beam me up Scotty! -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Behalf Of Richard A. Wilson Sent: Thursday, June 02, 2005 1:38 PM To: u2-users@listserver.u2ug.org Subject: Re: [U2] Finding last day of month 2100 2300 aren't leap years. I guess it depends on the application, antiques, historical dates etc. or future dates like when you will be fully vested in your retirement acct. g Rich Allen E. Elwood wrote: So...it won't be a problem until 2,400? I can live with that. And I'm gettin' up there in my years too. Just turned 49 last March. -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Behalf Of Richard A. Wilson Sent: Thursday, June 02, 2005 12:29 PM To: u2-users@listserver.u2ug.org Subject: Re: [U2] Finding last day of month you cant just use mod(x,4) logic the following from some website (dont remember where/when I found it) *** The rule for leap years is that all years divisable by 4 are leap years, except those years divisable by 100. The exception is that years divisible by 400 are leap years of course only us old foggies that used pick way back when remember this g Rich Allen E. Elwood wrote: I've always used the simple method. *Initialization section END.DATES = '31,28,31,30,31,30,31,31,30,31,30,31' END.DATES = CHANGE(END.DATES,',',@AM) *Main loop section LAST.DATE = END.DATESMONTH IF MONTH = 2 AND NOT(MOD(YEAR,4)) THEN LAST.DATE+=1 btw, there may be an error in this, my allergies are truly messing with my head today.and for the last weekGaaa Allen -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Behalf Of Dianne Ackerman Sent: Thursday, June 02, 2005 11:21 AM To: u2-users@listserver.u2ug.org Subject: Re: [U2] Finding last day of month That method can actually backfire, for example, if your starting date is 20040130, you'll end up on Feb 29 instead of Jan 31. What I would do is replace the day with 01 and add 1 to the month, then iconv and subtract 1 day. -Dianne Marco Manyevere wrote: 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. --- u2-users mailing list u2-users@listserver.u2ug.org To unsubscribe please visit http://listserver.u2ug.org/ --- u2-users mailing list u2-users@listserver.u2ug.org To unsubscribe please visit http://listserver.u2ug.org/ -- Richard A. Wilson Lakeside Systems Smithfield, RI, USA --- u2-users mailing list u2-users@listserver.u2ug.org To unsubscribe please visit http://listserver.u2ug.org/ --- u2-users mailing list u2-users@listserver.u2ug.org To unsubscribe please visit http://listserver.u2ug.org/ -- Richard A. Wilson Lakeside Systems Smithfield, RI, USA Voice 401-231-3959 Fax 206-202-2064 [EMAIL PROTECTED] www.lakeside-systems.com --- u2-users mailing list u2-users@listserver.u2ug.org To unsubscribe please visit http://listserver.u2ug.org/ --- u2-users mailing list u2-users@listserver.u2ug.org To unsubscribe please visit http://listserver.u2ug.org/
RE: [U2] Finding last day of month
I use the method my dad taught me when I was a kid, cuz I couldn't remember Thiry days hath September Hold 2 fists side-by-side, forefingers of right left hand touching. Each knuckle and each valley between knuckles represents a month. Start at the left pinky knuckle and count off months. If a month lands on a knuckle, it has 31 days. If it lands on a valley, it has 30 days. (Except February has that 28/29 thing going on.) _ _ _ _ _ _ _ _ /J\_/M\_/M\_/J\/A\_/O\_/D\_/ \ | F A J || S N | | || | | || | \ / \/ \ / \ / | left | | right | | fist | | fist | view from top QED --- u2-users mailing list u2-users@listserver.u2ug.org To unsubscribe please visit http://listserver.u2ug.org/
RE: [U2] Finding last day of month
Program that! Works with one hand too. Just hit the last knuckle a 2nd time before heading back. Easier to have 1 hand for pointing, pencil in the mouth to point marks up my hands. I was also taught this someplace along the line, and still use this method to this day. -Original Message- From: Stevenson, Charles [mailto:[EMAIL PROTECTED] Sent: Thursday, June 02, 2005 4:02 PM To: u2-users@listserver.u2ug.org Subject:RE: [U2] Finding last day of month I use the method my dad taught me when I was a kid, cuz I couldn't remember Thiry days hath September Hold 2 fists side-by-side, forefingers of right left hand touching. Each knuckle and each valley between knuckles represents a month. Start at the left pinky knuckle and count off months. If a month lands on a knuckle, it has 31 days. If it lands on a valley, it has 30 days. (Except February has that 28/29 thing going on.) _ _ _ _ _ _ _ _ /J\_/M\_/M\_/J\/A\_/O\_/D\_/ \ | F A J || S N | | || | | || | \ / \/ \ / \ / | left | | right | | fist | | fist | view from top QED --- u2-users mailing list u2-users@listserver.u2ug.org To unsubscribe please visit http://listserver.u2ug.org/ --- u2-users mailing list u2-users@listserver.u2ug.org To unsubscribe please visit http://listserver.u2ug.org/
RE: [U2] Finding last day of month
Oh dear, Oh dear! What ever happend to : DATE = '' FOR I = 31 TO 28 STEP -1 IF OCONV(ICONV(I:rest of date,'D'),'D') = I:rest of date THEN DATE = I:rest of date EXIT END NEXT I IF DATE = '' THEN Come to screeching HALT Of course for the people with backward dates (!!!), the I:rest of date would look like MONTH:I:YEAR with the appropriate separators, wouldn't it At 17:01 02/06/05 -0400, you wrote: I use the method my dad taught me when I was a kid, cuz I couldn't remember Thiry days hath September Hold 2 fists side-by-side, forefingers of right left hand touching. Each knuckle and each valley between knuckles represents a month. Start at the left pinky knuckle and count off months. If a month lands on a knuckle, it has 31 days. If it lands on a valley, it has 30 days. (Except February has that 28/29 thing going on.) _ _ _ _ _ _ _ _ /J\_/M\_/M\_/J\/A\_/O\_/D\_/ \ | F A J || S N | | || | | || | \ / \/ \ / \ / | left | | right | | fist | | fist | view from top QED --- u2-users mailing list u2-users@listserver.u2ug.org To unsubscribe please visit http://listserver.u2ug.org/ -- No virus found in this incoming message. Checked by AVG Anti-Virus. Version: 7.0.323 / Virus Database: 267.5.0 - Release Date: 02/06/05 -- No virus found in this incoming message. Checked by AVG Anti-Virus. Version: 7.0.323 / Virus Database: 267.5.0 - Release Date: 02/06/05 Regards, Bruce Nichol Talon Computer Services ALBURYNSW 2640 Australia http://www.taloncs.com.au Tel: +61 (0)411149636 Fax: +61 (0)260232119 If it ain't broke, fix it till it is! -- No virus found in this outgoing message. Checked by AVG Anti-Virus. Version: 7.0.323 / Virus Database: 267.5.0 - Release Date: 02/06/05 --- u2-users mailing list u2-users@listserver.u2ug.org To unsubscribe please visit http://listserver.u2ug.org/
RE: [U2] Finding last day of month
Aiming for SHORT (not clean, clear, or any of that other good stuff): PRINT D: ; INPUT D PRINT D[1,6]:MOD(D[5,2]+D[5,2]7,2)+30-(D[5,2]=2)*(2-MOD(D[1,4],4)#0) D?20040203 20040229 D?20030303 20030331 -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Marco Manyevere Sent: Thursday, June 02, 2005 10:31 AM To: u2-users@listserver.u2ug.org Subject: [U2] Finding last day of month 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? --- u2-users mailing list u2-users@listserver.u2ug.org To unsubscribe please visit http://listserver.u2ug.org/
RE: [U2] Finding last day of month
That incorrectly thinks 2100 is a leap. -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Ross Morrissey Sent: Friday, 3 June 2005 10:15 AM To: u2-users@listserver.u2ug.org Subject: RE: [U2] Finding last day of month Aiming for SHORT (not clean, clear, or any of that other good stuff): PRINT D: ; INPUT D PRINT D[1,6]:MOD(D[5,2]+D[5,2]7,2)+30-(D[5,2]=2)*(2-MOD(D[1,4],4)#0) D?20040203 20040229 D?20030303 20030331 -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Marco Manyevere Sent: Thursday, June 02, 2005 10:31 AM To: u2-users@listserver.u2ug.org Subject: [U2] Finding last day of month 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? --- u2-users mailing list u2-users@listserver.u2ug.org To unsubscribe please visit http://listserver.u2ug.org/ --- u2-users mailing list u2-users@listserver.u2ug.org To unsubscribe please visit http://listserver.u2ug.org/
Re: [U2] Finding last day of month
soapbox People! use the system why are you doing all this math in your program? The system knows what years are leap and what aren't. Let it do the work for you! /soapbox Will Johnson --- u2-users mailing list u2-users@listserver.u2ug.org To unsubscribe please visit http://listserver.u2ug.org/
RE: [U2] Finding last day of month
... or as a one line coding horror ;)
1 LAST.DAY =
REPLACE(CHANGE('31,28,31,30,31,30,31,31,30,31,30,31',',',@AM),2;28+(NOT(MOD((
THISDATE[1,4]),4)) AND (NOT(MOD((THISDATE[1,4]),400)) OR
MOD((THISDATE[1,4]),100(THISDATE[5,2]+0)
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Re: [U2] Finding last day of month
Ok smartass you say let's see you do it First of all, you can always get from the month you are in, to the next month by simply add 30 days to the middle of the month. And you never need to worry about leap years nor about 12th month logic Input Date ; * format mmdd Middleofthismonth = date[1,6]:'15' I.Middle = Iconv(Middleofthismonth,'D') I.Next = I.Middle + 30 NextMonth = Oconv(I.NEXT,'D2-') FirstDayNextMonth = NextMonth[1,2]:'01':NextMonth[5,2] I.FirstDay = Iconv(FirstDayNextMonth,'D') I.LastDayPreviousMonth = I.FirstDay - 1 Now can we scrunch this down? Input Date ; *format mmdd NextMonth = Oconv(Iconv(date[1,6]:'15','D') + 30,'D2') I.LastDayPreviousMonth = Iconv(NextMonth[1,2]:'01':NextMonth[5,2],'D') + 1 Do I win the big stuffed pickle? Will Johnson Fast Forward Technologies --- u2-users mailing list u2-users@listserver.u2ug.org To unsubscribe please visit http://listserver.u2ug.org/
Re: [U2] Finding last day of month
In a message dated 6/2/05 7:07:42 PM Pacific Daylight Time, [EMAIL PROTECTED] writes: Input Date ; * format mmdd Middleofthismonth = date[1,6]:'15' I.Middle = Iconv(Middleofthismonth,'D') I.Next = I.Middle + 30 NextMonth = Oconv(I.NEXT,'D2-') FirstDayNextMonth = NextMonth[1,2]:'01':NextMonth[5,2] I.FirstDay = Iconv(FirstDayNextMonth,'D') I.LastDayPreviousMonth = I.FirstDay - 1 Now can we scrunch this down? Input Date ; *format mmdd NextMonth = Oconv(Iconv(date[1,6]:'15','D') + 30,'D2') I.LastDayPreviousMonth = Iconv(NextMonth[1,2]:'01':NextMonth[5,2],'D') + 1 Before I get jumped on, there is an error in the above Input Date ; * format mmdd Middleofthismonth = date[1,6]:'15' I.Middle = Iconv(Middleofthismonth,'D') Should be changed to Input Date ; * format mmdd Middleofthismonth = date[5,2]:'-15-':date[1,4] I.Middle = Iconv(Middleofthismonth,'D') and then NextMonth = Oconv(Iconv(date[1,6]:'15','D') + 30,'D2') should be changed to NextMonth = Oconv(Iconv(date[5,2]:'-15-':date[1,4], 'D') + 30,'D2') Will Johnson Fast Forward Technologies --- u2-users mailing list u2-users@listserver.u2ug.org To unsubscribe please visit http://listserver.u2ug.org/
Re: [U2] Finding last day of month
[EMAIL PROTECTED] wrote on 06/02/2005 09:54:59 PM: Input Date ; *format mmdd NextMonth = Oconv(Iconv(date[1,6]:'15','D') + 30,'D2') I.LastDayPreviousMonth = Iconv(NextMonth[1,2]:'01':NextMonth[5,2],'D') + 1 That's an interesting approach, but it comes at the price of an extra ICONV, which is more expensive than a simple IF statement. Granted, the difference isn't significant for a small number of invocations, but after a while it adds up. By the way, you may want to check the logic of the scrunched version. It's not giving you what you want. --- u2-users mailing list u2-users@listserver.u2ug.org To unsubscribe please visit http://listserver.u2ug.org/
RE: [U2] Finding last day of month
Gee guys, mine was 4 lines. I think the request was for the 'shortest'... :-) Now, who's gonna put all these routines through a speed test to find out which is the most efficient? Me, I'm going back to watching Kull the conqueror -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Behalf Of [EMAIL PROTECTED] Sent: Thursday, June 02, 2005 7:46 PM To: u2-users@listserver.u2ug.org Subject: Re: [U2] Finding last day of month In a message dated 6/2/05 7:07:42 PM Pacific Daylight Time, [EMAIL PROTECTED] writes: Input Date ; * format mmdd Middleofthismonth = date[1,6]:'15' I.Middle = Iconv(Middleofthismonth,'D') I.Next = I.Middle + 30 NextMonth = Oconv(I.NEXT,'D2-') FirstDayNextMonth = NextMonth[1,2]:'01':NextMonth[5,2] I.FirstDay = Iconv(FirstDayNextMonth,'D') I.LastDayPreviousMonth = I.FirstDay - 1 Now can we scrunch this down? Input Date ; *format mmdd NextMonth = Oconv(Iconv(date[1,6]:'15','D') + 30,'D2') I.LastDayPreviousMonth = Iconv(NextMonth[1,2]:'01':NextMonth[5,2],'D') + 1 Before I get jumped on, there is an error in the above Input Date ; * format mmdd Middleofthismonth = date[1,6]:'15' I.Middle = Iconv(Middleofthismonth,'D') Should be changed to Input Date ; * format mmdd Middleofthismonth = date[5,2]:'-15-':date[1,4] I.Middle = Iconv(Middleofthismonth,'D') and then NextMonth = Oconv(Iconv(date[1,6]:'15','D') + 30,'D2') should be changed to NextMonth = Oconv(Iconv(date[5,2]:'-15-':date[1,4], 'D') + 30,'D2') Will Johnson Fast Forward Technologies --- u2-users mailing list u2-users@listserver.u2ug.org To unsubscribe please visit http://listserver.u2ug.org/ --- u2-users mailing list u2-users@listserver.u2ug.org To unsubscribe please visit http://listserver.u2ug.org/
Re: [U2] Finding last day of month
[EMAIL PROTECTED] wrote on 06/02/2005 10:34:09 PM: input oDate ; *format mmdd iDate = iconv(oDate,'dymd') loop until oconv(iDate+1,'dd) = 1 do iDate += 1 repeat lastDay = oconv(iDate,'dymd') Not sure if it's worth bragging about - but mine's smaller (146 v. 161 bytes). This will chew a lot of CPU cycles if you're working on the beginning of the month. Again, it doesn't make much of a difference on a small data sample, but if you're processing millions of records, it can add up. OK, I'll get off my performance soapbox now... Tim Snyder Consulting I/T Specialist , U2 Professional Services North American Lab Services DB2 Information Management, IBM Software Group [EMAIL PROTECTED] --- u2-users mailing list u2-users@listserver.u2ug.org To unsubscribe please visit http://listserver.u2ug.org/
RE: [U2] Finding last day of month
Oh please, Kieran. That's overboard.
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Kieran Clulow
Sent: Friday, 3 June 2005 11:43 AM
To: u2-users@listserver.u2ug.org
Subject: RE: [U2] Finding last day of month
... or as a one line coding horror ;)
1 LAST.DAY =
REPLACE(CHANGE('31,28,31,30,31,30,31,31,30,31,30,31',',',@AM),2;28+(NOT(MOD((
THISDATE[1,4]),4)) AND (NOT(MOD((THISDATE[1,4]),400)) OR
MOD((THISDATE[1,4]),100(THISDATE[5,2]+0)
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