Has anyone else hear about this and is going?
Greetings!
I hope this invitation is reaching the people in your organization
responsible for the IBM UniData or UniVerse database technology that you
are using in your company. If not please advise and I will resend, or
just forward to that
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
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
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
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
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
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
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
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
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
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
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
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
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
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.
* 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
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
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
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:
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
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
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
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:
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,
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]
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
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
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
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
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
I will be out of the office starting 03/06/2005 and will not return until
06/06/2005.
Pour les clients accidant au Support Technique des produits IBM U2,
veuillez renvoyer votre message sur : [EMAIL PROTECTED]
Merci
For non-french speaking customers : If your IBM U2 products Technical
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
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
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
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... 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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u2-users@listserver.u2ug.org
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
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 =
[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,
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]
[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
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 ;)
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