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
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')
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
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 =
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
All,
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@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
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
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
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
]
[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
.
===
-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
* 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
, 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
@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
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
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
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
@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
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]
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
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
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
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
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
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
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,
:[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
[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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