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02-08-2011 04:49 AM
02-08-2011 04:49 AM
Is there some way in VMS of getting the number of days since a certain date to today? To be precise the number of days since the 10th of April 1988 to today. I currently use a simple excel spreadsheet and enter the value myself, but if VMS can calculate this number for me it would be great.
Kind Regards,
Niall
Solved! Go to Solution.
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02-08-2011 04:58 AM
02-08-2011 04:58 AM
Re: Continuous day count
See HELP LEXICAL F$DELTA_TIME.
If You have a system older than VMS 7.3-2 and a fortran compiler, then You may use my version at
http://wwwvms.mpp.mpg.de/~huber/util/main/f$delta_time.HTML
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02-08-2011 05:25 AM
02-08-2011 05:25 AM
Re: Continuous day count
Yes I'm only running version 7.3. I was hoping that this could be done using a Lexical function of some sort. Is this possible?
Regards and Thanks,
Niall
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02-08-2011 05:30 AM
02-08-2011 05:30 AM
Re: Continuous day count
So either use my replacement program, or search for some earlier DCL command-files doing the calculation in pure DCL, I think at dcl.openvms.org there was one posted.
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02-08-2011 06:02 AM
02-08-2011 06:02 AM
Re: Continuous day count
http://dcl.openvms.org/stories.php?story=07/01/16/4051929
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02-08-2011 06:48 AM
02-08-2011 06:48 AM
Re: Continuous day count
This is what I enter and get nothing back at all, not even an error:
@time_difference.com; 16-JAN-2008:13:40:00 DELTA_TIME "TIME" 16-JAN-2007:13:40:00
Thanks again,
Niall
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02-08-2011 07:07 AM
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02-08-2011 07:30 AM
02-08-2011 07:30 AM
Re: Continuous day count
So this works, kind off. for some reason the day i was in 1988 does not, but one year later it does. Any ideas?
Prompt> @time_difference 10-Apr-1989:12:00:00 delta_time "TIME" 8-FEB-2011:00:00:00
Prompt> sh sym DELTA_TIME
DELTA_TIME == "-7974-12:00:00.00"
Prompt> @time_difference 10-Apr-1988:12:00:00 delta_time "TIME" 8-FEB-2011:00:00:00
%DCL-W-IVATIME, invalid absolute time - use DD-MMM-YYYY:HH:MM:SS.CC format
\8-FEB-2011 00:00:00.00-16383-0:0\
%DCL-W-IVATIME, invalid absolute time - use DD-MMM-YYYY:HH:MM:SS.CC format
\8-FEB-2011 00:00:00.00-32767-0:0\
Thanks,
Niall
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02-08-2011 07:39 AM
02-08-2011 07:39 AM
Re: Continuous day count
So to use it properly you'd have to go through the years in chunks of less than 9999 days.
See example below.
Google finds a good few examples, and the TIME_DIFFERENCE one on OpenVMS.org is fine, but a bit much.
A cure but noisy, general purpose example is:
http://www.convertunits.com/dates/from/Nov+17,+1858/to/Feb+8,+2011
Brute forcing through the years and days instead of a binary search works fine as well, or you can just use math as in other examples.
The example below uses the fact the '29' it odd, thus has the low bit set, and thus is TRUE, whereas 28 is false, to adjust for leap years.
Below two examples to just calculate the days since a reference date (not before 17-NOV-1858).
You can edit in your own reference date to speed it up a little if need be.
Turn into a subroutine as needed to calculate a difference.
Hein
$ type DAY_SINCE_17_NOV_1858.com
$
$ target = 123 ! This integer will become a string if Convert time works
$ target = F$CVT(P1,"ABSOLUTE","DATE")
$ IF F$TYPE(target) .NES. "STRING" THEN EXIT 16
$ year = 1858
$ days = 45 - 365
$ target_year = 'F$CVT( target,,"YEAR")
$
$year_loop:
$ year = year + 1
$ days = days + 365
$ IF F$CVTIME("1-MAR-''year' -1-",,"DAY") THEN days = days + 1 ! 28=false, 29=true
$ IF year.LT.target_year THEN GOTO year_loop
$
$ date = "1-JAN-''year'"
$day_loop:
$ IF date .EQS.target THEN GOTO done
$ date = F$CVT("''date' +1-","ABSOLUTE","DATE")
$ days = days + 1
$ goto day_loop
$
$ done:
$ write sys$output p1, " is ", days, " days since 17-NOV-1858 "
For OpenVMS 7.3-2 or better with F$DELTA
$
$ target = 123 ! This integer will become a string if Convert time works
$ target = F$CVT(P1,"ABSOLUTE","DATE")
$ IF F$TYPE(target) .NES. "STRING" THEN EXIT 16
$ days = 0
$ start_date = "17-NOV-"
$ start_year = 1858
$ target_year = 'F$CVT( target,,"YEAR")
$ target_date = target - "''target_year'"
$chunck_loop:
$ IF ( target_year - start_year ) .LE. 25 THEN GOTO last_chunck
$ start_next = start_year + 25
$ days = days + f$elem(0," ",f$delta( start_date + "''start_year'", start_date + "''start_next'" ))
$ start_year = start_next
$ goto chunck_loop
$
$last_chunck:
$ dddd = f$delta( start_date + "''start_year'", target)
$ days = days + f$elem(0," ", f$edit( dddd, "TRIM" ))
$ write sys$output p1, " is ", days, " days since 17-NOV-1858 "
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02-08-2011 07:50 AM
02-08-2011 07:50 AM
Re: Continuous day count
Just if it is fails with less than 9999, then it will probably fail also with looping it.
Maybe it possible to add a value to the returned DELTA_TIME. Say the number of days from 10-APR-1988 to 31-Jan-1999 and then add this number of days to the returned DELTA_TIME and use the start date to be 01-Jan-2000. How do i do that then... :)
Thanks,
Niall
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02-08-2011 07:59 AM
02-08-2011 07:59 AM
Re: Continuous day count
That's exactly what my second example does in a very generic format, using 25 year chunks since the beginning of (OpenVMS) time.
You could easily simplify picking a different reference date for which you know you will need only one more F$DELTA and stay in that range for the supported future or the script.
Your example...
$ write sys$output F$DELTA ( "10-APR-1988", "01-Jan-2000")
4283 00:00:00.00
$ write sys$output F$DELTA ( "01-Jan-2000", "08-Feb-2011")
4056 00:00:00.00
$ write sys$output F$DELTA ( "10-APR-1988", "08-Feb-2011")
8339 00:00:00.00
$ write sys$output 4283 + 4056
8339
Hein
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02-08-2011 07:59 AM
02-08-2011 07:59 AM
Re: Continuous day count
$ new_time = f$cvtime(f$fao(bformat,delta_value),,component)
%DCL-W-IVATIME, invalid absolute time - use DD-MMM-YYYY:HH:MM:SS.CC format
\8-FEB-2011 13:40:00.00-16383-0:0\
Seems the author is doubling the delta-time testing iteratively. And since the difference is >8192 , next is 163nn, giving an illegal combined time for DCL.
So this time_difference.com in fact is not working up to 9999 days, but only for something like 8192.
Or use my above mentioned program, it works definetly up to 9999 days:
f$delta_time/log temp 10-Apr-1988:13:40:00 8-FEB-2011:13:40:00
8339 00:00:00.00
BTW, if You are interested in just counting days, then there is another possibility:
use a "Julian Day number" routine like LIB$DAY for the two dates, and take the difference of the two resulting numbers: this works for dates back to the modified Julian Date day 0 of 17-NOV-1859.
Or even a general julian date program jd* like I have on
http://wwwvms.mpp.mpg.de/~huber/util/
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02-08-2011 08:25 AM
02-08-2011 08:25 AM
Re: Continuous day count
http://wwwvms.mpp.mpg.de/~huber/util/main/mjdv.for
it is as short as:
MPIW10_HUB>@day_diff 1-jan-1982 today
44970
55600
TDIFF = 10630 Hex = 00002986 Octal = 00000024606
$!day_diff.com:
$ mjdv 'p1'
$ d1=f$integer(mjdn)
$ mjdv 'p2'
$ d2=f$integer(mjdn)
$ tdiff=d2-d1
$ show symbol tdiff
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02-08-2011 08:38 AM
02-08-2011 08:38 AM
Re: Continuous day count
I can't make head nor tail from that last link you posted, can you paste the text in here.
I looked at your julian day counter, but it appears to be two days wrong or I miss counted somewhere. I expected 8340 not 8338. Below is me running it:
Prompt> @DAY_SINCE_10_APR_1988 8-FEB-2011:13:40:00
$ TYP DAY_SINCE_10_APR_1988.COM;
$ target = 123 ! This integer will become a string if Convert time works
$ target = F$CVT(P1,"ABSOLUTE","DATE")
$ IF F$TYPE(target) .NES. "STRING" THEN EXIT 16
$ year = 1988
$ days = 265 - 365
$ target_year = 'F$CVT( target,,"YEAR")
$
$year_loop:
$ year = year + 1
$ days = days + 365
$ IF F$CVTIME("1-MAR-''year' -1-",,"DAY") THEN days = days + 1 ! 28=false, 29=true
$ IF year.LT.target_year THEN GOTO year_loop
$
$ date = "1-JAN-''year'"
$day_loop:
$ IF date .EQS.target THEN GOTO done
$ date = F$CVT("''date' +1-","ABSOLUTE","DATE")
$ days = days + 1
$ goto day_loop
$
$ done:
$ write sys$output p1, " is ", days, " days since 10-Apr-1988 "
8-FEB-2011:13:40:00 is 8338 days since 10-Apr-1988
Thanks,
Niall
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02-08-2011 10:54 AM
02-08-2011 10:54 AM
Re: Continuous day count
http://wwwvms.mpp.mpg.de/~huber/util/main/mjdv.for ?
If You are using MS internet explorer, then yes, it does not deal right with .FOR files :-)
I attach it .
My run looks like
day_diff 10-APR-1988 today
8339
So it is correctly calculating days between dates, just add 1 in the day_diff.com above tdiff=d2-d1+1
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02-08-2011 02:30 PM
02-08-2011 02:30 PM
Re: Continuous day count
This procedure will work on V7.2. I believe it's the fastest possible implementation (left as an exercise to figure out how it works ;-)
I did have a version that worked for ranges greater than 10000 days, but I can't find it at the moment.
@SUBTIMES "10-APR-1988" "TODAY"
DELTA == "8340 00:00:00.00"
SUBTIMES.COM
$ ON WARNING THEN EXIT
$ tt=F$CVTIME(p1)
$ GOSUB x
$ b0=d
$ tt=F$CVTIME(p2)
$ GOSUB x
$ b1=d
$ d[32,32]=%XF0000000
$ d[0,32]=0
$ l0=F$CVUI(0,30,b0)
$ l1=F$CVUI(0,30,b1)
$ IF l0.LT.l1
$ THEN
$ d[0,30]=.NOT.((l1-l0).AND.%X3FFFFFFF)
$ d[30,30]=.NOT.(F$CVUI(30,30,b1)-F$CVUI(30,30,b0))
$ ELSE
$ d[0,30]=.NOT.((%X40000000+l1-l0).AND.%X3FFFFFFF)
$ d[30,30]=.NOT.(F$CVUI(30,30,b1)-F$CVUI(30,30,b0)-1)
$ ENDIF
$ delta==F$FAO("!%D",F$CVUI(32,32,F$FAO("!AD",8,d)))
$ SHOW SYM delta
$ EXIT
$ x:
$ d[32,32]=0
$ d[0,32]=0
$ b=60
$ Loop: d[b,1]=1
$ IF F$CVTIME(F$FAO("!%D",F$CVUI(32,32,F$FAO("!AD",8,d)))).GTS.tt THEN d[b,1]=0
$ b=b-1
$ IF b.GT.0 THEN GOTO loop
$ RETURN
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02-08-2011 03:08 PM
02-08-2011 03:08 PM
Re: Continuous day count
Actually, since you only want DAYS between dates (as opposed to a delta time between two absolute datetimes), it's much easier.
This procedure will calculate the number of days between any two dates from 17-NOV-1858 through to 18-JUN-9990. First parameter is the start date, second is end date (defaults to TODAY)
$ @dayssince 10-apr-1988
DAYS = 8340 Hex = 00002094 Octal = 00000020224
$ @dayssince 10-apr-1888
DAYS = 44864 Hex = 0000AF40 Octal = 00000127500
$ @dayssince 17-nov-1858 18-jun-9990
DAYS = 2970000 Hex = 002D5190 Octal = 00013250620
DAYSSINCE.COM
$ ON WARNING THEN EXIT
$ s=F$CVTIME(p1,"ABSOLUTE","DATE")
$ e=F$CVTIME(p2,,"DATE")
$ c=0
$ d=5000
$ G: IF F$CVTIME("''s'+''d'-0",,"DATE").LTS.e
$ THEN
$ c=c+d
$ s=F$CVTIME("''s'+''d'-0","ABSOLUTE")
$ GOTO G
$ ENDIF
$ x=d
$ F: t=F$CVTIME("''s'+''x'-0",,"DATE")
$ IF t.NES.e
$ THEN
$ d=(d+1)/2
$ IF t.LTS.e
$ THEN
$ x=x+d
$ ELSE
$ x=x-d
$ ENDIF
$ GOTO F
$ ENDIF
$ days=c+x
$ SHOW SYM days
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02-08-2011 04:43 PM
02-08-2011 04:43 PM
Re: Continuous day count
$ pipe cc days ; link days ; run days
8340 days
$
days.c
------
#include
#include
int main() {
static struct tm tm_beg;
int days;
tm_beg.tm_mday = 10; /* 10th */
tm_beg.tm_mon = 3; /* April */
tm_beg.tm_year = 88; /* 1988 */
tm_beg.tm_isdst = -1; /* don't know if this is STD or DST in your timezone */
days = (time(0) - mktime(&tm_beg)) / (24*3600) + 1;
printf("%d days\n", days);
}
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02-09-2011 12:01 AM
02-09-2011 12:01 AM
Re: Continuous day count
Just a question about the "+1" :
would one like the number of days between yesterday and today to be 2 ?
I prefer to leave off "+1".
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02-09-2011 10:03 AM
02-09-2011 10:03 AM
Re: Continuous day count
"
Just a question about the "+1" :
would one like the number of days between yesterday and today to be 2 ?
I prefer to leave off "+1".
"
It depends on whether you want to count current (partial) day as a day. If you don't, leave "+1" off. Probably, NOT counting partial day is more intuitive than counting it.
Thanks,
-Boris
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02-10-2011 09:51 PM
02-10-2011 09:51 PM
Re: Continuous day count
A little late, but you are correct.
I managed to paste a wrong version of the
code.
The leapyear test should be done before the increment. Sorry.
That makes it 8339.
It becomes 8340 if you want to count inclusive of start and end date for a duration.
The reason why this reply is so late is because I wanted to point out that it was not
my intention to edit the file for a specific start, but to just use an early enough date to handle start and end.
I used 17_NOV_1858 because that's the OpenVMS limit.
Next I tried 1-JAN-1900 because that's the EXCEL base date for use with DATEVALUE.
Handy for testing I thought....
Except that those amateurs at Microsoft got it wrong!
They have year 1900 as a Leap Year! Argh!
So I wasted a good bit of time getting confused by that + my own calculation order problem.
Excel::
text datevalue
1/1/1900 1
2/28/1900 59
3/1/1900 61
2/8/2011 40582
4/10/1988 32243
8339
I guess a handy base date is 1/1/1970.
That's the Unix base date ( UTC )
The code below gives:
$ @DAY_SINCE_1_JAN_1970.COM 11-feb-2011:04:40
11-FEB-2011:04:40 is 15016 days since 1_jan_1970. Unix Epoch Seconds: 1297399200
Confirmed with:
$ perl -e "print time"
1297399257
and also
$ @DAY_SINCE_1_JAN_1970.COM 1-jan-1970
1-JAN-1970 is 0 days since 1_jan_1970. Unix Epoch Seconds: 0
$ @DAY_SINCE_1_JAN_1970.COM 2-jan-1970
2-JAN-1970 is 1 days since 1_jan_1970. Unix Epoch Seconds: 86400
$ type DAY_SINCE_1_JAN_1970.COM
$
$ target = 123 ! This integer will become a string if Convert time works
$ target = F$CVT(P1,"ABSOLUTE","DATE")
$ IF F$TYPE(target) .NES. "STRING" THEN EXIT 16
$ year = 1969
$ days = -365
$ target_year = 'F$CVT( target,,"YEAR")
$
$year_loop:
$ year = year + 1
$ IF F$CVTIME("1-MAR-''year' -1-",,"DAY") THEN days = days + 1
$ days = days + 365
$ IF year.LT.target_year THEN GOTO year_loop
$
$ date = "1-JAN-''year'"
$day_loop:
$ IF date .EQS.target THEN GOTO done
$ date = F$CVT("''date' +1-","ABSOLUTE","DATE")
$ days = days + 1
$ goto day_loop
$
$ done:
$ seconds = days*86400 + 3600*F$CVT(p1,,"HOUR") + 60*F$CVT(p1,,"MINUTE") + F$CVT(p1,,"SECOND")
$ write sys$output p1, " is ", days, " days since 1_jan_1970. Unix Epoch Seconds: ", seconds
Cheers,
Hein
btw... the website I referred to reports:
http://www.convertunits.com/dates/from/Apr+10,+1988/to/Feb+8,+2011
"The total number of days between Sunday, April 10th, 1988 and Tuesday, February 8th, 2011 is 8,339 days.
This is equal to exactly 22 years, 9 months, and 29 days.
This does not include the end date, so it's accurate if you're measuring your age in days, or the total days between the start and end date. But if you want the duration of an event that includes both the starting date and the ending date, then it would actually be 8,340 days.
The total time span from 1988-04-10 to 2011-02-08 is 200,136 hours.
This is equivalent to 12,008,160 minutes.
You can also convert 8,339 days to 720,489,600 seconds."
Hein
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02-23-2011 07:49 AM
02-23-2011 07:49 AM
Re: Continuous day count
Niall
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02-24-2011 02:12 AM
02-24-2011 02:12 AM
Re: Continuous day count
"Except that those amateurs at Microsoft got it wrong!"
They claim it was Lotus who got it wrong:
http://support.microsoft.com/kb/214326
:-)
Oswald