I want to thank the following individual for showing me how to display
the
what I had thought to be the julian date. Little did I know that even
in school one does not learn the truth.
The answer for the UNIX system was: date +%j
Input from the field:
Woody Lee <fastest to respond with query and answer>
date +%j
Richard Turner <little history lesson>
<TRUE Julian date (i.e. numbers of days since noon on some day in 1857
or some year
like that) >
Jon Reeves <true to form - found it there>
<date +%j (Julian date within year -- consult date(1) man page for
other
formatting options)>
Andrew Weston <yes>
<Do you mean the date command: date +%j>
Lucio Chiappetti <Must have looked it up? or is a computer himself>
< If you mean the *TRUE* Julian Day Number, as defined by the
International Astronomical Union, i.e. the elapsed days from 12 UT
of 1 Jan 4713 B.C., you need the following subroutine :
C
C SUBROUTINE JAYDAY FROM MSSL XRAY PACKAGE
C
SUBROUTINE JAYDY(JY,JM,JD,JNL)
C CALENDAR TO JD DATE
INTEGER *4 JNL,ITEMP1
MY=JY
MM=JM
IF(MM-2)11,11,12
11 MM=MM+9
MY=MY-1
GOTO13
12 MM=MM-3
13 CONTINUE
ITEMP1=MY
JNL=15078+(1461*ITEMP1)/4+(153*MM+2)/5+JD
RETURN
END
If you mean the sequence number of the day in the current year (but
*PLEASE*
*PLEASE* *PLEASE* do *NOT* call it "julian date" ! It has nothing to do
with
it !) you can do
date +j
>
Once more I want to thank all those that responded.
Sincerely,
Andre' A.B. Paige
Received on Fri Sep 20 1996 - 17:27:59 NZST