Yes it's day-of-year.  Some Unix doc's list it as "Julian day-of-year".
Same difference...

Brian



                                                                                       
                       
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I'll show you one way to do it, if you quit calling that a Julian date. :)
( YYDDD is not a Julian date )

$> echo $DATE
2079

$> DATE=$(echo $DATE | awk '{ print substr("0000"$1,length("0000"$1)-4)
}')

$> echo $DATE
02079

Jared







"Ball, Terry" <[EMAIL PROTECTED]>
Sent by: [EMAIL PROTECTED]
03/21/02 11:08 AM
Please respond to ORACLE-L


        To:     Multiple recipients of list ORACLE-L <[EMAIL PROTECTED]>
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        Subject:        Sightly OT: Unix scripting


Environment:  Oracle 8.1.6.3 on Solaris 5.8

I am not a Unix Guru by any strech of the imagination, so I can use all
the
help I can get.  I am trying to write a shell script to execute sqlldr of
a
file whose name includes yesterday's 2 digit year and julian day in the
format: file_yyjjj.csv.  I can get today's date fine by setting a variable
to be DATE=`date '+%y%j'`

But when I try to get DATE -1 it strips the leading 0 (since it is
currently
02).  Does anyone have any suggestions as to how I can retain the leading
0
and still get yesterday's date?

TIA,

Terry

Terry Ball, DBA
Birch Telecom
Work: 816-300-1335
FAX:  816-300-1801

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