So many solutions ... so little time. Thanks everyone!
Thanks,
Jon Knight
-----Original Message-----
From: oracle-l-bounce_at_freelists.org [mailto:oracle-l-bounce_at_freelists.org]
On Behalf Of Connor McDonald
Sent: Wednesday, March 30, 2005 7:28 AM
To: oracle-l_at_freelists.org
Subject: RE: date minus one
I don't have a unix box handy, but I dug up this from my archives
TZ=aaa24 date +%Y%m%d
hth
connor
- "Graeme St. Clair" <Graeme.St.Clair_at_hds.com> wrote:
> Why wouldn't you just use Perl "time() - 86400" (# seconds in a day), and
> then transform as desired with gmtime or whatever? (In theory it might be
a
> second off if you did it at half a second past a midnight with a leap
> second...)
>
> Rgds, GStC.
>
>
> -----Original Message-----
> From: oracle-l-bounce_at_freelists.org [mailto:oracle-l-bounce_at_freelists.org]
> On Behalf Of Jared Still
> Sent: Tuesday, March 29, 2005 7:30 PM
> To: jknight_at_concordefs.com
> Cc: oracle-l
> Subject: Re: date minus one
>
> And yet one more.
>
> #!/bin/sh
>
> # ydate: A Bourne shell script that
> # prints yestarday's date
> # Output Form: Month Day Year
> # From Focus on Unix: http://unix.about.com
>
> # Set the current month day and year.
> month=`date +%m`
> day=`date +%d`
> year=`date +%Y`
>
> # Add 0 to month. This is a
> # trick to make month an unpadded integer.
> month=`expr $month + 0`
>
> # Subtract one from the current day.
> day=`expr $day - 1`
>
> # If the day is 0 then determine the last # day of the previous month.
> if [ $day -eq 0 ]; then
>
> # Find the preivous month.
> month=`expr $month - 1`
>
> # If the month is 0 then it is Dec 31 of
> # the previous year.
> if [ $month -eq 0 ]; then
> month=12
> day=31
> year=`expr $year - 1`
>
> # If the month is not zero we need to find
> # the last day of the month.
> else
> case $month in
> 1|3|5|7|8|10|12) day=31;;
> 4|6|9|11) day=30;;
> 2)
> if [ `expr $year % 4` -eq 0 ]; then
> if [ `expr $year % 400` -eq 0 ]; then
> day=29
> elif [ `expr $year % 100` -eq 0 ]; then
> day=28
> else
> day=29
> fi
> else
> day=28
> fi
> ;;
> esac
> fi
> fi
>
> # Print the month day and year.
> echo "$year/$month/$day"
> exit 0
>
>
>
> On Tue, 29 Mar 2005 12:55:06 -0600, Knight, Jon <jknight_at_concordefs.com>
> wrote:
> > Just curious how the rest of the world gets "yesterday" in UNIX.
> > We're running Solaris and we execute a sqlplus script with "select
> > sysdate-1 from dual;" and pipe it to tail to set an environment
variable.
> >
> > Is there a more UNIXy way, -or- maybe a java function. Any
> > suggestions welcome.
> >
> > TIA,
> > Jon Knight
> >
> > --
> > http://www.freelists.org/webpage/oracle-l
> >
>
>
> --
> Jared Still
> Certifiable Oracle DBA and Part Time Perl Evangelist
> --
> http://www.freelists.org/webpage/oracle-l
> --
> http://www.freelists.org/webpage/oracle-l
>
Connor McDonald
Co-author: "Mastering Oracle PL/SQL - Practical Solutions"
Co-author: "Oracle Insight - Tales of the OakTable"
web: http://www.oracledba.co.uk
web: http://www.oaktable.net
email: connor_mcdonald_at_yahoo.com
"GIVE a man a fish and he will eat for a day. But TEACH him how to fish,
and...he will sit in a boat and drink beer all day"
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Received on Wed Mar 30 2005 - 11:43:30 CST
