Hi,

How can I display date-7 in AIX?
I want to print a date of one week back.
Can I do it in AIX?

Thanks
Suresh

sureshht

Are you looking at doing this in a Unix shell script?

Because if you are, it's time for William to do his thing:

http://www.computing.net/unix/wwwbo...

Hi,
I am doing some work with logs on a weekly basis. I will check some files and finally print the error.If I dont find any error in a weeks time, I just want to print "no errors forund since "7 days back date".

Thanks
Suresh

sureshht

Did you even try shift_dates? Copy shift_dates into a directory. It's available from at least two sites mentioned in the other thread. Create another script that calls shift_dates:

#!/bin/ksh

# today's date
todays_date=$(date '+%Y%m%d')

# shift_dates is in the current directory. William's shift_date expects
# date in YYYYMMDD format
# seven days back
date7back=$(./shift_dates $todays_date -7)

echo $date7back

I tried this and it works fine.
But,I want date-7 in mm/dd/yyyy format.
Any idea?

Thanks
Suresh

sureshht

I see two choices: Either change the shift_date function's date argument or create a driver function that changes your format to the shift_date format, calls shift_date, and then changes the format back you your format. This program does that:

#!/bin/ksh

# this function requires
# $1 is date in mm/dd/yyyy format
# $2 is the number of days to shift
# return the date in mm/dd/yyyy format
function sd_driver {
typeset -i nodays

nodays=$2

# parse mm/dd/yyyy
set -- $(IFS="/"; echo $1)

newdate=$(./shift_date.ss $3$1$2 $nodays)
yy=$(echo $newdate|cut -b 1-4)
mm=$(echo $newdate|cut -b 5-6)
dd=$(echo $newdate|cut -b 7-8)
echo $mm/$dd/$yy
}

nd=$(sd_driver "02/28/2007" 1)
echo $nd

I was a little bored today, so I wrote this script to do julian - 365. I did no checking so if you put in a number that puts you back 2 years, you can modify it to work properly.

integer today=$(date +%j)
print "Enter the number of days to subtract: \c"
read daysToDelete

currentYear=$(date +%Y)
newJulian=$(print "$today - $daysToDelete" | bc )
echo $newJulian

for days in $(cal 2 $lastYear)
do
:
done

if [[ $newJulian -le 0 ]]
then
print "We're working with a date from last year"
(( lastYear = currentYear - 1 ))
if [[ $days -eq 29 ]]
then
print "Last year was a leap year"
julianDate=$(echo "366 $newJulian" | bc )
leapYear=1
year=$lastYear
else
print "Last year was a not a leap year"
julianDate=$(echo "365 + $newJulian" | bc )
year=$lastYear
fi
else
julianDate=$newJulian
year=$currentYear
fi

typeset -R2 year

if [[ $julianDate -eq 0 ]]
then
print "Date is $year/12/31"
exit 0
fi

if [[ $julianDate -le 31 ]]
then
print "Date is $year/01/$julianDate"
exit 0
fi

(( julianDate -= 31 ))

if [[ $leapYear -eq 1 && $julianDate -le 29 ]]
then
print "Date is $year/02/$julianDate"
exit 0
elif [[ $julianDate -le 28 ]]
then
print "Date is $year/02/$julianDate"
exit 0
fi

if [[ $leapYear -eq 1 ]]
then
(( julianDate -= 29 ))
else
(( julianDate -= 28 ))
fi

if [[ $julianDate -le 31 ]]
then
print "Date is $year/03/$julianDate"
exit 0
fi
(( julianDate -= 31 ))

if [[ $julianDate -le 30 ]]
then
print "Date is $year/04/$julianDate"
exit 0
fi
(( julianDate -= 30 ))

if [[ $julianDate -le 31 ]]
then
print "Date is $year/05/$julianDate"
exit 0
fi
(( julianDate -= 31 ))

if [[ $julianDate -le 30 ]]
then
print "Date is $year/06/$julianDate"
exit 0
fi
(( julianDate -= 30 ))

if [[ $julianDate -le 31 ]]
then
print "Date is $year/07/$julianDate"
exit 0
fi
(( julianDate -= 31 ))

if [[ $julianDate -le 31 ]]
then
print "Date is $year/08/$julianDate"
exit 0
fi
(( julianDate -= 31 ))

if [[ $julianDate -le 30 ]]
then
print "Date is $year/09/$julianDate"
exit 0
fi
(( julianDate -= 30 ))

if [[ $julianDate -le 31 ]]
then
print "Date is $year/10/$julianDate"
exit 0
fi
(( julianDate -= 31 ))

if [[ $julianDate -le 30 ]]
then
print "Date is $year/11/$julianDate"
exit 0
fi
(( julianDate -= 30 ))

print "Date is $year/12/$julianDate"
$

Thanks a lot guys

Suresh

sureshht