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Substract date (month) Problem

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# 1  
Old 02-10-2009
Substract date (month) Problem
#!/bin/ksh

month=`date | cut -c5-8`
year=`date | cut -c24-28`

echo "$month"
echo "$year"

--- This gives me output as Feb and 2009
but now I want to substract the 1 month from the current script and want output as

Jan 2009.


Please note I have searched a lot on forum and found option -v with date but it doesn't work wd on my mc.

Please help...!!!
# 2  
Old 02-10-2009
Kunal,

Check this link. You might find your answer. Smilie

Request you to go through the FAQ section and search for similar threads before creating a new one.

Regards,

Praveen
# 3  
Old 02-10-2009
Hi, praveen

as i said, date -v option not working from my end....

hence need help
# 4  
Old 02-10-2009
Try this
Code:

day=`date +%d`
month=`date +%m`
year=`date +%Y`
lmonth=`expr $month - 1`
if test "$lmonth" = "0"
then
lmonth=12
year=`expr $year - 1`
fi
echo "Last month was $lmonth/$year"
# 5  
Old 02-10-2009
try this option:

Code:
get_month_name() 
{
case $1 in
01) month_name="Jan"
02) month_name="Feb"
03) month_name="Mar"
04) month_name="Apr"
05) month_name="May"
06) month_name="Jun"
07) month_name="Jul"
08) month_name="Aug"
09) month_name="Sep"
10) month_name="Oct"
11) month_name="Nov"
12) month_name="Dec"
}

month=`date +%m`
year=`date +%Y`

month=$(( ${month} - 1 ))

if [[ ${month} -eq 0 ]]; then
month=12
year=$(( ${year} - 1 ))
fi
if [[ ${month} -lt 10 ]]; then # prefix a zero in front of
month=0${month}
fi
get_month_name "${month}"

echo ${month_name}${year}

HTH, Smilie

Regards,

Praveen
# 6  
Old 02-10-2009
Thanks a lot - Praveen....!!!

thanks again
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