how to retain zero


 
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# 1  
Old 03-14-2011
how to retain zero

Hi everyone


i have simple mathematical problem

Code:
 
bash> echo date +%m
bash> 03
 
bash> month=`date +%m`
bash> lmonth=$month-1
bash> echo $lmonth
bash>2


it gives me two but i want the answer of 02

please help me how to do this
# 2  
Old 03-14-2011
Code:
printf "%02d\n" $lmonth

# 3  
Old 03-14-2011
hii

thanks for reply

actually i need to append this lmonth variable with year
so how can we do this with printf command
# 4  
Old 03-14-2011
You could follow this recent thread.

Its for previous hour. You could do the same with previous month (for both the solution).
Easier if you have GNU date. just refer to the first solution.
# 5  
Old 03-14-2011
it's giving me this error

-bash$ date -d'1 hour ago' +%Y%m%d%H
date: illegal option -- d
usage: date [-u] mmddHHMM[[cc]yy][.SS]
date [-u] [+format]
date -a [-]sss[.fff]
# 6  
Old 03-14-2011
Try..

Code:
mon=$(date +%m)
arr=(12 1 2 3 4 5 6 7 8 9 10 11 12)
last_mon=${arr[$((mon - 1))]}
printf "%d%02d\n" $(date +%Y) $last_mon

# 7  
Old 03-14-2011
hii


thanks a lot for ur reply


i think of this code


Code:
 
typeset -RZ2 month
month=`date +%m`
year=`date +%Y`
month=$(($month-1))
if [ $month -lt 1 ]; then
        year=$(($year-1))
        month=12
fi
echo ${year}${month}

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