last month end date

vanathi · #1 (**permalink**) 03-19-2008

Hi,

How to get the lastmonth's end date? (have current date) through s shell script. I need to get it and pass it to a procedure.

Please advice.
Thanks in advance.

Yogesh Sawant · #2 (**permalink**) 03-19-2008

check if you can find a solution here

jim mcnamara · #3 (**permalink**) 03-21-2008

For future reference if someone reads this:
try cal

Code:

#!/bin/ksh
printf "%d %d" $(date "+%Y %m") | read year month
let month=$month-1
if [[ $month -eq 0 ]] ; then
   let year=$year-1
   let month=12
fi
cal $month $year | tr -s '\n' ' ' | awk '{print $NF}' | read day
printf "%d/%02d/%02d\n" $year $month $day

joeyg · #4 (**permalink**) 03-21-2008

Code:

if [[ $month -eq 0 ]] ; then
   let year=$year-1
   let month=1
fi

Shouldn't this be ?

Code:

if [[ $month -eq 0 ]] ; then
   let year=$year-1
   let month=12
fi

January is month 1, so when you subtract 1 you get 0. And you rightly trap for it, decrementing the year but setting the month to 1 for January. I think you should set to 12 for December.

curleb · #5 (**permalink**) 03-21-2008

Perl-based approach to deriving backwards in time...I haven't quite bothered to try going back to the future as of yet. But it also handles Leap Years...

Code:

$ pl_end_of_last_month_0=`perl -e '\
> $y= time - (86400 * (localtime(time))[3]); \
> printf "%04d%02d%02d\n", (localtime($y))[5] + 1900 ,(localtime($y))[4] + 1 ,(localtime($y))[3] ; ' ` 
$ echo $pl_end_of_last_month_0 
20070831
===

Code:

$ # Today... 
$ pl_today_0=`perl -e '\
> $y= time - (86400 * $ARGV[0]); \
> printf "%04d%02d%02d\n", (localtime($y))[5] + 1900 ,(localtime($y))[4] + 1 ,(localtime($y))[3] ; ' 0 ` 
$ echo $pl_today_0 
20070912
=== 
$ # Today minus 1... (um, yesterday...?) 
$ pl_today_1=`perl -e '\
> $y= time - (86400 * $ARGV[0]); \
> printf "%04d%02d%02d\n", (localtime($y))[5] + 1900 ,(localtime($y))[4] + 1 ,(localtime($y))[3] ; ' 1 ` 
$ echo $pl_today_1 
20070911
=== 
$ # Today minus a defined number... 
$ my_number=3 
$ pl_today_mynumber=`perl -e '\
> $y= time - (86400 * $ARGV[0]); \
> printf "%04d%02d%02d\n", (localtime($y))[5] + 1900 ,(localtime($y))[4] + 1 ,(localtime($y))[3] ; ' ${my_number} ` 
20070909
===

jim mcnamara · #6 (**permalink**) 03-21-2008

Quote:

Originally Posted by joeyg

Code:

if [[ $month -eq 0 ]] ; then
   let year=$year-1
   let month=1
fi

Shouldn't this be ?

Code:

if [[ $month -eq 0 ]] ; then
   let year=$year-1
   let month=12
fi

January is month 1, so when you subtract 1 you get 0. And you rightly trap for it, decrementing the year but setting the month to 1 for January. I think you should set to 12 for December.

You are right. I was being bombed with other things when I did that... my bad.

jim mcnamara · #7 (**permalink**) 03-21-2008

And perl/python/C is better solution.

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