Get last months month in alpha?

[Coolmar]

Hi,

Is there a simple way to get last month's month in alpha chars. For example, this month is "Feb", I need it to output last month "Jan"

Thanks,
S.

[A. Clay Stephenson · Accepted solution]

Here's one approach using caljd.sh:

#!/usr/bin/sh

caljd.sh $(caljd.sh) | read MO DUMMY YEAR
MON=$(caljd.sh -M -o $(caljd.sh -p 1 ${MO} 1 ${YEAR}))
echo "Month = ${MON}"

It works by determining the month day and year or the current month, then setting the day of the month to 1 and finding the Julian day of the previous day and then outputting the monthname. Change the "-o" in the last line to "-O" and the full month name will appear -- and full National Language Support works if LANG is set.

Here's caljd.sh; invoke as caljd.sh -u for full usage and examples.

[Sandman!]

Here's something that you could try though I would prefer to use caljd.sh

# let x=$(date +"%m")-1
# cal $(echo $x) $(date +%Y) | awk '{print substr($1,1,3);exit}'

[Denver Osborn]

THIS_MONTH=`date +%b`
[[ $THIS_MONTH = Jan ]] && LAST_MONTH=Dec
[[ $THIS_MONTH = Feb ]] && LAST_MONTH=Jan
[[ $THIS_MONTH = Mar ]] && LAST_MONTH=Feb
[[ $THIS_MONTH = Apr ]] && LAST_MONTH=Mar
..and so on..
echo Last month was $LAST_MONTH


-denver

[spex]

Hi Coolmar,

I had to do this one with pen and paper first...

$ echo "((($(date +%m)-1)+11)%12)+1"|bc

PCS

[James R. Ferguson]

Hi:

Another:

# perl -le '$m=shift or die;@mon=qw(Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Jan);
$n = $m == 1 ? 11:$m-2;print @mon[$n]' month_number

Regards!

...JRF...

[spex]

Oops... I missed the alpha part with my response above...

set -A mo Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec && echo ${mo[$(echo "((($(date +%m)-1)+11)%12)"|bc)]}

PCS

[Peter Nikitka]

Hi,

a simple ksh array solution:

set -A mo NULL Dec Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec
print ${mon[$(date +%m)]}

mfG Peter

[James R. Ferguson]

Hi (again):

I agree with Peter N. :

If you unconditionally want the previous month without providing a month number as an arguement, then my Perl solution can be reduced to:

# perl -le '@mon=qw(Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec);$n=(localtime)[4];$n = $n == 0 ? 11:$n-1;print @mon[$n]'

Regards!

...JRF...

[Hein van den Heuvel]

Along the same lines:

$ perl -e "print substr(qq(DecJanFebMarAprMayJunJulAugSepOctNov),3*(localtime)[4],3)"
Jan
$

Without listing the month names in perl, but much more work :-)

Not useful here, but the principle may help

perl -e "use Time::Local; @t=localtime; if($t[4]) { $t[4]-- } else {$t[5]--;$t[4]=12} $t=timelocal(@t); $m=(split(" ",scalar l
ocaltime($t)))[1]; print $m"

Jan


use Time::Local; Find source
@t=localtime; Now as array ($sec,$min,$hour,$mday,$mon,$year);
if($t[4]); Anything but Jan?
{ $t[4]-- } Just decrement the month
else {
$t[5]--;$t[4]=12} Month = 12
$t=timelocal(@t); Time in seconds
$m=(split(" ", 2nd word
scalar localtime($t)))[1]; Time as text
print $m" Finally!

Hein