date (how to get date of yesterday)

[Manuales]

Hi ..
if i run date i get:

$ a=`date "+%b %d"`
$ echo $a
Apr 30


how can i get one day before?
i mean, i need to get the value of yesterdad (according to current date) to keep it in a variable ...

in this case, today is "Apr 30"
I need get "Apr 30"

if today were "May 1" , i woould need to get "Apr 30" ..

how can i do that

plase let me know :)

Thanks in Advance.

[Yogeswar · Accepted solution]

#!/bin/ksh
# /usr/local/bin/when
# usage "when 7" to go back 7 days, "when 14" to go back 14, etc.
# useless to go back more than a month
set -A MONTHS Dec Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec
BACK=$1
THEN=$((`date +%d` - $BACK))
MONTH=`date +%m`
YEAR=`date +%Y`
if [ $THEN -le "0" ]
then
MONTH=$((MONTH-1))
if [ $MONTH -eq "0" ]
then
MONTH=12
YEAR=$((YEAR-1))
fi
set `cal $MONTH $YEAR`
SHIFT=$(( $THEN * -1 + 1 ))
shift $(($# - $SHIFT))
THEN=$1
fi
TMONTH=${MONTHS[MONTH]}
# uncomment next line for debugging
echo $THEN $MONTH $TMONTH $YEAR

[Manuales]

IT does not work :(



$ cat > fecha
#!/bin/ksh
# /usr/local/bin/when
# usage "when 7" to go back 7 days, "when 14" to go back 14, etc.
# useless to go back more than a month
set -A MONTHS Dec Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec
BACK=$1
THEN=$((`date +%d` - $BACK))
MONTH=`date +%m`
YEAR=`date +%Y`
if [ $THEN -le "0" ]
then
MONTH=$((MONTH-1))
if [ $MONTH -eq "0" ]
then
MONTH=12
YEAR=$((YEAR-1))
fi
set `cal $MONTH $YEAR`
SHIFT=$(( $THEN * -1 + 1 ))
shift $(($# - $SHIFT))
THEN=$1
fi
TMONTH=${MONTHS[MONTH]}
# uncomment next line for debugging
echo $THEN $MONTH $TMONTH $YEAR

$ ./fecha
sh: ./fecha: Execute permission denied.
$ chmod 755 fe*
$ ./fecha
./fecha[7]: 30 - : more tokens expected
$

[Manuales]

i did several tests and i does not work:

$ ./fecha
sh: ./fecha: Execute permission denied.
$ chmod 755 fe*
$ ./fecha
./fecha[7]: 30 - : more tokens expected
$ fecha `date`
./fecha[7]: 30 - Wed: bad number
$ fecha `date "+%b %d"`
./fecha[7]: 30 - Apr: bad number
$ fecha `date`
./fecha[7]: 30 - Wed: bad number
$

[James R. Ferguson]

Hi Manuales:

use Perl;

# perl -le 'print scalar localtime time - 86400'

# perl -MPOSIX -le 'print strftime "%m/%d/%Y",localtime(time-86400)'

If you use the second snippet, you can use the same formatting directives you use with the Unix 'date' command; e.g.:

# perl -MPOSIX -le 'print strftime "%b %d",localtime(time-86400)'
Apr 29

Regards!

...JRF...

[Manuales]

thanks a lot Krishna and James !!

$ perl -MPOSIX -le 'print strftime "%b %d",localtime(time-86400)'
Apr 29
$ perl -MPOSIX -le 'print strftime "%m/%d/%Y",localtime(time-86400)'
04/29/2008
$ perl -le 'print scalar localtime time - 86400'
Tue Apr 29 18:42:31 2008

[Yogeswar]

Try this One!

#! /usr/bin/ksh
# Get yesterday's date in YYYY-MM-DD format.
# With argument N in range 1..28 gets date N days before.




OFFSET=${1:-1}



case $OFFSET in
*[!0-9]* | ???* | 3? | 29) print -u2 "Invalid input" ; exit 1;;
esac



eval `date "+day=%d; month=%m; year=%Y`
typeset -Z2 day month
typeset -Z4 year



# Subtract offset from day, if it goes below one use 'cal'
# to determine the number of days in the previous month.
day=$((day - OFFSET))
if (( day <= 0 )) ;then
month=$((month - 1))
if (( month == 0 )) ;then
year=$((year - 1))
month=12
fi
set -A days `cal $month $year`
xday=${days[$(( ${#days[*]}-1 ))]}
day=$((xday + day))
fi



print $year-$month-$day
print $month/$day/${year#??}