Shell Programming and Scripting

Wont work on last day of month:

Code:

 # date -d "2008-01-31" | awk '{print $3 + 1,$2,$6}'
32 Jan 2008

ohhhhh no

Ya, you are right. I missed out that.

thanks.

I am not sure why the solutions given by 'Ikon' is not working. if i execute the script, it's returning the same date

#!/bin/ksh/
#myvar=`date '+%d %b %Y' --date="tomorrow"`
#myvar=`date '+%d %b %Y' --date='-1 days ago'`
myvar=`date '+%d %b %Y' --date="next day"`
echo $myvar

output is 09 JAN 2009

Also, if I used solution of 'SFYNC', i am getting following error

date: illegal option -- d
usage: date [-u] mmddHHMM[[cc]yy][.SS]
date [-u] [+format]
date -a [-]sss[.fff]

Please advise.

You are not using gnu date, your version does not support --date.

Try this:

Code:

TZ=EDT-24 date +"%d %b %Y"

Hi Ikon,

I am getting below output. How can I remove 'US/Eastern; word from the next line. Also could you please explain the code ?

TZ=EDT-24 date +"%d %b %Y"
echo $TZ

OUTPUT

10 Jan 2009
US/Eastern

It works kinda weird, IMO.

TZ=EDT-24 adds 24hours to the time
TZ=EDT+24 Removed 24hours
You probabally want to change TZ to what your time zone actually is.

I dont fully understand how it works. Because +1 doesnt add or subtract 1 hour.

Hi Ikon,

Actually I don't want time zone itself. I need only 'dd mon yyyy'