Hi guys,

I need to find tomorrows date in date fomat and should be in variable.
as I need to grep this date in a flat file

ie. if today's date is '09 JAN 2009'

output should be '10 JAN 2009'

unix/perl script will be fine.

date | awk '{print $1,$2,$3 + 1,$4}'


From....


simon@cartman:~$ date
Fri Jan 9 15:09:45 GMT 2009
simon@cartman:~$ date | awk '{print $1,$2,$3 + 1,$4}'
Fri Jan 10 15:09:46
simon@cartman:~$

Just realised....You would of course have to wrap this in a if statement to ensure the dates, months, years roll.

A better way would be to increment the unix time stamp by the correct integer (seconds in 24 hours), and then convert to the normal date.

Thanks for your quick reply

I am gettting the output as

Fri Jan 10 10:16:39 -- 10th Jan is SAT

but i need

10 Jan 2009 format.

If you have GNU date:

Hello All,

'snoop2048' solutions is working after some changes.

`date | awk '{print $3 + 1,$2,$6}'`

Thank you all for your quick response. Appreciated !!!

Have a great Weekend !!!

ohhhhh no

Ya, you are right. I missed out that.

thanks.