How do I derive a new date from a given date plus or minus a number of days in unix. For example:

25-dec-2003 + 4days to give me 29-dec-2003

Thank you very for your assistance

Unix machines keep time by storing the number of seconds from the epoch as a 32-bit integer. I would determine the current time (it's now 1073657881 o'clock), add the required number of seconds and convert back. Whilst it's not strictly a shell script [1], this should do it:

#!/path/to/perl -w
# Takes one argument, the number of days' offset
defined $ARGV[0] || exit 1;
$ARGV[0] =~ /^-?\d+$/ || exit 2;
$DBFORUMS=time;
$DBFORUMS+=($ARGV[0]*60*60*24);
print localtime($DBFORUMS) . "\n";
__END__

There's also a GNU extension to 'date' which returns the value of the number of seconds past the epoch. Depending on your system, you may have this utility and you can write a similar script in the shell.


[1] I have considered running the perl debugger as my login shell

Hi,

Please test this form

a=4 ; expr `date +%d` + $a

Depending on as you need use.

Gustavo.

Hi,
I have a little programm that do what you ask for. (infdate.x)
If you are intrested post a privat message. I can send it to you by Mail

DAYS_TO_BE_ADD=4 # or back then use -4

AKT_DATE=`date '+%d.%m.%Y'`
INFORMIX_DATE=`infdate.x -s $AKT_DATE`
INFORMIX_DATE=`expr $INFORMIX_DATE + $DAYS_TO_BE_ADD`
NEW_DATE=`infdate.x -s $INFORMIX_DATE`
echo "today ==> "$AKT_DATE
echo "new day ==> "$NEW_DATE

And this is the output
--------------------------------
today ==> 09.01.2004
new day ==> 13.01.2004

Hi,

I would like to use yopur small utility for date manipulation in Unix shell scripts. Please send me your program, if you dont mind.
My e-mail address: abhandari@deloitte.com

Thank you so much,

Arun Bhandari.