topic Date arithmetic within shell in Operating System - HP-UX

Date arithmetic within shell

James A. Donovan — Thu, 11 Jan 2001 01:17:25 GMT

Can anyone help me?

How do I perform arithmetic on dates within a shell script? I need to set a variable equal to something like "Now - 6 days", and, as if that wasn't enough, the final format of this date variable needs to be "MM/DD/YY HH:MM:SS". This is because the variable gets passed as a parameter to an executable which expects it in that format.

TIA,

Jim

Re: Date arithmetic within shell

James R. Ferguson — Thu, 11 Jan 2001 01:50:47 GMT

Hi Jim:

A C-program. Perl, awk or a shell function -- any of which you develop is needed for date offsets greater than 24-hours.

See this thread for an awk program (Al Langen's contribution) that does what you want:

http://forums.itrc.hp.com/cm/QuestionAnswer/1,1150,0x57eff841489fd4118fef0090279cd0f9,00.html

For a very clever computation of a dates +- 24-hours using the TZ variable, follow Rich Garland's contributions through the discussion referenced in this thread:

http://forums.itrc.hp.com/cm/QuestionAnswer/1,1150,0xf9fd5f260cafd4118fef0090279cd0f9,00.html

Regards!

...JRF...

Re: Date arithmetic within shell

James A. Donovan — Thu, 11 Jan 2001 02:19:08 GMT

Thanks James! I too found Al's code just after posting my plea for help.

I've tweaked it slightly so that it can handle subtracting any number of days from the date passed into it.

x=`date +%x`
z=6;export z
y=`echo $x|awk whatever.awk`

Returns a valid value for any value of z, where z does not result in a date from prior to the previous year. That is, if x="01/10/01", a z>=376 will return an invalid date.

BEGIN{ FS="/"
split( "Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec", month," ");
split( "31 28 31 30 31 30 31 31 30 31 30 31", days, " ");
}
{
year = $3
leapyear = year % 4;
if( leapyear == 0 )
days[2] = 29
else
days[2] = 28;
doy = 0;
for ( i = 1; i< $1; i++ )
doy = doy + days[i]
doy = doy + $2;
if ( doy <= ENVIRON["z"] )
{
doy = doy + 365;
year = year - 1;
leapyear = year %4;
if( leapyear == 0 )
{
days[2] = 29;
doy = doy + 1;
}
}
doy = doy - ENVIRON["z"];
for ( i = 1; i <= 12; i++ )
{
doy = doy - days[i];
if ( doy <= 0 )
{
doy = doy + days[i];
break;
}
}
printf( "%d/%d/%d", i, doy, year )
}

Re: Date arithmetic within shell

David DeMartini_1 — Thu, 11 Jan 2001 17:24:50 GMT

There is a REAL simple way to do this.

Here is a PERL example:

use Time::Local;
$yy = `date +%y`;
$mm = `date +%m`;
$day = `date +%d`;
chomp($day);
chomp($yy);
chomp($mm);
$start = timelocal("00", "00", "00", $day, ($mm-1), $yy);
$pcurr = $start - (7*86400); # get day 7 days prior
$fcurr = $start + (7*86400); # get day 7 days in future
print "\nDate: ".localtime($fcurr); # or pcurr for past

# you can also grab the elements and apply them to some hashes for conversions if you like.
($sec,$min,$hour,$mday,$mon,$year,$wday,$yday,$dst)=localtime($fcurr); # or use pcurr if you want the past

Basically it is three steps.

1. Get the current date/time in seconds (easy on UNIX)
2. Add the number of days * 86400
3. Convert the seconds to human readable date/time.

Have fun!