I have 2 k-shell variables each holding a date in the form "MM/DD/YYY
hh-mm-ss".
I want to find out how many seconds elapsed between these dates.
Other than k-shell, I can use sed, awk, perl
Can somebody tell me how I can go about with this problem.

Thanks
--sony

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sony.antony@gmail.com writes:
> I have 2 k-shell variables each holding a date in the form "MM/DD/YYY
> hh-mm-ss".
> I want to find out how many seconds elapsed between these dates.
> Other than k-shell, I can use sed, awk, perl
> Can somebody tell me how I can go about with this problem.

If you have GNU date, you can convert each date to a number of seconds
since 1970, *if* it's in a format that GNU date recognizes:

date +%s -d '01/17/2006 18:53:17'

If the format is really "MM/DD/YYY hh-mm-ss", then you have a couple
of problems: you'll need 4 digits for the year (I assume that was a
typo), and you'll want "hh:mm:ss" rather than "hh-mm-ss" (easily done
with sed).

Note that GNU date assumes 01/02/2003 is January 2, not February 1;
that matches your assumption, but it's not universal.

Or you could do the whole thing fairly easily in Perl.  Just parse the
format using a regular expression, extract the fields, adjust as
needed, and use the Time::Local module to convert to a raw timestamp.
For more information, see the PERL documentation; failing that, try
comp.lang.perl.misc.

--
Keith Thompson (The_Other_Keith) kst-u@mib.org  <http://www.ghoti.net/~kst>
San Diego Supercomputer Center             <*>  <http://users.sdsc.edu/~kst>
We must do something.  This is something.  Therefore, we must do this.

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On 2006-01-18, Keith Thompson wrote:
> sony.antony@gmail.com writes: 
>
> If you have GNU date, you can convert each date to a number of seconds
> since 1970, *if* it's in a format that GNU date recognizes:
>
>     date +%s -d '01/17/2006 18:53:17'
>
> If the format is really "MM/DD/YYY hh-mm-ss", then you have a couple
> of problems: you'll need 4 digits for the year (I assume that was a
> typo), and you'll want "hh:mm:ss" rather than "hh-mm-ss" (easily done
> with sed).

Or with the shell:

DATEVAR="01/11/2006 12-06-44"
IFS=' /:-'
set -f
set -- $DATEVAR

DV="$3-$1-$2 $4:$5:$6"
IFS=$' \t\n' ## bash/ksh93; sh: IFS=`printf " \t\n"`

## Using GNU date:
SECS=`date -d "$DV" +%s`


> Note that GNU date assumes 01/02/2003 is January 2, not February 1;
> that matches your assumption, but it's not universal.
>
> Or you could do the whole thing fairly easily in Perl.  Just parse the
> format using a regular expression, extract the fields, adjust as
> needed, and use the Time::Local module to convert to a raw timestamp.
> For more information, see the PERL documentation; failing that, try
> comp.lang.perl.misc.

For more information on date arithmetic, see the comp.unix.shell
FAQ <http://home.comcast.net/~j.p.h/cus-faq.html#6>, or see Chapter
8 of my book (see the URL in my .sig; that chapter is available
on-line).

--
Chris F.A. Johnson, author   |    <http://cfaj.freeshell.org>
Shell Scripting Recipes:     |  My code in this post, if any,
A Problem-Solution Approach  |          is released under the
2005, Apress                 |     GNU General Public Licence

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Thanks for the prompt responses.
Unfortunately Im on Solaris and the date does not have the -d option.
Im almost certain that I will have to use perl. But just wanted to ask
you guys before I start on that.
--sony

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sony.antony@gmail.com wrote:
> I have 2 k-shell variables each holding a date in the form "MM/DD/YYY
> hh-mm-ss".
> I want to find out how many seconds elapsed between these dates.
> Other than k-shell, I can use sed, awk, perl
> Can somebody tell me how I can go about with this problem.
>
> Thanks
> --sony

echo '01/13/2006 18-50-50 01/13/2006 18-50-55'|ruby -F'[ /-]' -nae'
p Time.local(*$F.values_at(8,6,7,9,10,11)) -
Time.local(*$F.values_at(2,0,1,3,4,5))'

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sony.antony@gmail.com writes:
> Thanks for the prompt responses.
> Unfortunately Im on Solaris and the date does not have the -d option.
> Im almost certain that I will have to use perl. But just wanted to ask
> you guys before I start on that.

Or you can install GNU date; it's part of the "coreutils" package.

I don't remember whether awk can do this.  (For Solaris, use "nawk";
the default "awk" is still an ancient version.)

--
Keith Thompson (The_Other_Keith) kst-u@mib.org  <http://www.ghoti.net/~kst>
San Diego Supercomputer Center             <*>  <http://users.sdsc.edu/~kst>
We must do something.  This is something.  Therefore, we must do this.

[ Post a follow-up to this message ]

Keith Thompson wrote:

> sony.antony@gmail.com writes:
> 
>
>
> Or you can install GNU date; it's part of the "coreutils" package.
>
> I don't remember whether awk can do this.

GNU awk can, but if he could get GNU awk, then you wouldn't think
getting GNU date would be a challenge either. Personally, I'd install
GNU awk since it'll be useful for future problems too.

(For Solaris, use "nawk";

or /usr/xpg4/bin/awk

> the default "awk" is still an ancient version.)
>

Right - affectionately known as "old, broken awk".

For the OP, I think you already got a couple of answers to your
question, but if not you might want to look at question 6 in the FAQ
(http://home.comcast.net/~j.p.h/cus-faq.html#6).

Ed.

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In article <1137548734.668168.268940@z14g2000cwz.googlegroups.com>,
<sony.antony@gmail.com> wrote:
>I have 2 k-shell variables each holding a date in the form "MM/DD/YYY
>hh-mm-ss".
>I want to find out how many seconds elapsed between these dates.
>Other than k-shell, I can use sed, awk, perl
>Can somebody tell me how I can go about with this problem.

If your ksh is recent enough, it understands dates pretty close to that form
:

$ printf "%T\n" "06/01/1995 11:35:20"
Thu Jun  1 11:35:20 1995

So:

$ newdate="01/18/2006 13-28-27" olddate="06/01/1995 11-35-20"
$ print $(( $(printf "%(%s)T" "${newdate//-/:}") - $(printf "%(%s)T" "$
{olddate//-/:}") ))
335587987

John
--
John DuBois  spcecdt@armory.com  KC6QKZ/AE  http://www.armory.com/~spcecdt/

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