Tomorrow's date - Dev Shed

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November 30th, 2006, 05:58 AM

nevil_m

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Tomorrow's date

What syntax is used to get tomorrows date. I am using HP unix ver 11.0

November 30th, 2006, 08:03 AM

SimonJM

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Are you just using shell scripting? And what aprt of the date are you after - dd/mm/yy or day name, etc.?

November 30th, 2006, 08:17 AM

suresh_kumarmp

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Try to use "calendar - reminder service"

I am not pretty sure :-)

Suresh

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November 30th, 2006, 08:41 AM

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You could install the GNU-coreutils on HP and use the date command with the --tomorrow switch or something to that effect.

Personally though I'd write a small program to do the job.

Code:

#include <time.h>
#include <stdio.h>

#define SECOND 60
#define MINUTE 60
#define HOUR 24

int main(int argc, char ** argv) {
        time_t today = time(NULL);
        time_t tomorrow = today + (time_t)(HOUR * MINUTE * SECOND);
        struct tm tim = *(localtime(&tomorrow));

        char date_tomorrow[100];
        strftime(date_tomorrow, 100, argv[1], &tim);
        printf("%s\n", date_tomorrow);
}

Save as date.c and compile with:

Code:

gcc date.c -o d

Code:

$ ./d
Fri Dec 01 14:35:05 2006
$ ./d '%Y-%m-%d %H:%M:%S'
2006-12-01 14:35:06

# as a constrast

$ date
Thursday November 30 14:35:08 GMT 2006
$ date '+%Y-%m-%d %H:%M:%S'
2006-11-30 14:35:15
$

November 30th, 2006, 02:08 PM

jim mcnamara

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Code:

#!/bin/ksh
tomorrow()
{
  perl -e '
     # today plus 86400 seconds is this time tomorrow
         $tomorrow = time + 86400;

         # this line gets the month name
         $month = (Jan,Feb,Mar,Apr,May,Jun,Jul,Aug,Sep,Oct,Nov,Dec)[(localtime $yesterday)[4]];

         # day of the month
         $day = (localtime $tomorrow)[3];
         $yr= (localtime $tomorrow)[5] + 1900;
         print "$month $day $yr", "\n";
  '
}

echo "Day after today is $( tomorrow )"

One way is to use a perl (or some other similarly-featured interpreted language like python) and create a small subroutine. I picked perl because most unix boxes have it.

November 30th, 2006, 11:09 PM

nevil_m

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Quote:

Originally Posted by SimonJM

Are you just using shell scripting? And what aprt of the date are you after - dd/mm/yy or day name, etc.?

Yes using shell scripting. Format requuired id dd/mm/yyyy

November 30th, 2006, 11:30 PM

nevil_m

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Quote:

Originally Posted by iota

You could install the GNU-coreutils on HP and use the date command with the --tomorrow switch or something to that effect.

Personally though I'd write a small program to do the job.

Code:

#include <time.h>
#include <stdio.h>

#define SECOND 60
#define MINUTE 60
#define HOUR 24

int main(int argc, char ** argv) {
        time_t today = time(NULL);
        time_t tomorrow = today + (time_t)(HOUR * MINUTE * SECOND);
        struct tm tim = *(localtime(&tomorrow));

        char date_tomorrow[100];
        strftime(date_tomorrow, 100, argv[1], &tim);
        printf("%s\n", date_tomorrow);
}

Save as date.c and compile with:

Code:

gcc date.c -o d

Code:

$ ./d
Fri Dec 01 14:35:05 2006
$ ./d '%Y-%m-%d %H:%M:%S'
2006-12-01 14:35:06

# as a constrast

$ date
Thursday November 30 14:35:08 GMT 2006
$ date '+%Y-%m-%d %H:%M:%S'
2006-11-30 14:35:15
$

I get an error while compiling as

(Bundled) cc: "date.c", line 8: error 1705: Function prototypes are an ANSI feature.
(Bundled) cc: "date.c", line 11: error 1716: Automatic aggregate initialization is an ANSI feature.

November 30th, 2006, 11:35 PM

nevil_m

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It did help

Quote:

Originally Posted by jim mcnamara

Code:

#!/bin/ksh
tomorrow()
{
  perl -e '
     # today plus 86400 seconds is this time tomorrow
         $tomorrow = time + 86400;

         # this line gets the month name
         $month = (Jan,Feb,Mar,Apr,May,Jun,Jul,Aug,Sep,Oct,Nov,Dec)[(localtime $yesterday)[4]];

         # day of the month
         $day = (localtime $tomorrow)[3];
         $yr= (localtime $tomorrow)[5] + 1900;
         print "$month $day $yr", "\n";
  '
}

echo "Day after today is $( tomorrow )"

One way is to use a perl (or some other similarly-featured interpreted language like python) and create a small subroutine. I picked perl because most unix boxes have it.

It did work. But month is appearing as Jan. Even the month is incrementing by 1. Can we get o/p in dd/mm/yyyy format ?

December 1st, 2006, 07:47 AM

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Quote:

Originally Posted by nevil_m

Hmm.. From what I've been reading on Google, HP's bundled compiler isn't ANSI compliant and is only used for compiling kernels. You'll need to use GCC or HP's ANSI C.

#10

December 1st, 2006, 09:40 AM

jim mcnamara

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The default HP compiler is basically K&R. It's marketing ploy to get you to buy a modern compiler...

Code:

#!/bin/ksh
tomorrow()
{
  perl -e '
     # today plus 86400 seconds is this time tomorrow
         $tomorrow = time + 86400;

         # this line gets the month name
                   
         $month = (localtime $tomorrow)[4];
         $month = (++$month < 10) ? "0$month" : "$month";         

         # day of the month
         $day = (localtime $tomorrow)[3];
         $day = ($day < 10) ? "0$day" : "$day";
         $yr= (localtime $tomorrow)[5] + 1900;
         printf "%s/%s/%s\n", $month, $day, $yr;
  '
}

echo "Day after today is $( tomorrow )"

This outputs 12/02/2006 on December 1, 2006

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