Date Arithmetic In Linux Shell Scripts

How to write a Linux shell scripts that backup previous month data files on monthly basis? Of course, it involves date time calculation! Unless you’re using GNU Coreutils / GNU date, doing date arithmetic in shell scripts will not be that straight-forward as a novice think!

I just can’t recall that SCO Unix and Concurrent Real Time Unix (RTU) have GNU date installed, when I was exposed to Unix for the first time in year 2000. But, I believe that GNU Coreutils is a standard package that available on most Linux distributions as of today.

How to do date arithmetic in Linux shell scripts using GNU date command?


GNU date is one of the many common Linux programs in GNU Coreutils package. Here are some of the examples of performing date time arithmetic with GNU date command, using -d or --date option switch:

date -d "last month" +%m

Return as last month in numeric format

date -d "2 month ago" +%b

Return as last month in abbreviated month name

date -d "-2 month" +%B

Return as last month in full month name

date -d "-1 hours" +%H

Returns previous hour in 24-hours format (00-23)

date -d "tomorrow"

Return the date and time of tomorrow

date -d "yesterday" +"%d %B %y"

Return the date of yesterday in dd MMM yy

date -d "last week" +%a

Return last week in abbreviated weekday name

date -d "$(date +%Y-%m-15) -1 month" +'Last month was %B!'

To avoid getting result of “July” when executing date -d "-1 month" +%B on 31 July 2006, because “2006-07-31 -1 month” might evaluate to 2006-07-01, because 2006-06-31 is an invalid date!

date -d "1970-01-01 00:00:01 UTC" +%s

Return the Unix Epoch time (the total seconds elapse since midnight of 1970-01-01), which is 1 secs in this case.

date -d "1970-01-01 1234567890 sec" +"%F %T %z"

Return the literal date format of the given Unix Epoch time.

Isn’t it easier for date time calculation related shell scripts with GNU date installed? As said just now, GNU date is not a standard command in Unix variants.

In case GNU date is not available, you might need to write a rather long shell scripts to accommodate some simple date time calculations. Potentially, it could end up as a messy or hard to maintain shell scripts!

In case GNU Coreutils / GNU Date is not installed, you can write a simple C program in few lines of source code to satisfy the date arithmetic in shell scripts! The compiled C program is highly portable among Unix / Linux variants, as the date arithmetic library (time.h) is a standard C library!


How to write and compile a simple date arithmetic C program?

To compile the C program source code using GCC compiler, the general syntax for these simple source code is

gcc -o program_name program_source_code.c

For C/C++ tutorial or standard C/C++ library reference, I’m referring to cplusplus.com

Calculate the nth-day from today


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

#define c_DaySecs 86400

int main (int argc, char ** argv)
{
   int Day;
   time_t TotSec;
   struct tm * TimeStruct;

   if ( argc != 2 )
   {
     printf("Syntax\n\t%s nth-Day\n",argv[0]);
     return 1;
   }
   else
   {
     Day = atoi( argv[1] );
     TotSec = time( NULL );

     TotSec = TotSec + ( Day * c_DaySecs );
     TimeStruct = localtime( &TotSec );

     printf("%i days is %s",Day,ctime(&TotSec));
     printf("%i days is %i-%i-%i\n", \
     Day, TimeStruct->tm_mday, \
     TimeStruct->tm_mon + 1, \
     TimeStruct->tm_year + 1900 );

     TimeStruct = NULL;

     return 0;
   }
}


Return the literal date of a given Unix Epoch time


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

int main (int argc, char ** argv)
{
   time_t TimeNow;
   struct tm * TimeStruct;

   if ( argc < 2 || argc > 3)
   {
      printf("Syntax\n\t%s SEC FLAG\n",argv[0]);
      return 1;
   }
   else
   {
      TimeNow = atoi( argv[1] );

      if (argv[2] == NULL || atoi( argv[2] ) == 0)
      {
         TimeStruct = localtime( &TimeNow );
         printf( "%s in Local Timezone is %s", \
            argv[1], asctime( TimeStruct ) );
      }
      else
      {
         TimeStruct = gmtime( &TimeNow );
         printf( "%s in GMT is %s", \
            argv[1], asctime( TimeStruct ) );
      }

      return 0;
   }
}


To convert a give literal date to Unix Epoch Time (total seconds elapsed since midnight of 1970-01-01)


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

int main (int argc, char ** argv)
{
   int DD, MM, YYYY = 0;
   time_t TimeNow;
   struct tm * TimeStruct;

   if ( argc != 4 )
   {
      printf("Syntax\n\t%s DD MM YYYY\n",argv[0]);
      return 1;
   }
   else
   {
      DD = atoi( argv[1] );
      MM = atoi( argv[2] );
      YYYY = atoi( argv[3] );
      TimeNow = time( NULL );
      TimeStruct = localtime( &TimeNow );

      TimeStruct->tm_year = YYYY - 1900;
      TimeStruct->tm_mon = MM - 1;
      TimeStruct->tm_mday = DD;

      TimeNow = mktime( TimeStruct );

      printf( "%ld\n", TimeNow );
      return 0;
   }
}


Howto, Linux, Arithmetic, Coreutils, Date, Epoch, Linux, Shell Script, Time


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