convert date variable to doy variable


 
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# 1  
Old 12-16-2008
convert date variable to doy variable

Hello:

I have a script that requires the use of DOY (as a variable) instead of YY/MM/DD. I already obtained these parameters from downloaded files and stored them as variables (i.e. $day, $month, $year).

Is there any simple script that I could incorporate in mine to do this job?

Regards,

Antonio
# 2  
Old 12-16-2008
Code:
date '+%j'

# 3  
Old 12-16-2008
To convert a string to date and output the "Julian" date use the following:

Code:
 # date -d "2008-01-02" +%j
002

This is assuming your using GNU date.
# 4  
Old 12-16-2008
Just one more question. I was able to get the DOY. I included it in my script by making the following:

$comm = "date -d $year-$month-$day +%j";
system("$comm");

For 2008/12/16 the scripts prints in the screen 351.

However, at the end of the day I need to capture the answer, 351 for instance, as a variable. I am sorry but I am new at linux.

Regards,
# 5  
Old 12-16-2008
Shell:
Code:
DOY=`date -d $year-$month-$day +%j`

Perl:
Code:
$DOY=`date -d $year-$month-$day +%j`;


Last edited by Ikon; 12-16-2008 at 06:35 PM..
# 6  
Old 12-16-2008
Is this a Perl script? Sounds like it if you are using the system command.

Code:
$ cat ./date_test.pl
#!/usr/bin/perl

use strict;
use warnings;
use POSIX qw(strftime);
use Time::Local;

#If you just want the julian of the current system date
my $x;

$x = strftime("%j", localtime());

print "x = $x\n";

#If you want the julian of a specific date, let's say 2008-12-10
my $testdate;
my $yyyy = 2008;
my $mm   = 12;
my $dd   = 10;

$testdate = timelocal(0, 0, 0, $dd, $mm-1, $yyyy);
$x = strftime("%j", localtime($testdate));
print "x = $x\n";

exit 0;


$ ./date_test.pl
x = 351
x = 345

$ date -d "2008-12-16" +%j
351

$ date -d "2008-12-10" +%j
345

# 7  
Old 12-17-2008
If you are using ksh93, support for date calculations and conversions is built in.
Code:
$ doy=$(printf "%(%j)T" "2008-12-16")
$ print $doy
351
$ doy=$(printf "%(%j)T")
$ print $doy
352
$

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