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Shell Programming and Scripting

Date manipulation

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# 1  
Old 10-21-2011
Question Date manipulation
In my shell script I take date as a input parameter from command line in the format "21 Oct 2011" which would be
Code:
 date +'%d %b %Y'

Now i need to do two things here.
1) Validate the date entered by user
2) Calculate yesterday's date from the input. So in this case it should be: "20 Oct 2011"

Can someone please help me do this? I know I could do this quickly in perl or some other scripting lang, but here i only have bash shell scripting option

Thanks in advance.
Last edited by davidtd; 10-21-2011 at 06:16 PM.. Reason: spell correction
# 2  
Old 10-21-2011
Hi davidtd,

Try:
Code:
$ cat script.pl
use warnings;
use strict;
use POSIX;
use Time::Local;

@ARGV == 1 or die qq[Usage: perl $0 "date"\n];

my %TAB_MONTH = ( 
        jan     =>      0,
        feb     =>      1,
        mar     =>      2,
        apr     =>      3,
        may     =>      4,
        jun     =>      5,
        jul     =>      6,
        aug     =>      7,
        sep     =>      8,
        oct     =>      9,
        nov     =>      10,
        dec     =>      11
);

my ($mday, $month, $year) = split /\s+/, $ARGV[0];

timelocal( 0, 0, 0, $mday, $TAB_MONTH{ lc $month } || -1 , $year - 1900) or die qq[ERROR: Incorrect format of date\n];
my $utc_date = POSIX::mktime( 0, 0, 0, $mday, $TAB_MONTH{ lc $month }, $year - 1900 );
my $errno = POSIX::errno();
$utc_date -= (60 * 60 * 24);

(undef, undef, undef, $mday, $month, $year) = localtime $utc_date;
my %rTAB_MONTH = reverse %TAB_MONTH;
printf qq[%d %s %d\n], $mday, $rTAB_MONTH{ $month }, $year + 1900;
$ perl script.pl "21 Oct 2011"
20 oct 2011
$ perl script.pl "46 Oct 2011"
Day '46' out of range 1..31 at script.pl line 25
 $ perl script.pl "21 Gbd 2011"
Month '-1' out of range 0..11 at script.pl line 25

Regards,
Birei
# 3  
Old 10-22-2011
Java
Thanks, but this looks little complicated for my script... I think i will drop the validation requirement but can some one help me with Calculating yesterday's date from the input. So in this case how can i get previous days date in this case: "20 Oct 2011"
# 4  
Old 10-22-2011
maybe you can have a look at post #9 of :
https://www.unix.com/shell-programmin...zz-help-2.html
# 5  
Old 10-22-2011
Code:
$ uname -sr
SunOS 5.9
$ date +'%d %b %Y'
22 Oct 2011
$ TZ=CST+24 date +'%d %b %Y'
21 Oct 2011
$

# 6  
Old 10-22-2011
In some other country whose timezone differs from GMT, you should take into account the offset a reliable calculation could be something like




Code:
# -- calculate time using GMT -- 
Uday=`TZ=GMT date +%d`
Uhr=`TZ=GMT date +%H`
Lday=`date +%d`
Lhr=`date +%H`   
if [ $Lday = $Uday ]
then Offset=`expr $Uhr - $Lhr` 
else Offset=`expr $Uhr + 24 - $Lhr` 
fi   
Diff=`expr 24 + $Offset` echo -n "yesterday was " TZ=GMT+$Diff date +%Y-%m-%d   
Diff=`expr 24 - $Offset` echo -n "tomorrow will be " TZ=GMT-$Diff date +%Y-%m-%d

# 7  
Old 10-24-2011
Thanks all for some good tips. But here is how i resolved my problem.
My issue was, date is in the format like: "24 Oct 2011" which is "date +'%d %b %Y'" in shell script, and I wanted to find the previous day which should be : "23 Oct 2011". Below is the code I executed to get this result.

Code:
input="24 Oct 2011"
temp1=$(echo $input | awk '{print $1}')
temp2=$(expr $temp1 - 1)
temp3=$(echo $input | awk '{print $2,$3}')
preDay=$temp2" "$temp3

Resulting output is:
Code:
23 Oct 2011
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