Script with input date, output day

Question

I want to write a bash script that reads a keyboard input (date dd:mm:yy), stores it in a variable then echoes back the Day it corresponds to, ie

Input : 03/08/2015

Output : Wednesday

I'm having some problems with formatting this, any help is much appreciated!!

Answer 1

date accepts input with its -d flag, and prints it. However, input formatted like yours is treated like MM/DD/YYYY. But this shouldn't be too hard.

First, let's get input to variable:

read INPUT

Now, we need to change input from DD/MM/YYYY to MM/DD/YYYY. You can do that with sed and awk(you can also do that with awk:

INPUT=`echo $INPUT | awk -F '/' '{t=$1;$1=$2;$2=t;gsub(" ", "/");print;}'`

Now that INPUT is in proper format, feed it to date and make it only print the day:

date -d $INPUT +%A

Put it all together:

#!/bin/bash
read INPUT
INPUT=`echo $INPUT | awk -F '/' '{t=$1;$1=$2;$2=t;gsub(" ", "/");print;}'`
date -d $INPUT +%A

Answer 2

Perl has some nice built-in datetime modules:

$ perl -MTime::Piece -E '
     print "Input date (dd/mm/YYYY): ";
     chomp( $date = <> );
     $datetime = Time::Piece->strptime($date, "%d/%m/%Y");
     say $datetime->strftime("%e %B is a %A");
'
Input date (dd/mm/YYYY): 3/8/2015
 3 August is a Monday