Take input arguments and pass them to date

Question

So I'm trying to write a shell script to take an input as below:

bash whichday Jan 10 2017

and it should take that input and print out what day it occurs on. Below is the script I'm trying to use:

#!/bin/bash
if [ $# -ne 3 ]
    then
    echo "Please enter the month day year (example: Jan 5 2002)"

else
    str=\"$1" "$2" "$3\"
    echo $str
    date -d $str +%A
fi 

but everytime I run this, I get the following output:

"Jan 10 2017"
date: extra operand ‘2017"’
Try 'date --help' for more information.

When I do

date -d "Jan 10 2017" +"%A"

on command line, it outputs the correct day (Tuesday), but the script, which should be executing the exact same command, doesn't work. Please help, I don't understand why this is happening.

Answer 1

Just set

str="$1 $2 $3"

and then

date -d "$str" +'%A'

By escaping the double quotes, you making them part of the value of $str, and date can't parse them.

You may also bypass the str variable completely by using $*:

echo "$*"
date -d "$*" +'%A'