Function to get previous business day given a date in Linux - linux

Question

Given an input date, I want to write a bash function that will output the previous business day. 
By this I mean the preceding weekday (Monday through Friday);
I don't need it to take holidays into account. 
So, for example, given "Jan 2, 2018" the result should be "Jan 1, 2018"
(even though that is a holiday),
but given "Jan 1, 2018" the result should be "Dec 29, 2017"
(because Dec 30 and 31 were Saturday and Sunday). 
I don't require any particular format;
just something that is human-readable and acceptable to date -d.
I have tried the following but the input date does not seem to be correctly taken into account:
function get_previous_busday()
{
DAY_OF_WEEK=`$1 +%w`
if [ $DAY_OF_WEEK -eq 0 ] ; then
LOOKBACK=-2
elif [ $DAY_OF_WEEK -eq 1 ] ; then
LOOKBACK=-3
else
LOOKBACK=-1
fi
PREVDATE=date -d "$1 $LOOKBACK day"
}
I want to apply it for today:
PREVDATE=$(get_previous_busday $(date))
echo $PREVDATE
and for yesterday:
PREVDATE=$(get_previous_busday (date -d "$(date) -1 day"))
echo $PREVDATE
But it is not working:
main.sh: line 3: Fri: command not found
main.sh: line 4: [: -eq: unary operator expected
main.sh: line 6: [: -eq: unary operator expected
main.sh: line 11: -d: command not found
main.sh: command substitution: line 20: syntax error near unexpected token `date'
main.sh: command substitution: line 20: `get_previous_busday (date -d "$(date) -1 day"))'

Answer 1

A function to do what you want is:
get_previous_busday() {
if [ "$1" = "" ]
then
printf 'Usage: get_previous_busday (base_date)\n' >&2
return 1
fi
base_date="$1"
if ! day_of_week="$(date -d "$base_date" +%u)"
then
printf 'Apparently "%s" was not a valid date.\n' "$base_date" >&2
return 2
fi
case "$day_of_week" in
(0|7) # Sunday should be 7, but apparently some people
# expect it to be 0.
offset=-2 # Subtract 2 from Sunday to get Friday.
;;
(1) offset=-3 # Subtract 3 from Monday to get Friday.
;;
(*) offset=-1 # For all other days, just go back one day.
esac
if ! prev_date="$(date -d "$base_date $offset day")"
then
printf 'Error calculating $(date -d "%s").\n' "$base_date $offset day"
return 3
fi
printf '%s\n' "$prev_date"
}
For example,
$ get_previous_busday
Usage: get_previous_date (base_date)
$ get_previous_busday foo
date: invalid date ‘foo’
Apparently "foo" was not a valid date.
$ get_previous_busday today
Fri, Nov 30, 2018 1:52:15 AM
$ get_previous_busday "$(date)"
Fri, Nov 30, 2018 1:52:51 AM
$ PREVDATE=$(get_previous_busday $(date))
$ echo "$PREVDATE"
Fri, Nov 30, 2018 12:00:00 AM
$ get_previous_busday "$PREVDATE"
Thu, Nov 29, 2018 12:00:00 AM
$ PREVPREVDATE=$(get_previous_busday "$PREVDATE")
$ printf '%s\n' "$PREVPREVDATE"
Thu, Nov 29, 2018 12:00:00 AM
$ get_previous_busday "$PREVPREVDATE"
Wed, Nov 28, 2018 12:00:00 AM

Answer 2

There are a couple of ways to get the offset needed to get back to a business day.
You can write a case statement:
case $dow in
0|7) backday=2;; # For Sunday (either named 0 or 7) go back 2 days
1) backday=3;; # For monday go back three (3) days.
*) backday=1;; # For the rest, just one day.
esac
You can use math:
backday=$(( ((dow%7)>1) ? 1 : (dow%7)+2 ))
Or a lookup array:
a=(0 1 2 3 4 5 6 7)
b=(2 3 1 1 1 1 1 2)
backday=${b[dow]}
For any alternative, you may use this (without error detection) function,
and some tests:
#!/bin/bash
get_previous_busday() { dow=$(date -d "$*" '+%w')
backday=$(( ((dow%7)>1) ? 1 : (dow%7)+2 ))
prev_date="$(date -d "$* -$backday day")"
printf '%s\n' "$prev_date"
}
get_previous_busday "$(date)"
get_previous_busday $(date -d "-1 day")
get_previous_busday $(date -d "-2 day")
get_previous_busday $(date -d "-3 day")
get_previous_busday $(date -d "-4 day")
Will print:
Fri Nov 30 10:03:45 UTC 2018
Fri Nov 30 10:03:45 UTC 2018
Fri Nov 30 10:03:45 UTC 2018
Thu Nov 29 10:03:45 UTC 2018
Wed Nov 28 10:03:45 UTC 2018