unix shell script to get nth business day - shell

Question

Referencing the solution posted on this unix.com thread for getting the Nth business day of the month, I tried to get the 16th business day of the month using the following code, but it doesn't work.
currCal=`/usr/bin/cal`
BUSINESS_DAYS=`echo $($currCal|nawk 'NR>2 {print substr($0,4,14)}' |tr "\n" " ")`
The error when executing this is:
nawk: syntax error at source line 1 context is
NR>2 {print >>> substr(test. <<< sh,4,14)}
nawk: illegal statement at source line 1
I'm guessing it takes $0 as the script name, causing the syntax error. Please help.

Answer 1

There seem to be a few issues with what you have above.
First, I agree with #John1024 that in order to get the nawk error you've posted, you must actually be running:
BUSINESS_DAYS=`echo $($currCal|nawk "NR>2 {print substr($0,4,14)}" |tr "\n" " ")`
with double quotes around the nawk script.
Furthermore, once you resolve the nawk error, you're going to run into issues with how you are using currCal. You get the actual output of the cal command into the currCal variable, but then are using the variable value (that is the output of cal) as a command before the | rather than echoing it into the pipe or something similar.
This brings up an additional question of why you're using echo on the result of a subshell command (the $() part) within another subshell (the outer ``s).
Finally, the two lines you show above only get a list of the business days in the current month into the BUSINESS_DAYS variable. They do not output/save the 16th such day.
Taking all of the above into consideration (and also changing to use the $() subshell syntax consistently), you might want one of the following invocations:
If you really need to cache the current month's calendar and want to pull multiple days:
currCal="$(/usr/bin/cal)"
BUSINESS_DAYS="$(echo "${currCal}" | \
nawk 'NR>2 {print substr($0,4,14)}' | \
tr "\n" " ")"
DAY=16
DAYTH_DAY="$(echo "${BUSINESS_DAYS}" | nawk -v "day=${DAY}" '{ print $day }')
If this is just a one-and-done:
DAY=16
DAYTH_DAY="$(/usr/bin/cal | \
nawk 'NR>2 {print substr($0,4,14)}' | \
tr "\n" " " | \
nawk -v "day=${DAY}" '{ print $day }')"
One more note: the processing here can be simplified if done entirely in awk(/nawk), but I wanted to stick to the basic framework you had already chosen.
Update per the request in the comment:
A pure POSIX awk version:
DAY=16
DAYTH="$(cal | awk -v "day=${DAY}" '
(NR < 3) { next ; }
/^.[0-9 ]/ { $1="" ; }
/^ / || (NF == 7) { $NF="" ; }
{ hold=hold $0 ; }
END { split(hold,arr," ") ; print arr[day] ; }')"
Yes, simplified is a matter of opinion, and I'm sure someone can make this more concise. Explanation of how this works:
Skip the header of the cal output:
(NR < 3) { next ; }
For weeks that have a date on the Sunday, trim the date of that Sunday:
/^.[0-9 ]/ { $1="" ; }
For weeks that start after Sunday (first week of a month) or weeks that have a full seven days, trim the date of Saturday for that week:
/^ / || (NF == 7) { $NF="" ; }
Once the lines only have the dates of weekdays, curry them into hold:
{ hold=hold $0 ; }
At the end, split hold on spaces so we can grab the Nth day:
END { split(hold,arr," ") ; print arr[day] ; }')"

Answer 2

No awk, just software tools:
set -- $(cal -h | rev | cut --complement -b-5,20- | rev | tail -n +3) ; \
shift 15 ; echo $1
Output:
22
The output of cal is tricky to parse because:
It's right justified.
It's space delimited.
One or two digit dates means two or one delimiting spaces.
More leading spaces for first days of month.
Parsing won't quite work without the -h option, (turn off 'today' highlighting).