I have two digit month value (01 to 12). I need to get the three letter month abbreviation (like JAN, FEB, MAR etc.) I am able to get it in mixed case using the following command:
date -d "20170711" | date +"%b"
The output is "Jul" I want it to be "JUL". Is there a standard date option to get it?
^ use upper case if possible
Result:
$ date +%^b
JUL
Bonus: how I got this answer:
man date Enter /case Enter n
^! This answer just needs to take into account the variable input of the month number.
You could just pipe it to tr(1):
date -d "20170711" +"%b" | tr '[:lower:]' '[:upper:]'
date -d '20170711' '+%^b' and date -d '20170711' '+%b' | tr '[:lower:]' '[:upper:]' works well.
You can also do this using parameter expansion:
$ month=$(date -d '20170711' '+%b')
$ echo "${month^^}"
JUL
Since you're dealing with a fairly static piece of information (barring more intercalary events), just use built-in shell commands:
function capdate() {
case "$1" in
(01) printf "JAN";;
(02) printf "FEB";;
(03) printf "MAR";;
(04) printf "APR";;
(05) printf "MAY";;
(06) printf "JUN";;
(07) printf "JUL";;
(08) printf "AUG";;
(09) printf "SEP";;
(10) printf "OCT";;
(11) printf "NOV";;
(12) printf "DEC";;
(*) printf "invalid"; return 1;;
esac
}
Sample run:
$ m=$(capdate 01); echo $?, $m
0, JAN
$ m=$(capdate Caesar); echo $?, $m
1, invalid
Adjust the text if your locale has different date +%b names.
case statement :p
Commented
Jul 12, 2017 at 12:30
Another solution, using awk
date -d "20170711" | date +"%b" | awk '{print toupper($0)}'
date to format it in the same command? date -d 20170711 +%b ?
date -d "20170611" | date +"%b")datecommand is simply discarded; the seconddatecommand prints the current month regardless