Issue
I am using a bash script that expects command line arguments preceded by a dash followed by a file name:
./script -abc file_name
In the script, I try to print out all of the arguments to the script using:
echo "$@"
which displays:
-abc file_name
which is expected. Everything prints fine except for when I use the letter "n" by itself, i.e.:
./script -n file_name
The resulting echo statement inside the script only prints out the file_name argument and not the "-n" argument at all. This behavior only occurs with the letter "n" by itself. When I try "-nx" or 'n' with any other letter(s) the echo statement prints out fine...
Why does this echo statement not work with "-n" by itself?
Solution
The -n or -e are valid options for the echo command.
Therefore it's consumed by the echo command and only the remaining text is shown.
That's a known problem of echo and to avoid this you can use printf.
printf "%s\n" "$*"
Answered By - jeb Answer Checked By - Cary Denson (WPSolving Admin)