Process all arguments except the first one (in a bash script)
I have a simple script where the first argument is reserved for the filename, and all other optional arguments should be passed to other parts of the script.
Using Google I found this wiki, but it provided a literal example:
echo "${@: -1}"
I can't get anything else to work, like:
echo "${@:2}"
or
echo "${@:2,1}"
I get "Bad substitution" from the terminal.
What is the problem, and how can I process all but the first argument passed to a bash script?
Use this:
echo "${@:2}"
The following syntax:
echo "${*:2}"
would work as well, but is not recommended, because as @Gordon already explained, that using * , it runs all of the arguments together as a single argument with spaces, while @ preserves the breaks between them (even if some of the arguments themselves contain spaces). It doesn't make the difference with echo , but it matters for many other commands.
If you want a solution that also works in /bin/sh try
first_arg="$1"
shift
echo First argument: "$first_arg"
echo Remaining arguments: "$@"
shift [n] shifts the positional parameters n times. A shift sets the value of $1 to the value of $2 , the value of $2 to the value of $3 , and so on, decreasing the value of $# by one.
http://wiki.bash-hackers.org/scripting/posparams
It explains the use of shift (if you want to discard the first N parameters) and then implementing Mass Usage (look for the heading with that title).