Remove last argument from argument list of shell script (bash)

问题

This question concerns a bash script that is run in automator osx. I am using automator actions to get and filter a bunch of file references from the finder. Then I append to that list the name of the parent folder, also via an automator action. Automator then feeds these arguments to an action called "run shell script". I am not sure exactly how automator invokes the script but the argument list looks like this when echoed with: echo "$@"

/Volumes/G-Raid/Online/WAV_TEST/Testbok 50/01/01000 43-001.wav /Volumes/G-Raid/Online/WAV_TEST/Testbok 50/02/02000 43-002.wav /Volumes/G-Raid/Online/WAV_TEST/Testbok 50/03/03000 43-003.wav /Volumes/G-Raid/Online/WAV_TEST/Testbok 50

In this case path to 3 files and a folder.

In the shell script I launch an application called ripcheckc* with the args passed from automator minus the last argument(the folder) in the list.

I use this to remove the last argument:

_args=( "$@" )
unset _args[${#_args[@]}-1]

And this is echo $_args:

/Volumes/G-Raid/Online/WAV_TEST/Testbok 50/01/01000 43-001.wav /Volumes/G-Raid/Online/WAV_TEST/Testbok 50/02/02000 43-002.wav /Volumes/G-Raid/Online/WAV_TEST/Testbok 50/03/03000 43-003.wav

Same as before but without the folder.

Now, if I run ripcheckc with "$@" as argument it works (but fails later on because of that last path in the argument list) If I use ${_args[@]} the application will just abort silently. When I echo $@ and _args the output looks identical except for the last argument.

My question is - what is the difference between $@ and $_args that make the first valid input and the second not?

*The application is ripcheckc

I hope my question makes sense.

EDIT: Solved.

回答1:

Assuming that you already have an array, you can say:

unset "array[${#array[@]}-1]"

---

For example, if your script contains:

array=( "$@" )
unset "array[${#array[@]}-1]"    # Removes last element -- also see: help unset
for i in "${array[@]}"; do
  echo "$i"
done

invoking it with: bash scriptname foo bar baz produces:

foo
bar

回答2:

I have used this bash one-liner before

set -- "${@:1:$(($#-1))}"

It sets the argument list to the current argument list, less the last argument.

---

How it works:

• $# is the number of arguments
• $((...)) is an arithmetic expression, so $(($#-1)) is one less than the number of arguments.
• ${variable:position:count} is a substring expression: it extracts count characters from variable starting at position. In the special case where variable is @, which means the argument list, it extracts count arguments from the list beginning at position. Here, position is 1 for the first argument and count is one less than the number of arguments worked out previously.
• set -- arg1...argn sets the argument list to the given arguments

So the end result is that the argument list is replaced with a new list, where the new list is the original list except for the last argument.

回答3:

You can also get all but the last argument with

"${@:0:$#}"

which, honestly, is a little sketchy, since it seems to be (ab)using the fact that arguments are numbered starting with 1, not 0.

Update: This only works due to a bug (fixed in 4.1.2 at the latest) in handling $@. It works in version 3.2.

来源：https://stackoverflow.com/questions/20398499/remove-last-argument-from-argument-list-of-shell-script-bash