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

Question

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-Ra…

Accepted Answer

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
```

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

Solution

Problem