Extract parameters before last parameter in "$@"

Question

I'm trying to create a Bash script that will extract the last parameter given from the command line into a variable to be used elsewhere. Here's the script I'm working on: ``` #!/bin/bash # compact - archive and compact file/folder(s) eval LAST=\$$# FILES="$@" NAME=$LAST # Usage - display usage if no parameters are given if [[ -z $NAME ]]; then echo "compact ... .tar.gz" exit fi # Check if an archive name has been given if [[ -f $NAME ]]; then echo "File exists or you forgot to enter a filename. Exiting." exit fi tar -czvpf "$NAME".tar.gz $FILES ``` Since the first parameters could be of any number, I have to find a way to extract the last parameter, (e.g. compact file.a file.b file.d files-a-b-d.tar.gz). As it is now the archive name will be included in the files to compact. Is there a way to do this?

Accepted Answer

To remove the last item from the array you could use something like this:

```
#!/bin/bash

length=$(($#-1))
array=${@:1:$length}
echo $array
```

Even shorter way:

```
array=${@:1:$#-1}
```

But arays are a Bashism, try avoid using them :(.

Extract parameters before last parameter in "$@"

Solution

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