Access positional parameters out of order in a for loop

Question

I have a script which makes use of multiple (i.e. positional) arguments from the command line, and within it a for loop that iterates through these arguments:

for i in "$@"; do 
...
done

This means that the arguments are populated into $i one by one, in the order that they're specified on the command line.

This seems to mean that it's impossible to select an argument that is not currently populating $i.

For example, in the following line in my script, I need to reference two different variables, which outside of a for loop would be 1ドル, 2ドル and 1ドル respectively:

ffmpeg -i "$i" -i "$i" ... "${i%.*}.mp4"

The script is executed in the following format:

./script.sh image.jpg *.flac

...but as I understand things, this problem would apply even to scripts that don't work with globs.

Is it possible to access positional arguments from a for loop as needed, rather than simply in the order they are specified in?

• glenn jackman: Please show some sample filenames, and the resulting ffmpeg invocations you want to run. Does the ... part involve any filenames?
• Tom Yan: Sounds like you want to create a variable with the value of 1ドル and then shift 1? You can also do array slicing on $@. See stackoverflow.com/questions/2701400/…
• harrymc: In bash you can select arguments out of order and in loops, by using indirect variable expansion, ${!i}, that expands to the content of the parameter whose name is stored in the i variable. Or for example ${@:3} for all arguments from the third to the end. See link.

Answer 1

I was able to figure this out in the end, although I don't consider this a perfect answer to the question because it doesn't satisfy the requirement of being able to reference the arguments from within the for loop.

In this sense I see this answer as more of a workaround, and I'm more than happy for someone else to come along and directly answer the question in a more comprehensive way. In any case, it's a workaround that works for my current script.

 img="1ドル"
 shift
 for i in "$@"; do
 ...
 ffmpeg -i "$img" -i "$i" ... "${i%.*}.mp4"
 ...
 done

As far as I can tell, this works by assigning the first argument to a variable $img outside of the for loop. Each instance of $img in the script is processed and then discarded by shift (which is the same as shift 1) before moving on to the next.

Once all $img arguments are removed from the arguments list, the for loop uses the $@ to loop through the remaining arguments as normal.

The script can then be called with something like:

./script.sh customimage.png *.flac

Or for multiple instances of the $img variable:

./script.sh customimage.png custom-image2.jpg *.flac