Passing paths and filenames to a program from bash script

Question

I use a utility called mp3wrap which combines multiple mp3s into one. Say I have a directory "album 1" containing three files "track 1.mp3", "track 2.mp3" and "track 3.mp3". I will run the following:

mp3wrap album\ 1.mp3 album\ 1/*.mp3

which will combine all three tracks in the "album 1" directory (the second argument) into a new file "album 1.mp3" (the first argument).

I have a bunch of directories like this, "album 2", "album 3" etc., and I would like to write a script to automate the process. I tried this:

for i in album*
do
  mp3wrap $i.mp3 $i/*.mp3
done  

but i get the error

1.mp3: No such file or directory

It seems my script expands $i to "album 1.mp3" which is passed as two separate arguments. After some searching and reading I've ended up with this:

for i in album*
do
  mp3wrap "${i}.mp3" "${i}/*.mp3"
done  

Now I don't get an error message, instead I get the "usage" page from the mp3wrap program. So clearly it's not receiving the arguments in the same way as when I run the command from the command line, and I don't know how to check what the actual command line looks like when executed by the bash script.

What am I doing wrong? Any help appreciated!

Answer 1

You are close to it. Here is the correct code:

for i in album*
do
    mp3wrap "${i}.mp3" "${i}"/*.mp3
done

Variables and arguments should almost always be quoted to avoid your first issue. On the contrary, shell globing characters like * and ? should not, else they lose their special meaning.