Ygtjki

I would like to batch rename filenames turning
A-B-C#2-D.wav to A-B-C#1-D.wav. So for example:

A-B-C#2-D.wav
A-B-C#8-G.wav
A-B-C6-E.wav

becomes

A-B-C#1-D.wav
A-B-C#7-G.wav
A-B-C5-E.wav

So the number in the third substring should be decreased by one.
(# is the part of filenames instead of comment; both filenames with and without # are possible.)

With zsh:

autoload zmv # best in ~/.zshrc
zmv -n '(*[^0-9])(<1->)(*.wav)(#qn)' '$1$(($2-1))$3'

(remove the -n if satisfied)

The (#qn) is for the list to be numerically sorted so that A-B-C#9-D.wav be renamed before A-B-C#10-D.wav for instance.

However, if there were both a A-B-C#9-D.wav and A-B-C#10-D.wav, zmv would flag the fact that one file would be renamed to the name of an existing file and abort the command. You'd need to add the -f option to still force it.

Or with zsh (for the numeric glob order) and perl's rename:

rename -n 's/d+/$&-1/e' ./*[1-9]*.wav(#qn)

(same caveat and same -f option with some variants of rename).

With a bash-specific loop (for the regular-expression testing conditional =~):

for file in ?-?-*-?.wav
do
 [[ $file =~ ^([^[:digit:]]+)([[:digit:]]+)(-..wav)$ ]] &&
 echo mv -- "$file" "$BASH_REMATCH[1]$((10#$BASH_REMATCH[2] - 1))$BASH_REMATCH[3]"
done

This uses a wildcard to pick up the desired wav files; matching files will have one character (letter) between the dashes and before the .wav at the end. Each filename is run through the regular expression, which separates it into three pieces:

1. 
   ^([^[:digit:]]+) -- the leading portion; everything up to the first digit


2. 
   ([[:digit:]]+) -- the digit(s)


3. 
   (-..wav) -- the trailing portion

If the match succeeds, then we rename the file, using the leading portion, the digits minus one, and the trailing portion.

Remove the echo when the output looks correct.