bash array count always returns 1

Question

I searched all over for this, but the terms are apparently too general. I'm writing a script to search a group of folders for .mp3 files. Some folders don't have mp3's so they have to be excluded.

I created an array to hold the uniq'd folder names. This find command will get the folders I need.

Folders=$(sudo find /my/music/ -type f -name "*.mp3" | cut -d'/' -f7 | sort -u)

When I try to count the number of folders in the array, I always get 1

echo ${#Folders[@]}

echo ${Folders[@]} prints them out on separate lines so I thought they were separate array elements. Can anyone explain what is going on? You might have to jiggle the field number in the cut command to reproduce locally.

Answer 1

Folders is not an array but a variable.

You need:

Folders=( $(sudo find /my/music/ -type f -name "*.mp3" | cut -d'/' -f7 | sort -u) )

i.e. enclose the command substitution with (). Now ${#Folders[@]} would give you the number of elements of array Folders.