why function variable not working in command substitution inside a function?

Question

im running this in bashrc file

function simplefunc () {
    output=$(ls -1 "$HOME")
    linecount=$(${output} | wc -l)

    echo "${linecount}"
    echo "${output}"
}

getting this error

Desktop: command not found
0
Desktop
Documents
Downloads
Music
Pictures
Public
snap
SoftMaker
Templates
venv
Videos

i tried these too

putting "local" before variable or

# linecount=$(output | wc -l)
# echo "$(${output} | wc -l)"

Also [don't parse `ls` output.](https://mywiki.wooledge.org/ParsingLs) Depending on what exactly you want, try e.g. `(cd "$HOME"; printf '%s\n' * ) | nl` — tripleee, Aug 19 '21 at 11:19

score 0 · Accepted Answer · answered Aug 19 '21 at 11:11

0

I think you should change the third line to:

linecount="$(echo "${output}" | wc -l)"
# OR alternatvely:
# linecount="$(wc -l < <(echo "$output"))"

answered Aug 19 '21 at 11:11

nino

1 Answers1