CODEWITHSUNDEEP
zsh
Jiahao Cai
Jiahao Cai

Reputation: 1250

Unexpected no such file error in zsh function

I have defined this function in zsh, but it cannot work on my computer.

function m() {
    local path=/Users/james/Music/cloud_music_link
    local cmd="ls $path | sed -n $1p"
    echo `$cmd`
    # afplay `$cmd`
}

It always said:

m:3: no such file or directory: ls /Users/james/Music/cloud_music_link | sed -n 10p

and when I copy the ls /Users/james/Music/cloud_music_link | sed -n 10p and run it in zsh, everything is ok, why would this happen?

And the folder cloud_music_link is a soft link, I am not sure if this matters.

Upvotes: 1

Views: 674

Answers (2)

Jiahao Cai
Jiahao Cai

Reputation: 1250

Finally I have figured it out, it is because the name of the variable path, I changed it to p and everything works OK. The name path will conflict with zsh.

This code works:

function m() {
    local p=/Users/james/Music/cloud_music_link/
    song="$(ls $p | sed -n $1p)"
    echo $song
    killall afplay 
    afplay "$p$song"&
}

Upvotes: 0

Charles Duffy
Charles Duffy

Reputation: 295639

You want to pass the file (or subdirectory) at a given position inside of your given directory first to echo, and then to aplay? Easily done, in a way compatible with both bash and zsh:

m() {
    local num=${1:-1}
    local dir=/Users/james/Music/cloud_music_link
    local -a files=( "$dir"/* )
    local file=${files[$(( num - 1 ))]}
    echo "$file"
    afplay "$file" &
}

See:

Upvotes: 1

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