zsh capture and pipe output to another command

Question

I had the zsh script

printf '%s\0' www/scripts6/lib/* | xargs -0 -I{} -P 50 babel {} > {}

and it worked (i think.). But I needed to actually copy the files to the scripts directory instead of scripts6. Also, I needed to add an extra path to the list of paths manually, so I tried doing this:

paths=$(printf '%s\0' www/scripts6/lib/* | sed -e 's@www/scripts6/@@g')
paths=$paths'main.js'$'\0'
echo -n $paths | xargs -0 -I {} -P 50 babel www/scripts6/{} > www/scripts/{}

That was a BIG ACCOMPLISHMENT for me. Anyway, the script makes the file named {} in the scripts directory (and nothing in the scripts/lib directory). This leads me to believe that only the first {} in the code gets replaced by the argument, and not the second one (and I have no explanation for the absence of files in the scripts/lib directory).

How can I make the script work as expected?

Answer 1

Redirection happens before xargs executes. so your first example also creates a file named {}, because xargs never got the chance to replace the last {}.

Also, while you do try to separate the filenames by nulls to treat filenames "correctly", the effort is in vain once you use echo. echo in zsh interprets the C-style escapes(\t, \n, etc) by default. You would need the -E option to disable that.

Try this and see if it does what you want

setopt monitor
for f in www/scripts6/lib/*(e*'REPLY=${REPLY#www/scripts6/}'*) main.js; do;
  while (( $#jobstates >= 51 )); do 
    sleep 1
  done
  echo -E babel $f \> www/scripts/$f &
done

If the output seems ok, remove the echo -E and replace \> with >