[Bug]: Piping in command substitution does not work

[damejeras]

What happened?

I have been using fabric in my zsh function

# Assistant
function _assist() {
  output=$(echo -n "$@" | fabric --pattern shell_command)
  echo "$output"
  vared -p "Do you want to evaluate the command? (y/n) " -c choice
  if [[ $choice == y ]]; then
    eval "$output"
  fi
  print -s "$output"  # Add the command to history
}

It was working on my old setup (not sure which version of fabric I was using previously), but I installed @latest through go install and it doesn't work anymore. If I run my function I get the response to my prompt telling that the input was not provided. I was debugging the issue and it seems that echo output does not get piped to fabric. If I run this without command substitution ($(...)) it works as expected. I believe there is a bug in how fabric handles pipes within command substitutions.

Version check

• Yes I was.

Relevant log output

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Relevant screenshots (optional)

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