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Re: Listing the files of a package output
From: |
Konrad Hinsen |
Subject: |
Re: Listing the files of a package output |
Date: |
Tue, 20 Jun 2023 21:35:50 +0200 |
Hi Guillaume,
> As the directory of the "out" output will have a shorter name, you could
> sort the names by size and take the first one:
>
> --8<---------------cut here---------------start------------->8---
> for name in $(guix build coreutils); do printf "%d\t%s\n" "${#name}"
> "${name}"; done | sort -n | head -n 1 | cut -f 2
> --8<---------------cut here---------------end--------------->8---
Indeed, thanks! A but lengthy by my taste, but it does the job!
Cheers,
Konrad.