Listing the files of a package output

Listing the files of a package output

[Konrad Hinsen]

Hi Guix,

this looks like a beginner's question, but I can't figure out how to do
this: list the files in a given package output, from the command line (I
do know how to do this in Guile).

For a package with a single output, I can do

  ls -R $(guix build hello)

For a package output other than "out", I can do

  ls -R $(guix build coreutils | grep debug)

But what about coreutils:out?

Cheers,
  Konrad

Re: Listing the files of a package output

[Guillaume Le Vaillant]

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Konrad Hinsen <konrad.hinsen@fastmail.net> skribis:

> Hi Guix,
>
> this looks like a beginner's question, but I can't figure out how to do
> this: list the files in a given package output, from the command line (I
> do know how to do this in Guile).
>
> For a package with a single output, I can do
>
>   ls -R $(guix build hello)
>
> For a package output other than "out", I can do
>
>   ls -R $(guix build coreutils | grep debug)
>
> But what about coreutils:out?
>
> Cheers,
>   Konrad

Hi.
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---

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Re: Listing the files of a package output

[Konrad Hinsen]

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.