Re: Command for printing derivations that will be built?

[Efraim Flashner <efraim@flashner.co.il>]

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On Tue, May 09, 2023 at 07:04:59PM +0000, Rodrigo Morales wrote:
> From what I understand in the output of `guix package -u' (see below),
> whenever I execute it, some derivations that exist in `/gnu/store' are
> passed to `guix build' and so they are built. Am I correct?
> 
> If I'm correct, I would like to know how I can print the list that is
> reported by `guix package -u' below "The following derivations will be
> built". I know I can execute `guix package -u' and press `C-c' in the
> next 5 seconds. However, I'd like to do this the right way. That is,
> write a command that only prints that list and does nothing more.
> 
> The reason I'm asking is because sometimes I want to build specific
> derivations in order to have those packages advertised in my local
> network in the order that I feel it's the best for me.
> 
> ,----
> | $ guix package -u
> | The following packages will be upgraded:
> |    (... list of packages ...)
> |
> | The following derivations will be built:
> |    (... list of derivations in /gnu/store ...)
> |
> | building (... a derivation ...)
> | - 'unpack' phase
> `----

One option is to run `guix package -u . -n'. '-u .' will do all the
packages, and '-n' is the short flag for --dry-run, which won't actually
build any of the packages but will show you what will be built.

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Efraim Flashner   <efraim@flashner.co.il>   פלשנר אפרים
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