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From: Albert Krewinkel <>
To: pandoc-discuss-/
Subject: Re: Better way to get default datadir
Date: Fri, 10 Nov 2023 14:49:13 +0100	[thread overview]
Message-ID: <> (raw)
In-Reply-To: <ea89a95c-fd70-429e-ae1f-db343f8bdd19n-/>

DG <> writes:

> I have a script installs templates and default files on multiple platforms.
> In order to get the default datadir I use sed :
> pandoc -v | sed -rn "s/User data directory: (.*)$/\1/p"
> It works, but it is (a) not very elegant and (b) it depends on sed and
> might fail on windows. My question is: What am I missing? Is there a
> more direct way to get the default datadir?

First, here's a way that DOESN'T work, but maybe it should?

    pandoc lua -e 'print(PANDOC_STATE.user_data_dir)'

This only prints `nil`, because the value is not set when pandoc is
called as a Lua interpreter. But it could be argued that the default
data dir should be returned if it exists.

The alternative is to use a single-line file


and to pass it to pandoc as a Lua filter, e.g.

    echo "" | pandoc -L datadir.lua

Some shells support "process substitution". In those cases we can write
a one liner

    echo "" | pandoc -L <(echo 'print(PANDOC_STATE.user_data_dir)')

or even just

    pandoc -L <(echo 'print(PANDOC_STATE.user_data_dir)') <<< ''

But that's not portable.

Feel free to raise an issue on the bug tracker to get future support for
the potentially simpler version mentioned above.

Albert Krewinkel
GPG: 8eed e3e2 e8c5 6f18 81fe  e836 388d c0b2 1f63 1124

  parent reply	other threads:[~2023-11-10 13:49 UTC|newest]

Thread overview: 3+ messages / expand[flat|nested]  mbox.gz  Atom feed  top
2023-11-10  9:28 DG
     [not found] ` <ea89a95c-fd70-429e-ae1f-db343f8bdd19n-/>
2023-11-10 13:49   ` Albert Krewinkel [this message]
     [not found]     ` <>
2023-11-12 10:09       ` BPJ

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