Re: Better way to get default datadir

[Albert Krewinkel <albert+pandoc-9EawChwDxG8hFhg+JK9F0w@public.gmane.org>]

DG <dominik.gerstorfer-Re5JQEeQqe8AvxtiuMwx3w@public.gmane.org> 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

    print(PANDOC_STATE.user_data_dir)

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