[PythonCAD] DWG R2004 SUPPORT IN PYTHONCAD
Marco Alicera
marco.alicera at gmail.com
Mon Sep 25 21:41:24 CEST 2006
Eric Wilhelm-2 wrote:
>
> # from Art Haas
> # on Thursday 03 August 2006 01:25 pm:
>
>>I'd like to be able to support the AC1018 format, but I don't have a
>>timeline as to when that will happen. My work over the last couple of
>>months has been concentrated on improving the GUI responsiveness,
>> fixing bugs, and adding an occasional new feature. I regret that
>> still the code needed for directly importing older DWG/DXF files is
>> not written as well as the lack of code for newer version of these
>> files.
>
> We had a start on implementing DXF -> pythoncad conversion in
> vectorsection. The trouble with DWG is that it is as convoluted as DXF
> but also in a very bit-juggly binary format (e.g. some values take 2-3
> bits and various other not-8 widths.) The first issue makes it
> difficult because of the conceptual mapping of entities. The second is
> just really hard to code and debug.
>
> It's difficult, but not impossible. If anyone is interested in working
> on it, I'm more than happy to help you get started. The first goal for
> vectorsection would be a standalone dxf<->pythoncad conversion. From
> there, we would handle dwg<->pythoncad (where the conceptual mapping
> stuff would be done and we would basically only need a working DWG
> class to drop-in as a replacement for the DXF stuff.) After that,
> pythoncad could use the standalone converters directly or steal some
> code.
>
> hi Eric,
>
> i am the one interested in helping with the implementing DXF -> pythoncad
> conversion in
> vectorsection.
>
> i am an absolute begginer in python, i used to program in pascal many
> years ago and i belive i can still do something.
>
> i will need some time to get used with python and the pythoncad projetc,
> but i have the time.
>
> I'm not sure if i am in the mailing list, am i? please let me know.
>
> best regards
>
> Marco Alicera
>
>
> --Eric
> --
> Hot dogs: just another condiment.
> --Heart-attack Man
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