[GRASS-user] Re: [GRASS-dev] Helpp needed to call Grass from
inside Python Plugin
Glynn Clements
glynn at gclements.plus.com
Wed Mar 24 22:20:45 EDT 2010
Bishwarup Banerjee wrote:
> I have tried to iimplement the example in my way. The code is as follows
> import core as grass
This should be:
import grass.script as grass
or:
import grass.script.core as grass
> def main():
> input_map = options['H:/Shared_data/test1.tif']
> output_map = options['Test_24']
This is bogus. The keys in the "options" dictionary are the names of
the options defined in the g.parser comments (assuming that you have
them; if you don't, grass.parser() won't work).
> But I get the following error, I cant make out where I am going wrong. I
> have set all the Env variables as suggested in the wiki page. Please help me
> to implement it.
>
>
> F:\GRASS-6-SVN\etc\python\grass\script>python test.py
> Traceback (most recent call last):
> File "test.py", line 18, in <module>
> options, flags = grass.parser()
> File "F:\GRASS-6-SVN\etc\python\grass\script\core.py", line 359, in parser
> os.execvp("g.parser.exe", [name] + argv)
> File "F:\Quantum GIS\python\os.py", line 353, in execvp
> _execvpe(file, args)
> File "F:\Quantum GIS\python\os.py", line 389, in _execvpe
> func(fullname, *argrest)
> OSError: [Errno 22] Invalid argument
I have no idea what's causing this.
I suggest modifying the parser() function to print the arguments which
are being passed to os.execvp():
> File "F:\GRASS-6-SVN\etc\python\grass\script\core.py", line 359, in parser
> os.execvp("g.parser.exe", [name] + argv)
--
Glynn Clements <glynn at gclements.plus.com>
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