[GRASS-user] Write python grass program with osgeo4w

[Moskovitz, Bob]

Hello list,
 
A coworker and I have been trying to use the first example in  <http://download.osgeo.org/grass/grass6_progman/pythonlib.html> http://download.osgeo.org/grass/grass6_progman/pythonlib.html.  We discovered that "import grass.script as grass" has to be changed to "import grass".  Below is the example and the error we are getting is this:
 
C:\>m.test.py
Traceback (most recent call last):
  File "C:\OSGeo4W\apps\grass\GRASS-~1.0SV/scripts\m.test.py", line 31, in <module>
    options, flags = grass.parser()
  File "C:\OSGeo4W\apps\grass\GRASS-~1.0SV\etc\python\grass.py", line 208, in parser
    os.execvp("g.parser", [name] + argv)
  File "C:\OSGeo4W\apps\Python25\lib\os.py", line 353, in execvp
    _execvpe(file, args)
  File "C:\OSGeo4W\apps\Python25\lib\os.py", line 389, in _execvpe
    func(fullname, *argrest)
OSError: [Errno 22] Invalid argument
 
 
Any ideas what is going on?
 
Bob
 
 
 
#====================================================
#!/usr/bin/env python
 
#%Module
#% description: Checks if vector map is 3D
#% keywords: vector
#%End
#%option
#% key: map
#% type: string
#% gisprompt: old,vector,vector
#% key_desc: name
#% description: Name of vector map 
#% required: yes
#%end
 
import sys
import grass
 
def main():
    info = grass.parse_command('v.info',
                               flags = 't',
                               map = options['map'])
    if info['map3d'] == '1':
        print 'Vector map is 3D'
    else:
        print 'Vector map is 2D'
 
    return 0
 
if __name__ == "__main__":
    options, flags = grass.parser()
    sys.exit(main())

 

Bob Moskovitz 
Research Analyst I 
Seismic Hazard Zonation Project
California Geological Survey 
http://www.conservation.ca.gov/cgs/shzp<?xml:namespace prefix = o ns = "urn:schemas-microsoft-com:office:office" />

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