currently executing python file
Here is a way to get the currently executing Python file:
print (lambda x:x).func_code.co_filename
Probably there is a simpler way, but I haven't found it. The best alternative I found is to use the special attribute __file__, but it doesn't work for SConscript files:
print (lambda x:x).func_code.co_filename print __file__
results is
/home/olpa/p/.../SConscript.py /usr/lib/python2.4/site-packages/SCons/Script/__init__.pyc
Categories:
python