Hi,
If I have a python script that is executed via a symlink, is there a way that I can find the path to the script rather than the symlink? I've tried using the methods suggested in this question, but they always return the path to the symlink, not the script.
For example, when this is saved as my "/usr/home/philboltt/scripts/test.py" :
#!/usr/bin/python
import sys
print sys.argv[0]
print __file__
and I then create this symlink
ln -s /usr/home/philboltt/scripts/test.py /usr/home/philboltt/test
and execute the script using
/usr/home/philboltt/test
I get the following output:
/usr/home/philboltt/test
/usr/home/philboltt/test
Thanks! Phil