I have a script named test1.py which is not in a module. It just has code that should execute when the script itself is run. There are no functions, classes, methods, etc. I have another script which runs as a service. I want to call test1.py from the script running as a service.
print "I am a test" print "see! I do nothing productive."
# Lots of stuff here test1.py # do whatever is in test1.py
I'm aware of one method which is opening the file, reading the contents, and basically eval'ing it. I'm assuming there's a better way of doing this. Or at least I hope so.
The usual way to do this is something like the following.
def some_func(): print 'in test 1, unproductive' if __name__ == '__main__': # test1.py executed as script # do something some_func()
import test1 def service_func(): print 'service func' if __name__ == '__main__': # service.py executed as script # do something service_func() test1.some_func()
This is possible in Python 2 using
See the documentation for the handling of namespaces, if important in your case.
In Python 3, this is possible using (thanks to @fantastory)
However, you should consider using a different approach; your idea (from what I can see) doesn't look very clean.