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[Question #250174]: __file__ or equivalent?

 

New question #250174 on Sikuli:
https://answers.launchpad.net/sikuli/+question/250174

I would like to locate a file relative to the executing sikuli script.

In normal Python I would do this based on the __file__ predefined value, but it does not seem to be defined.

Is there some other way to get "path to this script" or do I have to hard-code the path?

Thanks,
DC

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