@Brentmeister The original post wanted to know the scripts name (and path).
For this you can use getBundlePath(). So if you have a fixed structure like the above, every script would be able to find and use any path in this construct using the standard features of os.path module. -- You received this bug notification because you are a member of Sikuli Drivers, which is subscribed to Sikuli. https://bugs.launchpad.net/bugs/567303 Title: Use of "__file__" gives NameError Status in Sikuli: Fix Released Bug description: Trying to use __file__ python functionality results in Sikuli throwing a NameError exception. Example Code: script_name = __file__ I think I might get why this happens. It's the same behavior as if using __file__ in a python command line interpreter session. But, it would be nice for this to work inside of Sikuli the same as if running command "python testfile.py" from command line. So, I am requesting that Sikuli is modified to recognize __file__ as valid python code that gives the current script name. Or, alternatively to implement a new Sikuli specific command to get the name of the current running file. Thanks! Current Setup: Sikuli 0.9.9 Windows XP, 32-bit To manage notifications about this bug go to: https://bugs.launchpad.net/sikuli/+bug/567303/+subscriptions _______________________________________________ Mailing list: https://launchpad.net/~sikuli-driver Post to : [email protected] Unsubscribe : https://launchpad.net/~sikuli-driver More help : https://help.launchpad.net/ListHelp
