返回文件的文件路径，而不是当前目录

Question

I have a Python/Sikuli script where I need to store the location of the file in a variable.

I have the IDE on one location, and the file is stored in a different directory.
IDE: C:\\Sikuli\\runIDE.cmd
File: C:\\Script\\Files\\HelloWorld2.sikuli\\HelloWorld.py

Now I use os.getcwd() to fetch the location of the IDE.
But how do I return the full directory path of the file "HelloWorld.py"?

I also used os.path.abspath("HelloWorld2.py") and os.path.dirname(os.path.realpath('HelloWorld2.py')) .
But also that didn't work.
Can anyone help me with this?

Answer 1

In regular Python, using __file__ returns the full path of the directory that contains the executed file:

os.path.dirname( os.path.abspath( __file__ ) )

However, in Sikuli, the __file__ reference is undefined:

from sikuli import *
import os

print( os.path.dirname( os.path.abspath( __file__ ) ) )

When run in Sikuli v2.0.4, we see that __file__ is unsupported :

[error] script [ test2 ] stopped with error in line 6
[error] NameError ( name '__file__' is not defined )
[error] --- Traceback --- error source first
line: module ( function ) statement 
6: main (  <module> )     print( os.path.dirname( os.path.abspath(__file__) ) )
[error] --- Traceback --- end --------------

Instead, use getBundle() , such as:

from sikuli import *
import os

print( os.path.dirname( getBundlePath() ) )

Answer 2

Sikuli seems to have some issue's with the __file__
What does work with Sikuli and Python is:

os.path.dirname(getBundlePath()) 

Edit for the question below:
To get the names of the files that are in the directory you can do:

path = os.path.dirname(getBundlePath()) 
directory = os.listdir(path) 
for file in directory: 
    print(file)