Suppose you have the following:
$ more a.py
import os
class A(object):
def getfile(self):
return os.path.abspath(__file__)
-
$ more b.py
import a
class B(a.A):
pass
-
>>> import b
>>> x=b.B()
>>> x.getf开发者_开发百科ile()
'/Users/sbo/tmp/file/a.py'
This is clear. No surprise from this code. Suppose however that I want x.getfile() to return the path of b.py without having to define another copy of getfile() under class B.
I did this
import os
import inspect
class A(object):
def getfile(self):
return os.path.abspath(inspect.getfile(self.__class__))
I was wondering if there's another strategy (and in any case, I want to write it here so it can be useful for others) or potential issues with the solution I present.
CW as it's more a discussion question, or a yes/no kind of question
sys.modules[self.__class__.__module__].__file__
Was able to get this working in python 3 as follows:
import os
class Parent:
def __init__(self, filename=__file__):
self.filename = filename
def current_path(self):
pth, _ = os.path.split(os.path.abspath(self.filename))
return pth
Then in a different module...
from practice.inheritance.parent import Parent
class Child(Parent):
def __init__(self):
super().__init__(__file__)
...accessing current_path()
from either Parent
or Child
returns the respective module path as expected.
>>> from practice.inheritance.parent import Parent
>>> parent = Parent()
>>> print(parent.current_path())
/Users/davidevans/PycharmProjects/play35/practice/inheritance
>>> from practice.inheritance.subpackage.child import Child
>>> child = Child()
>>> print(child.current_path())
/Users/davidevans/PycharmProjects/play35/practice/inheritance/subpackage
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