打开一个相对于python模块路径的文件

Question

A little bit of an explanation of my situation before my question.

I created a module called foo. foo's location in the file system is

/runtime/foo

and the script "test.py" which imports foo by using sys.path.append() is located in.

/runtime/bar/test.py

Foo's structure is as follows

foo
    mtr.txt
    __init__.py
    datasources.py

I would like a class within datasources.py to open the file mtr.txt. However, I am not able to do this without explicting giving a path within datasources. In this case it would be

/runtime/foo/mtr.txt

My code will work if I do give it this path, but this is something that should be doable, but I can't find the answer.

I have tried the following commands within a class in datasources.py.

open("mtr.txt")
open("./mtr.txt")

and a few other things using the os.path.dirname()

Is there a way to open the file 'mtr.txt' without giving its full path within datasources.py?

Thanks

Answer 1

Use pkg_resources

import pkg_resources
data = pkg_resources.resource_string(__name__, "mtr.txt")

There are more methods provided by this package, check the docs for Resource Manager API .