使用脚本位置的相对路径时如何在python中打开文件

Question

I have a directory Structure as such

MainFolder/unitTest/test.py
MainFolder/run.py
MainFolder/model/file.py
MainFolder/config/config.conf

Inside file.py i am trying to open a file

f = open('config/config.conf', 'r')

Normally the file.py is called from run.py. The current working directory is that of the run.py. so if i want to access the config file . i go to config folder and then to the file 'config/config.conf' If i want to run it from the unit test i need to go up one level and then into the config folder. '../config/config.conf'

Is there a way that when opening a file , it starts looking for the file from where the file.py is located so that regardless of where the file.py is called from i can access the file as ../config/config.conf

This is going to be hosted on a remote server so i cant give it the full path...

Answer 1

What you need may be to organize your codes in Python modules . It helps you design/access/manage each file in a Pythonic way.

To solve your problem, you can work around by setting MainFolder as the current working directory with os.chdir at the beginning of the file.py . It means that you need to have the full path of MainFolder either by using os.path.getcwd in run.py to find out or storing it in config file when deployed.

Answer 2

Sure, just create a file, eg pwd.py in your project root directory.

import os

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

then in your modules:

from pwd import PWD

f = open('%s/config/config.conf' % PWD, 'r')