execfile in it's working directory from other file

Question

I have structure like this:

c
|-myMain.py
|-.....\someDir
              |-startup.py
              |-subDir
                     |-x.py

I've found few similar questions, and answers bring me to this:

#myMain.py
import os
if __name__ == '__main__':
    os.chdir("c:\\......\\someDir")
    execfile("startup.py")

#startup.py
from subDir import x
if __name__ == '__main__':
    x.doSomething()

Problem is that import fails in startup.py when I run myMain.py:

ImportError: No module named subDir

but, it works when I run startup.py directly. Any help would be appreciated.

FWIW, you could avoid having to do the `chdir()` to a hardcoded fully qualified path with something like `STARTUP_FILE = os.path.join(os.path.dirname(__file__), 'someDir', 'startup.py')` followed by `execfile(STARTUP_FILE)`. — martineau, Nov 23 '12 at 16:11

score 1 · Accepted Answer · answered Nov 23 '12 at 15:14

1

In order for python to see subDir as a valid python structure, you must make it a package. You make it a package by including a __init__.py file in that directory. The file can be empty.

Once subDir has such a file, the statement from subDir import x should work.

answered Nov 23 '12 at 15:14

Martijn Pieters

1,048,767
296
4,058
3,343

Ok, it works now, but I've also had to add path to someDir as @martineau suggested – Aleksandar Nov 26 '12 at 11:54

score 0 · Answer 2 · answered Aug 29 '17 at 21:56

0

Future readers:

Pradyumna's answer from here solved this issue for me:

import sys, change "sys.path" by appending the path during run time,then import the module that will help

[i.e. sys.path.append(execfile's directory)]

answered Aug 29 '17 at 21:56

Shmuelt

119
8

execfile in it's working directory from other file

2 Answers2