CODEWITHSUNDEEP
pythongoogle-app-engine
Dev Sahoo
Dev Sahoo

Reputation: 12101

How to open a file in the parent directory in python in AppEngine?

How to open a file in the parent directory in python in AppEngine?

I have a python file module/mod.py with the following code

f = open('../data.yml')
z = yaml.load(f)
f.close()

data.yml is in the parent dir of module. The error I get is

IOError: [Errno 13] file not accessible: '../data.yml'

I am using AppEngine SDK 1.3.3.

Is there a work around for this?

Upvotes: 35

Views: 70888

Answers (5)

tonisives
tonisives

Reputation: 2528

@ThatsAmorais answer in a function

import os

def getParent(path: str, levels=1) -> str:
    """
    @param path: starts without /
    @return: Parent path at the specified levels above.
    """
    current_directory = os.path.dirname(__file__)

    parent_directory = current_directory
    for i in range(0, levels):
        parent_directory = os.path.split(parent_directory)[0]

    file_path = os.path.join(parent_directory, path)
    return file_path

Upvotes: 0

Selbstliebe
Selbstliebe

Reputation: 11

I wrote a little function called get_parent_directory() which might help getting the path of the parent directory:

import sys
def get_parent_directory():
    list = sys.path[0].split('/')[:-1]
    return_str = ''
    for element in list:
        return_str += element + '/'
    return return_str.rstrip('/')

Upvotes: 1

anjandash
anjandash

Reputation: 827

🔰🔰 alternative solution 🔰🔰

You can also use the sys module to get the current working directory.
Thus, another alternative to do the same thing would be:

import sys
f = open(sys.path[0] + '/../data.yml')

Upvotes: 5

ThatsAMorais
ThatsAMorais

Reputation: 659

Having encountered this question and not being satisfied with the answer, I ran across a different solution. It took the following to get what I wanted.

  1. Determine the current directory using os.path.dirname:

    current_directory = os.path.dirname(__file__)

  2. Determine the parent directory using os.path.split:

    parent_directory = os.path.split(current_directory)[0] # Repeat as needed

  3. Join parent_directory with any sub-directories:

    file_path = os.path.join(parent_directory, 'path', 'to', 'file')

  4. Open the file:

    open(file_path)

Combined together:

open(os.path.join(os.path.split(os.path.dirname(__file__))[0], 'path', 'to', 'file')

Upvotes: 11

Marcelo Cantos
Marcelo Cantos

Reputation: 185972

The open function operates relative to the current process working directory, not the module it is called from. If the path must be module-relative, do this:

import os.path
f = open(os.path.dirname(__file__) + '/../data.yml')

Upvotes: 58

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