create a file outside the directory in python

Question

I want to create a file outside the current working directory in python. Here's my directory structure.

|--myproject
|   |-- gui
|   |   |-- modules
|   |   |   |-- energy
|   |   |   |   |-- configuration
|   |   |   |   |   |-- working_file.py
|   |-- service
|   |   |-- constants
|   |   |   |-- global_variables.json

I'm current working in /myproject/gui/energy/configuration/working_file.py and I want to create a file under /myproject/service/constants named global_variables.json

I tried

with open("../../../../../service/constants/global_variables.json", 'w') as file_handler:
        content = json.load(file_handler)

1) check current working directory 2) where do you want to create it 3) check where it was actually created 4) profit (fix the path) — Karoly Horvath, Dec 29 '13 at 11:44
@KarolyHorvath- Well, the problem is the file isnt getting created at all, that's why I posted. Hence would have followed your steps on my own, — Praful Bagai, Dec 29 '13 at 11:47
so you supplied an invalid path. surely you can come up with a valid one (for test), right? — Karoly Horvath, Dec 29 '13 at 11:48
You say you want to create a file but you are actually reading a file, which you have opened in `w` (write mode). So what do you want to do? — Burhan Khalid, Dec 29 '13 at 12:06
Why are you opening a file in write mode and then trying to read from it? — user2357112, Dec 29 '13 at 12:22
@BurhanKhalid - Can you please provide your suggestions on http://stackoverflow.com/questions/20870603/django-content-type-table-auth-permission — Praful Bagai, Jan 01 '14 at 16:45

score 6 · Answer 1 · edited May 23 '17 at 11:52

6

Relative paths are resolved from the current working directory and not from the directory where the script is located. If the file you're trying to create needs to be on a specific directory, use an absolute path (e.g. /absolute/path/to/myproject/service/constants/global_variables.json).

If you can't know this absolute path, refer to this SO question

edited May 23 '17 at 11:52

Community

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answered Dec 29 '13 at 11:53

svvac

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score 3 · Answer 2 · edited May 23 '17 at 11:59

~~Python doesn't interpret ../, it will look for a directory named ".." in the cwd.~~

You either have to hardcode the path:

with open("/path/to/myproject/service/constants/global_variables.json", 'w') as file_handler:
    content = json.load(file_handler)

Or find the full path to the current executing script:

EDIT: I was wrong, python does interpret "..", what's happening here is that's is start is the cwd not your script.

$ echo 'Hello world' > text_file.txt
$ mkdir test/
$ cd test
$ python
[...]
>>> open('../text_file.txt').read()
'Hello world\n'

score 0 · Answer 3 · answered Dec 29 '13 at 12:03

Are you sure the path is correct?

Assuming that the current path is: ../myproject/gui/modules/energy/configuration

The path that you mention is:

"..(a)/..(b)/..(c)/..(d)/..(e)/service/constants/global_variables.json"  

(a) = energy/
(b) = modules/
(c) = gui/
(d) = myproject/
(e) = ../

your services directory is I think in the myproject directory not in the one before it. Not sure if that is your question ... Is that your question?

score 0 · Answer 4 · answered Dec 29 '13 at 13:02

What you can do is find current file path and script directory path from that path, like this

dir = os.path.dirname(__file__)

And then you can add or join the place where you want to create file respective to this path

jsonfilepath = "../../../../../service/constants/global_variables.json"

reljsonfilepath = os.path.join(dir, jsonfilepath)

f = open (reljsonfilepath, 'w')

Please check as this is untested code.

score 0 · Answer 5 · answered Dec 29 '13 at 15:19

For the given below Project Structure:

.
`-- myproject
    |-- gui
    |   `-- energy
    |       `-- configuration
    |           `-- test.py
    `-- services
        `-- constants
            `-- out.txt


import os

## Finding absolute path of the current module
drive, tcase_dir = os.path.splitdrive(os.path.abspath(__file__))

## It's good if we always traverse from the project root directory
## rather than the relative path
## So finding the Project's root directory
paths = tcase_dir.split(os.sep)[:-4]
base_dir = os.path.join(drive,os.sep,*paths)


## Known Sub-Directories
SERVICES_DIR = r'services'
CONSTANTS_DIR = r'constants'

## absolute path to the ../myproject/service/constants/ directory
constants_abs_path = os.path.join(base_dir, SERVICES_DIR, CONSTANTS_DIR)

with open(os.path.join(constants_abs_path, r'out.txt'), 'r') as fp:
    ## Do the file Operations here ##

create a file outside the directory in python

5 Answers5