I am trying to execute a celery task that among other things, communicates with a specific Django view with a pipe.
I have been trying all day to import the celery tasks file (tasks.py) from the Django views (views.py) without any success.
I have also checked the file permissions but this was not the case.
I have added to sys.path the path of the file desired to import (tasks.py), and then import it, but I keep getting an ImportError. However, when trying to import another script on the same folder (script1), the import succeeds.
views.py
...
sys.path.append('/home/celery')
import script1 #SUCCEEDS
# Trying to import "/home/celery/tasks.py" here
import tasks #FAILS
...
tasks.py
...
#app.task
def start_operation(client,**kwargs):
# Pipe comes here
...
The celery directory structure is the following:
/home
├── celery
├── script1.py
tasks.py
And the Django proyect directory structure:
/myapp
├── myapp
├── views.py
Thanks in advance.
Related
I am working on a Django app in which I want to import some class/function from generator.py into my views.py to process an user-submitted input. My folder structure looks like this:
project/
models/
__init__.py
generator.py
web/
django/
__init__.py
settings.py
urls.py
wsgi.py
django_app
__init__.py
views.py
etc.
Inside views.py, I have
from ...models.generator import Generator
When I try to run server, what I get is:
ValueError: attempted relative import beyond top-level package
I've seen many answers, but most are about manipulating sys.path or changing PYTHONPATH. I'm not sure how and where to do either cause I'm rather new to Django.
Can someone tell me exactly which commands to run to allow the import to be done?
Import of python is based on sys.path and cannot be outside the top-level dir. More information.
That's mean, if you append top of BASE_DIR path in sys.path you can import. But it's a tricky way.
def import_function():
import sys
sys.path.append(os.path.dirname(BASE_DIR))
module = __import__('models.generator') # import code
sys.path.pop()
return module
g_module = import_function()
g_module.Generator
For some reason when I try to execute my custom command:
flask create_user
I get an error whenever I try import anything from my src folder:
File "/app/backend/flask_app/command.py", line 17, in seed_db
import src.models
ModuleNotFoundError: No module named 'src'
/
__init__.py
command.py
run.py
src/
__init__.py
models/
__init__.py
user_model.py
command.py
import click
from flask import Flask
from flask.cli import with_appcontext
app = Flask(__name__)
#app.cli.command()
#with_appcontext
def create_user():
import src.models as models
print("hello")
app.cli.add_command(seed_db)
I found the solution.
I had to delete the root dir __init__.py file.
I found this answer from the famous Miguel Grinberg:
https://github.com/miguelgrinberg/flasky/issues/310#issuecomment-340641813
These files (__init__.py) make python believe the main directory of the package is one directory above, so that is the directory that is added to the Python path. Usually you want your top-level directory to not be a Python package, so that the current directory goes to the path.
__
It is a matter of python way of import modules, not about Flask command.
Since command.py file is at the same level of the models folder you can just write:
import models.user_models as users
After that you can acces the user_models module from users reference:
import models.user_models as users
users.some_function_name()
I have a Django project (Python 2.7.15) with the following structure:
mysite/
manage.py
mysite/
__init__.py
settings.py
urls.py
wsgi.py
polls/
__init__.py
admin.py
apps.py
migrations/
__init__.py
models.py
tests.py
views.py
utils.py
utils/
__init__.py
filters.py
In my utils/filters.py file I have a class MyFilter. From polls/admin.py, however, when I try to run from utils.filters import MyFilter, I get ImportError: No module named filters. How can I import my custom filter inside the polls app without renaming the polls/utils.py module or the utils package?
NOTE: This it's not a circular import problem. This happens even if I don't import anything from utils/filters.py. It's a name conflict between utils/ and polls/utils.py. Python tries to find filters.MyFilter inside polls/utils.py and it doesn't find it so it throws the error. I just want to figure out a way to bypass this conflict and force python to look for filters.MyFilter inside the utils/ package in the project root.
In Python 2, import utils is ambiguous because it can be a relative or an absolute import.
If you enable the Python 3 behaviour by adding the following import to the top of your module,
from __future__ import absolute_import
then from utils.filters import MyFilter will be treated as an absolute import and will work.
Once you have added the future import, you would need to use an explicit relative import import .utils if you wanted to import polls/utils.py from polls/admin.py.
Under my Django project I have created a directory for an aiohttp service.
1) How is the best way to structure it?
This is my current structure:
myproject/
myservice/
__init__.py
service.py
utils.py
myproject/
__init__.py
settings.py
urls.py
uwsgi.py
manage.py
2) If my service needs to import some settings from myproject.settings, how can I do it? Should I move service.py under the root?
I get:
ImportError: No module named 'myproject'
Your server.py script, which uses aiohttp should be next to your manage.py file.
You should probraly also add:
import os
os.environ.setdefault('DJANGO_SETTINGS_MODULE', 'myproject.settings')
from django import setup
setup()
I have a django application I've added celery. In django application I have a package named 'parser', 'api'. I configured the celery as I followed the following tutorial: First steps with Django. In parser package I have 'models.py'. Do you 'task.py' package 'api'. When I try to do 'from parser import models' in api package . I get the following error: No module named models
I looked and found that the following import file: lib/python2.7/lib-dynload/parser.x86_64-linux-gnu.so
webapp/
manage.py
api/
__init__.py
models.py
views.py
tasks.py
...
parser/
__init__.py
models.py
views.py
...
settings/
__init__.py
base.py
celery.py
dev.py
live.py
local.py
urls.py
wsgi.py
In case I need 'models.py' of parser package. Command you use to start the celery is following: celery -A settings worker --loglevel=info. When I run celery in manage.py then take the right file: python manage.py celery -A settings worker --loglevel=info
api/task.py
from __future__ import absolute_import, division, print_function, unicode_literals
import time
from celery import task
from parser.models import FileUploadProcess # Error import
#task()
def test_task(param1):
print("Test task called. Param: {}".format(param1))
return 42
#task()
def parse_file(file_candidate, candidate_id):
FileUploadProcess(candidate_id=candidate_id, is_process=True).save()
# parse file
time.sleep(15)
FileUploadProcess.objects.filter(candidate_id=candidate_id).update(is_process=False)
Can somehow tell me Imports right package?
'from parser import models'
You need to use is so:
from parser.models import ClassName
where ClassName is name of class you want to import
or just
import parser.models as models