32👍
You need to write a custom management command, instead of doing these weird acrobatics.
Create a module called management (in other words, create a directory management
and inside it create an empty __init__.py
file) inside the directory of any app that you have listed in INSTALLED_APPS
. So if you have myapp
, you would create:
myapp
|
> management
| | __init__.py
> models.py
> views.py
Then in the management directory, create another module commands
and in it create a file which is the name of your command; for example my_command.py
, like this:
myapp
|
> management
| | __init__.py
| | commands
| | | __init__.py
| | | my_command.py
> models.py
> views.py
In my_command.py
, write this boilerplate code:
from django.core.management.base import BaseCommand, CommandError
from myapp.models import MyModel
class Command(BaseCommand):
help = 'Does some magical work'
def handle(self, *args, **options):
""" Do your work here """
self.stdout.write('There are {} things!'.format(MyModel.objects.count()))
Once you save the file, you’ll be able to do python manage.py my_command
and it will have access to all your models and settings.
If you need to run it as a daemon, Daniel Roseman wrote django-initd
which does exactly that. Once you have installed it:
from django.core.management.base import CommandError
from daemon_command import DaemonCommand
from myapp.models import MyModel
class Command(DaemonCommand):
help = 'Does some magical work'
def loop_callback(self, *args, **options):
""" Do your work here """
self.stdout.write('There are {} things!'.format(MyModel.objects.count()))
Once you have that done from the github readme:
The important parts of such a process are these:
* it comes up automatically on server startup
* it logs errors and information to a named location, which is configurable
* if the process dies, it restarts itself straight away
[...]
Run the command as normal, but pass one of --start, --stop or --restart to
work as a daemon. Otherwise, the command will run as a standard application.
11👍
The below script should work provided that the layout of your project (and the path to your settings file) looks like this:
/var/www/cloudloon/horizon/openstack_dashboard/settings.py
#!/usr/bin/env python
import os, sys
sys.path.append("/var/www/cloudloon/horizon")
os.environ["DJANGO_SETTINGS_MODULE"] = "openstack_dashboard.settings"
from django.contrib.auth.models import User
I believe the problem you’re seeing is due to the arrangement of your project, or that you need to append another directory level to the sys.path
call in the script.
Edit:
Looking at your github project, horizon
and openstack_dashboard
are on the same directory level. What you’d want to do is set your sys.path
to one level higher:
sys.path.append("/var/www/cloudloon")
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5👍
This worked for me:
from django.conf import settings
import myfolder.settings as app_settings
settings.configure(INSTALLED_APPS=app_settings.INSTALLED_APPS,DATABASES=app_settings.DATABASES)
import django
django.setup()
from myapp123.models import Sample
for s in Sample.objects.all():
print(s)
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1👍
My solution (inspired by http://blog.gabrielsaldana.org/using-django-models-in-external-python-scripts/):
import os
import sys
sys.path.append( os.path.join(os.path.dirname(__file__), 'MyDjangoProject'))
os.environ.setdefault("DJANGO_SETTINGS_MODULE", "MyDjangoProject.settings")
import django
from django.conf import settings
django.setup()
from myapp.models import MyModel
for s in MyModel.objects.all():
print(s)
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