Accessing django database from python script - python

I'm trying to access my Django database from within a regular Python script. So far what I did is:
import os
import django
from django.db import models
os.environ.setdefault("DJANGO_SETTINGS_MODULE", "m_site.settings")
django.setup()
from django.apps import apps
ap=apps.get_model('py','Post')
q=ap.objects.all()
for a in q:
print(a.title)
There is an application in Django called py where I have many posts (models.py, which contains class Post(models.Model)).
I would like to have the possibility to access and update this database from a regular Python script. So far script above works fine, I can insert, update or query but it looks like it is a separate database and not the database from Django's py application. Am I doing something wrong or missing something? Any help would be very appreciated.

Consider writing your script as a custom management command. This will let you run it via manage.py, with Django all properly wired up for you, e.g.
python manage.py print_post_titles
Something like this should be a good start:
from django.core.management.base import BaseCommand
from py.models import Post
class Command(BaseCommand):
help = 'Prints the titles of all Posts'
def handle(self, *args, **options):
for post in Post.objects.all():
print(a.title)

Related

Is it possible to get django info without running it

I have a django model, the whole code is completed. but I want to access my model info. a code like this to get field names.
for f in myModel._meta.fields:
print(f.get_attname())
is it possible to do it from an external python script without running django server?
other possible automated ways of doing this and saving results to a file are also appreciated.
try1
because Im using docker I ran it up. and from django container I started python shell
>>> from django.conf import settings
>>> settings.configure()
>>> import models
it gave django.core.exceptions.AppRegistryNotReady: Apps aren't loaded yet.
try2
by #Klaus D advice in comments I tried management command. so I created
users/
__init__.py
models.py
management/
__init__.py
commands/
__init__.py
_private.py
modelInfo.py
structure. in modelInfo.py I did
from django.core.management.base import BaseCommand, CommandError
from users import views2
def savelisttxtfile(the_list, path_, type_='w', encoding="utf-8"):
with open(path_, type_, encoding=encoding) as file_handler:
for item in the_list:
file_handler.write("{}\n".format(item))
class Command(BaseCommand):
def handle(self, *args, **options):
dic=[]
for f in views2.ChertModel._meta.fields:
print(f.get_attname())
dic.append(f.get_attname())
savelisttxtfile(dic,"F:\projects\sd.txt")
and from another python file I tried
os.chdir(r'F:\projects\users\management\commands')
from subprocess import run
import sys
run([sys.executable, r'F:\projects\users\management\commands\modelInfo.py'])
and it returned
CompletedProcess(args=['C:\\ProgramData\\Anaconda3\\python.exe', 'F:\projects\users\management\commands\modelInfo.py'], returncode=1)
and the results were not save in sd.txt
thanks to #klaus D and management command documentation I made this structure
users/
__init__.py
models.py
management/
__init__.py
commands/
__init__.py
_private.py
modelInfo.py
and in modelInfo.py I did
from django.core.management.base import BaseCommand, CommandError
from users import views2
def savelisttxtfile(the_list, path_, type_='w', encoding="utf-8"):
with open(path_, type_, encoding=encoding) as file_handler:
for item in the_list:
file_handler.write("{}\n".format(item))
class Command(BaseCommand):
def handle(self, *args, **options):
dic=[]
for f in views2.ChertModel._meta.fields:
print(f.get_attname())
dic.append(f.get_attname())
savelisttxtfile(dic,"F:\projects\sd.txt")
and to run it I went to manage.py location and executed python manage.py modelInfo to launch it.
Regarding your "try1" it seems to be a little bit trickier to start a python shell like python manage.py shell than what you propose there.
Fortunately you can do this:
python manage.py shell < your_script.py
and your script will be executed as if typed directly into the "django shell". Keep in mind that you still need to import your models relative to your project, i.e. from myapp.models import mymodel.

Automatically create django model instances at startup with empty database

My django project requires some model instances to be created at startup if they do not exist.
I currently create the required model instances I need in an app config.
class MyAppConfig(AppConfig):
name = 'my_app'
def ready(self):
create_required_objects()
def create_required_objects():
from my_app.models import MyObject
for name in MyObject.reserved_names:
if not MyObject.objects.filter(name=name).exists():
MyObject.objects.create(name=name, not_editable=True)
This works perfectly when the sqlite database is initialized, however if I clear the database and then try to run the sever, I get the following error:
django.db.utils.OperationalError: no such table: my_app_object
I would like to be able to clear the database (preferably by just removing db.sqlite3) and run the server.
Use post_migrate signal to create a new instance when migrating to the new database:
Like:
from django.db.models.signals import post_migrate
from my_app.models import MyObject
def create_required_objects(sender, **kwargs):
for name in MyObject.reserved_names:
if not MyObject.objects.filter(name=name).exists():
MyObject.objects.create(name=name, not_editable=True)
class MyAppConfig(AppConfig):
name = 'my_app'
def ready(self):
post_migrate.connect(create_required_objects ,sender=self)
This code automatically generates the user after migrating to the database.
You can use model_bakery to populate some temp data. You may need to do makemigrations and migrate to set up all tables in your database, you can follow this workflow.
python manage.py makemigrations
python manage.py migrate
In terms of populating data, you can try the following code:
from model_bakery import baker
from my_app.models import MyObject
baker.make(MyObject)
Add baker.make(MyObject) in your create_required_objects function after installation of model_bakery:
pip install model_bakery

How to access Python class in another folder when building Django management command?

I'm trying to turn a Python script into a Django management command. My script is in an application folder called sites. Folder structure:
project
|--sites
|scanner.py
|--management
|__init__.py
|--commands
|__init__.py
|getdeals.py
I'm trying to have getdeals.py run as a management command. It finds objects in my Site model and then uses them to create an instance of the SiteDeals class, which is in the scanner.py file.
getdeals.py:
from django.core.management.base import BaseCommand
from sites.models import Site
class Command(BaseCommand):
help = "Scans all sites for deals"
def handle(self, *args, **options):
site_set = Site.objects.all()
for site in site_set:
scraper = SiteDeals(site)
When I run python manage.py getdeals it says NameError: name 'SiteDeals' is not defined.
I thought of taking the code from handle and writing it as a main() function in scanner.py, and then accessing it from getdeals.py, but can't work out how to access it that way.
How do I access SiteDeals from the scanner.py file, given that it is in another folder from my management/commands folder, so that I can pass my objects to it?
You need to import SiteDeals, just as you import the Site model. Try:
from sites.scanner import SiteDeals

Run python script from django

I added some python file in my django project but that python file did not execute. How can i run that python scrip in django.
OK it seems you want to run a script outside of the HTTP request/response cycle,
I'd recommend you make a Django admin command, because a script will need to e.g. have access to the database environment to update records
from django.core.management.base import BaseCommand
from yourapp.models import Thing
import your_script
class Command(BaseCommand):
help = 'Closes the specified poll for voting'
def handle(self, *args, **options):
# Do your processing here
your_script.process(Thing.objects.all())
With this you can call ./manage.py process_thing and it'll run independently
More on Django admin commands here, https://docs.djangoproject.com/en/1.8/howto/custom-management-commands/
If the processing is triggered programmatically e.g. from user requests,
you'll have to setup a queue and have a task job created for each request, I'd try Celery, more on that here http://celery.readthedocs.org/en/latest/django/first-steps-with-django.html

Where to put Django startup code?

I'd like to have these lines of code executed on server startup (both development and production):
from django.core import management
management.call_command('syncdb', interactive=False)
Putting it in settings.py doesn't work, as it requires the settings to be loaded already.
Putting them in a view and accessing that view externally doesn't work either, as there are some middlewares that use the database and those will fail and not let me access the view.
Putting them in a middleware would work, but that would get called each time my app is accessed. An possible solution might be to create a middleware that does all the job and then removes itself from MIDDLEWARE_CLASSES so it's not called anymore. Can I do that without too much monkey-patching?
Write middleware that does this in __init__ and afterwards raise django.core.exceptions.MiddlewareNotUsed from the __init__, django will remove it for all requests :). __init__ is called at startup by the way, not at the first request, so it won't block your first user.
There is talk about adding a startup signal, but that won't be available soon (a major problem for example is when this signal should be sent)
Related Ticket: https://code.djangoproject.com/ticket/13024
Update: Django 1.7 includes support for this. (Documentation, as linked by the ticket)
In Django 1.7+ if you want to run a startup code and,
1. Avoid running it in migrate, makemigrations, shell sessions, ...
2. Avoid running it twice or more
A solution would be:
file: myapp/apps.py
from django.apps import AppConfig
def startup():
# startup code goes here
class MyAppConfig(AppConfig):
name = 'myapp'
verbose_name = "My Application"
def ready(self):
import os
if os.environ.get('RUN_MAIN'):
startup()
file: myapp/__init__.py
default_app_config = 'myapp.apps.MyAppConfig'
This post is using suggestions from #Pykler and #bdoering
If you were using Apache/mod_wsgi for both, use the WSGI script file described in:
http://blog.dscpl.com.au/2010/03/improved-wsgi-script-for-use-with.html
Add what you need after language translations are activated.
Thus:
import sys
sys.path.insert(0, '/usr/local/django/mysite')
import settings
import django.core.management
django.core.management.setup_environ(settings)
utility = django.core.management.ManagementUtility()
command = utility.fetch_command('runserver')
command.validate()
import django.conf
import django.utils
django.utils.translation.activate(django.conf.settings.LANGUAGE_CODE)
# Your line here.
django.core.management.call_command('syncdb', interactive=False)
import django.core.handlers.wsgi
application = django.core.handlers.wsgi.WSGIHandler()
You can create a custom command and write your code in the handle function. details here https://docs.djangoproject.com/en/dev/howto/custom-management-commands/
Then you can create a startup script that runs the django server then executes your new custom command.
If you are using mod_wsgi you can put it in the wsgi start app
Here is how I work around the missing startup signal for Django:
https://github.com/lsaffre/djangosite/blob/master/djangosite/models.py
The code that is being called there is specific to my djangosite project, but the trick to get it called by writing a special app (based on an idea by Ross McFarland) should work for other environments.
Luc

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