I am writing a app for django which i am planning to publish. This app requires a Bolean Setting variable CONSUMER_REGISTRATION.
Aim of getting this variable is to decide whether to define ConsumerRegistrationModel or not.
This is what I did.
from django.db import models
from django.conf import settings
if getattr(settings, 'CONSUMER_REGISTRATION', False):
class ConsumerRegistration(models.Model):
...
Its working fine. The only Issue i am facing that developers will need to run makemigrations and migrate commands each time they change the variable in settings.
1- Can this work be automated ?. So if they change the variable then some code in django auto run the makemigrations and migrate commands.
2- Or is it perfectly fine to leave this work on developers ??
3- Also I want to ask that is it a good aproach to do this in django ?
The accepted answer doesn't really provide a way to do what the OP is asking, which is to conditionally declare a model.
People may have various reasons for wanting to do this, from not declaring a model at all, to declaring models differently based on settings (it is implied that if you are doing this: you are intend to run the same code base in different places using different settings).
One solution is to put the model in its own app, and conditionally include the app based on a setting:
# Set this your per-project settings:
CONSUMER_REGISTRATION = True
# Set this in the generic settings
INSTALLED_APPS = [...]
if CONSUMER_REGISTRATION:
INSTALLED_APPS.append('consumer_registration') # Models will be loaded.
There's nothing wrong with creating an app which just contains a model.
With regards to "automating" it, the table will be created if migrations are run when the setting is true. It will not delete the table if it is changed to false.
You could simply define the model without any conditionals and tweak your app logic so that instances of ConsumerRegistration model are only interacted with (i.e. created, updated etc.) when the 'CONSUMER_REGISTRATION' flag is set to True.
Running migrations every single time the value of 'CONSUMER_REGISTRATION' is changed would make much more mess than leaving ConsumerRegistration table empty.
As indicated by #dahrens, you could isolate the ConsumerRegistration model along with relevant logic in a separate app, which would only be installed as needed by developers.
Related
TL;DR:
Is it possible from within a django project to detect which command is currently running?
DETAILS:
We have a Django project (running 3.2.7) running which migrations are a little mixed up.
There is a config model which has been developed way after the initial migrations has been done. This config model is basically a Django ORM model mapped to a DB object and has some caching features (which I will not go into details).
The actual code running on servers never experience any problems, since config model is already provisioned on DB. But right now I'm trying to dockerize the whole project and I need to apply migrations from scratch. When I try to apply the migrations to a fresh db,
It reads settings.py, which contains ROOT_URLCONF variable, that is being directed to another .py file
this .py file contains urlpatterns list, which contains references to various django models
Some of these models includes reference to the config model that I've mentioned earlier. When they are imported, those models are trying to pull configs from this config model, which is not initialized yet on the DB, causing psycopg2.errors.UndefinedTable: relation "core_config" does not exist error.
Model is so deeply integrated in the system so I cannot modify the depending models to use anything non dependent on this model.
My main idea is to detect if makemigrations command is running and returning an empty urlpatterns object if this is the case.
Not sure if this is the best approach, but this is the best I've been able to come up with. If you have any other solution, I'm all ears.
So... I've done a lot of research on this... there are answers, but not complete or appropriate answers. I have an in-use and in-production django "project" in which the "main" application is called "pages" ... for reasonably dumb reasons. My problem is now to add mezzanine ... which has a sub-module mezzanine.pages (seems to be required .... but I'm pretty sure I need it).
mezzanine.pages apparently conflicts with "pages" ...
Now ... my pages contains a slew of non-trivial models including one that extends user (One-to-One ref), and many references to other app's tables (fortunately only outbound, ForeignKey). It also has management/commands and about 20 migrations of it's own history.
I gather I either have to changes pages to mypages or is there another route (seemingly changing mezzanine.pages seems wrong-headed).
for reference, The project is on Django 1.8 right now, so the preferred answer includes migrations.
I've worked on this since I posted it, and the real answer is what I've synthesized from multiple sources (including other stack exchange posts).
So... Everything changed in Django before I started using it. After 1.7, the 'migrations' bit was internalized and posts including the word "South" are about how the world was before 1.7. Further, the complication in my case dealt with the issue of migrations in that the project was already active and had real data in production.
There were some posts including a GITHub chunk of code that talked about migrating tables from one App to another App. This is inherently part of the process, but several posts noted that to do this as a "migration" you needed the Migration.py to be in another App. Maybe even an App created for the purpose.
In-the-end, I decided to approach the problem by changing the label in the Application class of apps.py in the application in question. In my case, I am changing "pages" to "phpages" but the directory name of my app is still pages. This works for me because the mezzanine app's "pages" sub-App is back in the python library and not a conflict in the filesystem. If this is not your situation, you can solve it with another use of label.
So... Step-by-step, my procedure to rename pages to phpages.
Create apps.py in the pages sub-directory. In it put:
class PagesConfig(AppConfig):
name = "pages"
label = "phpages"
verbose_name = "Purple Hat Pages"
Key among these is label which is going to change things.
In __init__.py in the pages sub-directory, put default_app_config = "pages.apps.PagesConfig"
In your settings.py change the INSTALLED_APPS entry for your app to 'pages.apps.PagesConfig', ...
All of your migrations need to be edited in this step. In the dependencies list, you'll need to change 'pages' to 'phpages'. In the ForeignKeys you'll need to also change 'pages.Something' to 'phpages.Something' for every something in every migration file. Find these under pages/mitrations/nnnn_*.py
If you refer to foreign keys in other modules by from pages.models import Something and then use ForeignKey(Something), you're good for this stop. If you use ForeignKey('pages.Something') then you need to change those references to ForeignKey('phpages.Something'). I would assume other like-references are the same.
For the next 4 steps (7, 8, 9 and 10), I built pagestophpages.sql and added it to the pages sub-directory. It's not a standard django thing, but each test copy and each production copy of the database was going to need the same set of steps.
UPDATE django_contecnt_type SET app_label='phpages' WHERE app_label='pages';
UPDATE django_migrations SET app='phpages' WHERE app='pages';
Now... in your database (my is PostgreSQL) there will be a bunch of tables that start with "pages". You need to list all of these. In PostgreSQL, in addition to tables, there will be sequences for each AutoField. For each table construct ALTER TABLE pages_something RENAME TO phpages_something; For each sequence ALTER SEQUENCE pages_something_id_seq RENAME TO phpages_something_id_seq;
You should probably backup the database. You may need to try this a few times. Run your SQL script through your database shell. Note that all other changes can be propagated by source code control (git, svn, etc). This last step must be run on each and every database.
Obviously, you need to change pages and phpages to your stuff. You may have more than one table with one auto field and it may not be named something.
Another thing of note, in terms of process, is that this is probably a hard point in your development where everything needs be in sync. Given that we're playing with editing migrations and changing names, you need a hard stop in development so that everything that's going to be changed (dev box, test box, staging box, production box ... and all of their databases) is at the same revision and schema. YMMV.
This is also solving the problem by using the label field of class Application. I choose this method in deference to changing the directory name because it involved fewer changes. I chose not to change the name field because that did not work for me. YMMV.
I must say that I'm a little disappointed that myapp/pages conflicts with mezzanine.pages. It looks like some of the reasons are due to the pages slug being used in the database table name (and off top of my head, I don't see a good solution there). What I don't see that would make sense is the equivalent to "from mezzanine import pages as mpages" or somesuch. The ability to alias imported apps (not talking about apps in my own file tree). I think this might be possible if I sucked in the app into my own file tree --- but this doesn't seem to be a sanctioned act, either.
I like the Django ORM. It's simple, easy to use, and reasonably powerful.
I'm currently developing some internal sites for the VFX company I work for, for which I've used Django. In the meantime, we are developing other python applications and libraries to be used in various contexts in production. There's a number of places in which our core library needs to be interacting with some databases, and using an ORM like Django would really help things. I'm aware of other options like SqlAlchemy or PeeWee, but I'd like to see if Django will work since I use it on the websites and I like its API better.
Using Django as an ORM in a library is tricky (as I explored in a previous question), because Django expects to be used as a website with "apps". In a library, I might want to define any number of data models, which would exist in appropriate places in the library but not inside any Django app (as we're not using any other parts of the framework). So far so good.
I can create a baseclass for my models anywhere in the library as follows:
from django.db import models
from django.apps import apps
import django.conf
django.conf_settings.configure(
DATABASES = ...
)
apps.populate((__name__,))
class LibModel(models.Model):
class Meta:
abstract = True
app_label = __name__
Then anywhere in the library I can create my own models with this baseclass. Since I'm not relying on the "app" for the database names, I need to state them explicitly.
class SpecificModel(LibModel):
# fields go here
class Meta(LibModel.Meta):
db_table = "specific_model_table_name"
This gets around my concern of having to simulate the structure of an "app". The name property in the base class supplies Django with all it needs, and then Django quits whining about not finding an app. The other model files can live wherever they want.
However, there is a glaring use case where this all falls apart. Say that my Django web application wants to use some functionality from the company core python library, which now uses the Django ORM for various things. Since I make a call to django.conf.settings.configure in the library, Django is going to scream about defining the settings more than once when it tries to run the main application.
So basically, a library using the Django ORM is incompatible with Django. Wonderful.
Is there any way around this? I mean, it's a lovely ORM - is it really this impossible to use in a standalone modular way? Is the Django architecture utterly singleton in nature, making this impossible?
*Not a duplicate
I'm trying to have a company python library that uses Django as an ORM. Some of the things that could depend on it might be Django websites themselves. How do I get around Django's singleton insistence on only setting the settings config once? Or is it possible? None of these answers address this!
You can check if django has already been configured.
from django.apps import apps
from django.conf import settings
if not apps.ready:
settings.configure()
django.setup()
When starting Django application - core python library can be configured as separate app an be loaded on startup.
Also, check this answer on dynamic app loading at runtime.
A simple answer is how to initialize Django in a standalone application and do it compatible with Django applications.
import os
import django
if not 'DJANGO_SETTINGS_MODULE' in os.environ:
os.environ['DJANGO_SETTINGS_MODULE'] = 'mysettings'
# # or without DJANGO_SETTINGS_MODULE directly
# from django.conf import settings
# settings.configure(DATABASES=... other...)
django.setup()
# this shouldn't be before DJANGO_SETTINGS_MODULE or settings.configure(...)
from myapp.models import MyModel
# this shouldn't be called before django.setup()
queryset = MyModel.objects.filter(...)
This is more compatible with Django then the answer by Oleg Russkin (where a risk of cyclic dependency at django.setup() is possible if the code is called inside inside a setup() initiated by another similar code or a normal Django project started by manage. It is similar to manage.py where django.setup() is also called internally by execute_from_command_line(sys.argv). The setup initializes all modules related to INSTALLED_APPS all urls modules and consequently all views etc. Many of them are lazy, but still. If any code called by setup() depends on this then neither the condition not apps.ready doesn't help. The setup() is not reentrant and the startup fails.)
Much more general answer
An important concept of Django is to support writing reusable parts of code ("applications" in Django terminology, that can be developed and tested independently. There can be also a tree of dependencies, but uncontrolled mutual dependencies should be avoided if possible) Reusable applications are expected that they can be easier combined to whole project ("project" in Django terminology is with all settings necessary to run it by Python.)
The only unavoidable and useful "singleton" in Django ORM are database connections django.db.connections and django.conf.settings especially INSTALLED_APPS. Only one connection should be used to the same database from the same process or thread.
Django is very configurable. An extreme example: It is possible to write a single file project where all code like settings, models, URL configs and views is defined in one file. That extreme that is probably useful only for some special tests or very short demos or as an exercise. It is even possible to define a project by one file with two "reusable" applications :-)
Django supports also "legacy databases" where the database structure is shared with existing non Django applications and models can be created by inspectdb command. Table names in such models are explicit and don't contain the app name. On the other hand the app name prefix is useful to prevent a possible conflict of the same table names from independent "applications". An important decision is if you can use it as a "legacy" database or a normal Django database.
You can decide between following two solutions or to combine them:
Use e.g. foo_models or bar.models and import all models from them e.g. to app.models and add only that "app" to INSTALLED_APPLICATIONS. This can be viable if it is only for one company and never otherwise and central name assigment is possible. (easiest but little naive)
Use some coarse separation of namespaces to several apps. You should probably use not more than one app with simple names without app name prefix.
Think ahead about migrations**. They will be probably very complicated and very soon impossible if you will create later more projects for the same database and different subsets of database tables without separating them to more apps and without app namespace.
There is really no "singleton" in Django ORM except of django.db.connections itself. If you use more databases you can direct some tables to a specific database by DATABASE_ROUTERS, even with two different models that use the same table name without a prefix.
I added some models in my models.py and I want to add an admin class to use a wysiwyg-editor in text-fields.
Well, I know that Django itself doesn't support migrations and I've used South, but it doesn't work either.
South doesn't "see" the change.
Could it be, that South just detects changes to fields, but not if I add a new class?
How can I tweak Django to detect such changes?
syncdb and South are only concerned with descendants of Model in apps listed in INSTALLED_APPS. Everything else is handled by Django directly.
You seem to be very confused, unfortunately. Of course Django reads the code in models.py - otherwise what would be the point of it? Django uses that code initially to define the model SQL when doing syncdb, but it doesn't modify existing database tables in subsequent calls to syncdb - hence the need for South.
But naturally, Django uses models.py and admin.py and all the other Python code to define its own configuration and state. (And note that admin classes are not defined in models.py but in admin.py.)
If you are not seeing changes, you will need to restart your server.
I'm fairly sure that if you follow the steps as outlined in the tutorial to create an admin app it'll just work. Migration isn't an issue as the admin app creates new tables rather than altering the existing one.
I need a place to run an initialization code that is application specific (like connecting to signals).
When I put the code to __init__.py module of an application I've ended up with a circular import of the models.
Is there a way to fire a function when the framework is setup and before any request is executed?
I use quite old version of django 96.6, but I'm also interested in the solutions for the current version.
Regarding the duplication of other questions:
Here is how the question differ from the duplicates suggested by S.Lott in comments:
Correct place to put extra startup code in django?
Django need to be fully initialized when the function is ran. So code in manage.py won't work.
Where should I place the one-time operation operation in the Django framework?
The function initialize the connection between my applications. So the code must be ran in each thread that will actually handle the requests.
Comments to current solutions:
I can't use urls as most of my apps don't have any urls exposed. They just listen to signals and store additional information in the database.
Signals, specifically, are recommended to be put in the models.py of your app.
Try models.py or urls.py and let us know if you have any luck.
The best place for stuff like this... anywhere, just import it in your urls.py file (for obvious reasons urls are loading before any requests).
If you don't provide urls, then you really need to put it in models.py, that's just the way it is.
Now, on to your problems: You want to define it in its own module, great, do that. To avoid a circular import, use django.db.models.get_model to return the model dynamically for you. You can provide an initialisation function for your signals module to import the relevant model and connect the relevant signals. This function would then be called at the end of models.py, being run only ever once and after your model is initialised.
There's still a chance that this wont work (if the models aren't yet ready when you set it up), but give it a try and let us know.
For me, the following works:
In init.py:
from . import models
from . import signals
signals.py imports from models, but not vice versa. signals.py contains module code that is run immediately when it is imported and is thus run when the django server starts.