My models won't show up in the admin panel, and I suspect it's because admin.py isn't being discovered.
Also, all the models work as expected in the application; I can import two.MyModelB and use it in the shell, in my web-app, etc.
root/settings.py:
INSTALLED_APPS = [
'django.contrib.admin',
'one',
'two', etc]
root/urls.py:
from django.contrib import admin
print("### Running admin-autodiscover ###")
admin.autodiscover()
Note that I don't usually use admin.autodiscover(), but it was a suggestion I came across when I was searching for solutions to this. Including it (or not) makes no difference either way.
root/admin.py:
from django.contrib import admin
from django.conf import settings
from one.models import MyModelA
from two.models import MyModelB
print("### This should be discovered! ###")
class MyModelAdmin(admin.ModelAdmin):
readonly_fields = ('id',)
# Register your models here.
admin.site.register(MyModelA, MyModelAdmin)
admin.site.register(MyModelB, MyModelAdmin)
superuser confirmed:
console output:
(env) PS E:\Django> python manage.py runserver
Watching for file changes with StatReloader
Performing system checks...
### Running admin-autodiscover ###
System check identified no issues (0 silenced).
Django version 4.1.5, using settings 'root.settings'
Starting development server at http://127.0.0.1:8000/
The statement from urls.py is output, but not the one from admin.py. I assume that indicates there's some malfunction with django.contrib.admin, but it's far beyond my knowledge.
Curiously, making admin.py files elsewhere ... cause them to be discovered:
root/random/folder/admin.py:
from django.contrib import admin
print("Is this one discovered?")
console:
(env) PS E:\Django> python manage.py runserver
Is this one discovered?
Is this one discovered?
Watching for file changes with StatReloader
Performing system checks...
### Running admin-autodiscover ###
System check identified no issues (0 silenced).
Django version 4.1.5, using settings 'root.settings'
Starting development server at http://127.0.0.1:8000/
So Django finds that admin.py - twice, once automagically and once because I explicitly call admin.autodiscover(). But the admin.py sitting literally right next to settings.py goes ignored.
I can presumably work around this by moving my model declarations into some other random folder's admin.py ... but I would really like to keep it where it is, because my project is large and I don't want to put core settings in obscure locations (if I can help it).
Related
Under my Django project there are a few apps and all of them have unit tests. One of them that I'm working right now is supposed to be included only in dev/stage environments, so I'm enabling it using a environment variable.
When this variable is present it is added to INSTALLED_APPS and it is working just fine, the problem is that Django is executing the tests for this app even when it is not in INSTALLED_APPS, and it fails with the following message:
ImportError: Failed to import test module: debug.tests.unit.test_services`
...(traceback information)...
RuntimeError: Model class debug.models.Email doesn't declare an explicit app_label and isn't in an application in INSTALLED_APPS.
When I define the app_label in the class Meta of models in this app the error is different, it says it can't find a table, I assume that this is because the app is not in INSTALLED_APPS, so it's migrations are not executed.
OperationalError: no such table: debug_email
I'm not sure why Django executes the tests for all apps, but not it's migrations.
Am I missing something from Django configuration for tests?
https://docs.python.org/3/library/unittest.html#unittest.TestLoader.discover says:
If load_tests exists then discovery does not recurse into the package, load_tests is responsible for loading all tests in the package.
So in the lowest __init__.py in your app which you don't always want run:
from django.apps import apps
def load_tests(loader, tests, pattern):
from django.conf import settings
if apps.is_installed("your_dev_app"):
# Actually load the tests - thanks to #barney-szabolcs
return loader.discover(start_dir=dirname(abspath(__file__)), pattern=pattern)
You need to return the discovered tests in load_tests.
So, adding to #DaveLawrence's answer, the complete code is:
# your_dev_app/__init__.py
from django.apps import apps
from os.path import dirname, abspath
def load_tests(loader, tests, pattern):
"""
loads tests for your_dev_app if it is installed.
"""
from django.conf import settings
if apps.is_installed("your_dev_app"):
return loader.discover(start_dir=dirname(abspath(__file__)), pattern=pattern)
When you run:
python manage.py test
the command will look per default recursive for all files with the pattern test*.py in the working directory. It isn't affected by INSTALLED_APPS in settings.py.
You can specify a certain app to test it:
python manage.py test app_label
or specify a path:
python manage.py test myapp/tests
If you want to exclude some tests you can tag them and use the option --exclude-tag.
Run python manage.py test --help to get information on all options.
The official documentation gives a lot of information on the different possibilities how to run the tests.
EDIT:
If you have apps that are required only in the development environment, but not in the production, you could split your settings.py. One possible solution would be to outsource all development settings into a file local_settings.py and exclude it from versioning or from the production branch, i.e. don't push it in the production environment.
local_settings.py
DEBUG = True
INSTALLED_APPS += (
# Django Debug Toolbar would be for example
# used only in development
'debug_toolbar',
'your dev app',
)
settings.py
try:
from .local_settings import *
except ImportError:
pass
I've been following the DjangoProject tutorial. When I run python manage.py startapp newapp while in the same directory as manage.py. In the newapp directory I see init.py, models.py, tests.py, and views.py but not admin.py file. Where is admin.py?
I am running Django 1.5.1 in Windows 8
You have to create an admin.py file.
you don't necessarily need an admin.py file,
just import the admin module in your models.py file,
from django.contrib import admin
and for each model do the following:
admin.site.register(model1)
admin.site.register(model2)
However, this is not best practice, but since it's just a tutorial, it will work.
You also need to uncoment the relevant lines in the urls.py file
I think I had the same frustrations following the DjangoProject tutorial - however, when I cross-referenced it with with the DjangoBook tutorial (for the same version, I believe, 1.5.1), I found that an admin.py file was not necessarily created after a python manage.py startapp xyz command -- moreover, I also uncommented all of the admin options in urls.py, views.py, and settings.py - so a bit of a mix of what Neal and Ibrahim said
You have to create your own admin.py file in the app if you want it. Indeed, this file is optionnal and isn't created by startapp.
If you want a default template to begin your admin.py, it should be:
from django.contrib import admin
from models import Model1, Model2
class Model2Admin(admin.ModelAdmin):
list_display = ('title', 'content', 'date')
# Just an example, chekc docs and tutorials for more info.
admin.site.register(Model1)
admin.site.register(Model2, Model2Admin)
The reason there is no default admin.py is because you don't have any models yet when you create your new application; so there is nothing to add to the admin section.
Further, you may not want to admin all the models in your application; or you may be creating an application that does not need any admin hookups; or you may not be using the admin application at all in your project.
Since django cannot decide this for you, there is no default admin.py generated.
To create one, if you are following the tutorial - simply keep reading and in part two you'll create the admin.py file when you learn about the admin contrib app and how to integrate it with your custom models.
(Updating my question with more information.)
My django app is running fine on my dev server.
I have a view that pulls from the database using the below line that works fine:
from myapp.models import MyTable
However, if I add the above 'from/import' to another module (see below structure, it's the module named 'problem_module.py') I'm writing where I want to pull from the sqlite3 database, I get this error.
raise ImportError("Could not import settings '%s' (Is it on sys.path?): %s" % (self.SETTINGS_MODULE, e))
ImportError: Could not import settings 'myfolder.settings' (Is it on sys.path?): No module named myfolder.settings
I've read and tried various solutions recommended when people get this error, but I missing something because i'm unable to resolve it.
I'm using Django 1.4 and have the lay-out as recommended.
mysite/
manage.py
mysite/
__init__.py
settings.py
urls.py
wsgi.py
myapp/
__init__.py
models.py
admin.py
views.py
indevelopment/
__init__.py
problem_module.py
I figured out what was happening and why after going through the traceback carefully and looking at the django source code. Here is what happens.
When you run:
python manage.py runserver
the environment variable gets set properly assuming you already changed this small little file or just don't pay attention to it because django 1.4 automatically configures it for you.
os.environ.setdefault("DJANGO_SETTINGS_MODULE", "myapp.settings")
However, because this setting of os.environ is under a:
if __name__ = "__main__"
expression, it only gets run if call that file directly, as you do with:
python manage.py runserver
Otherwise, if you are running a file that needs that environment variable - say testing a module in Eclipse - , the os.environ needs to get set in another place (shell, etc).
All the that I got generally pointed to this but I needed the context.
But as a little adjustment (yes, not a good idea as it couples) on the source code you can also hardcode it in manually in/django/conf/__init__.py
Specifically to see where it happens, the change below works:
# in module: /django/conf/__init__.py
class LazySettings(LazyObject):
def _setup(self):
try:
# Comment out the call to os.environ and hardcode in your app settings
# settings_module = os.environ[ENVIRONMENT_VARIABLE]
# WARNING: bad practice to do this. ;.
settings_module = "myapp.settings"
Have you changed/set DJANGO_SETTINGS_MODULE?
Try export DJANGO_SETTINGS_MODULE=mysite.settings and start your dev server.
modify your manage.py:
os.environ.setdefault("DJANGO_SETTINGS_MODULE", "mysite.settings")
PyCharm sometimes override DJANGO_SETTINGS_MODULE to empty string. Try to debug your manage.py and see if it realy changes after setdefault() call.
If its not either change pycharm settings or use os.environ['DJANGO....']='my_settings'..
or hack files at .idea/. .idea/workspaed.xml contains
env name="DJANGO_SETTINGS_MODULE" value="" in this case
I'm currently running some Django tests and it looks that DEBUG=False by default. Is there a way to run a specific test where I can set DEBUG=True at the command line or in code?
For a specific test inside a test case, you can use the override_settings decorator:
from django.test.utils import override_settings
from django.conf import settings
class TestSomething(TestCase):
#override_settings(DEBUG=True)
def test_debug(self):
assert settings.DEBUG
Starting with Django 1.11 you can use --debug-mode to set the DEBUG setting to True prior to running tests.
The accepted answer didn't work for me. I use Selenium for testing, and setting #override_settings(DEBUG=True) makes the test browser always display 404 error on every page. And DEBUG=False does not show exception tracebacks. So I found a workaround.
The idea is to emulate DEBUG=True behaviour, using custom 500 handler and built-in django 500 error handler.
Add this to myapp.views:
import sys
from django import http
from django.views.debug import ExceptionReporter
def show_server_error(request):
"""
500 error handler to show Django default 500 template
with nice error information and traceback.
Useful in testing, if you can't set DEBUG=True.
Templates: `500.html`
Context: sys.exc_info() results
"""
exc_type, exc_value, exc_traceback = sys.exc_info()
error = ExceptionReporter(request, exc_type, exc_value, exc_traceback)
return http.HttpResponseServerError(error.get_traceback_html())
urls.py:
from django.conf import settings
if settings.TESTING_MODE:
# enable this handler only for testing,
# so that if DEBUG=False and we're not testing,
# the default handler is used
handler500 = 'myapp.views.show_server_error'
settings.py:
# detect testing mode
import sys
TESTING_MODE = 'test' in sys.argv
Now if any of your Selenium tests encounters 500 error, you'll see a nice error page with traceback and everything. If you run a normal non-testing environment, default 500 handler is used.
Inspired by:
Where in django is the default 500 traceback rendered so that I can use it to create my own logs?
django - how to detect test environment
Okay let's say you want to write tests for error testcase for which the urls are :-
urls.py
if settings.DEBUG:
urlpatterns += [
url(r'^404/$', page_not_found_view),
url(r'^500/$', my_custom_error_view),
url(r'^400/$', bad_request_view),
url(r'^403/$', permission_denied_view),
]
test_urls.py:-
from django.conf import settings
class ErroCodeUrl(TestCase):
def setUp(self):
settings.DEBUG = True
def test_400_error(self):
response = self.client.get('/400/')
self.assertEqual(response.status_code, 500)
Hope you got some idea!
Nothing worked for me except https://stackoverflow.com/a/1118271/5750078
Use Python 3.7
breakpoint()
method.
Works fine on pycharm
You can't see the results of DEBUG=True when running a unit test. The pages don't display anywhere. No browser.
Changing DEBUG has no effect, since the web pages (with the debugging output) are not visible anywhere.
If you want to see a debugging web page related to a failing unit test, then do this.
Drop your development database.
Rerun syncdb to build an empty development database.
Run the various loaddata scripts to rebuild the fixtures for that test in your development database.
Run the server and browse the page.
Now you can see the debug output.
I added a model to admin via admin.site.register, and it does not show up in admin. Since admin is so "It just works", I have no idea of how to debug this. Pointers?
After adding and registering your admin:
# app/admin.py
class YourModelAdmin(admin.ModelAdmin):
pass
admin.site.register(YourModel, YourModelAdmin)
Make sure your app is in your project settings.py:
# settings.py
INSTALLED_APPS = (
# other apps ...
'app',
)
Sync your project for that model if you have not done so already:
python manage.py syncdb
Restart your server, CTRL-C:
python manage.py runserver
In such a situation is also a good practice to check if the user logged in to the admin panel has rights to manage such a model. If they do then you could change your code to access the functions as root.
When in doubt, shut down server, syncdb, start server.
I have the experience, that sometimes after changing an admin.py the dev-sever won't be restarted. in that case touch settings.py helps.
I think the checklist in Thierry's answer is almost definitive, but make sure that urls.py contains admin.autodiscover() to load INSTALLED_APPS admin.py modules.
# urls.py
from django.conf.urls.defaults import *
from django.contrib import admin
admin.autodiscover()
urlpatterns = patterns('',
('^admin/', include(admin.site.urls)),
)
More info in the django docs.
Have you added the application to your installed apps? That has happened to me both one and two times. :) Otherwise it would be useful for us to see the code to help you.
Also make sure there are no syntax errors in your admin.py or anything. That can cause an app to fail to be registered with the AdminSite.
I've faced the same problem, but it was a little tricky than yours.
Consider, that you have a project with, say, five or even more apps. For me it is more obvious to register all models in just one admin.py file, so I have decided to do it in one place - core directory. Of course, it was not an app, so none of models showed up on admin page.
comment out the some lines in urls.py see docs for more details
admin.autodiscover()
urlpatterns = patterns('',
('^admin/', include(admin.site.urls)),
)