Disclaimer: i am new to python but have drupal programming experience
I am reading the Definitive Guide to Django (http://www.djangobook.com/en/1.0/chapter07/). After issuing
python manage.py startapp books
python creates the books package with views.py inside. Later in the tutorial, we enter the following into that views.py file:
# Create your forms here.
from django import forms
from django.forms import form_for_model
from models import Publisher
PublisherForm = forms.form_for_model(Publisher)
TOPIC_CHOICES = (
('general', 'General enquiry'),
('bug', 'Bug report'),
('suggestion', 'Suggestion'),
)
class ContactForm(forms.Form):
topic = forms.ChoiceField(choices=TOPIC_CHOICES)
message = forms.CharField(widget=forms.Textarea())
sender = forms.EmailField(required=False)
def clean_message(self):
message = self.cleaned_data.get('message', '')
num_words = len(message.split())
if num_words < 4:
raise forms.ValidationError("Not enough words!")
return
Unless I typed something wrong or misunderstood something (either is likely), we now (seemingly) have a collision between forms (books/forms.py) and django forms. So, which does Python refer to above in
message = forms.CharField(widget=forms.Textarea())
This statement:
from django.forms import form_for_model
Really only pulls the name form_for_model into the module-global namespace, the already existing name forms is not affected. Even this:
import django.forms
would not be a problem, because it only makes the name available in its fully-qualified form, django.forms (which is unambigios and does not collide with forms).
Unlike PHP, Python only imports into the current namespace, and each module is its own namespace; imports in other modules do not affect the current namespace, and vice versa.
Although this example does not cause a conflict as pointed out if there is a conflict you can use from ____ import ___ as ___ to specify your own import name
http://docs.python.org/reference/simple_stmts.html#import
Related
I want to customize the field texture in django’s new “readonly mode”: E.g. foreign keys shall be displayed as links.
In general I identified the following options:
Implement custom fields for every model – results in code duplication
Reimplement django’s display_for_field method
Basically I could copy & paste the django.contrib.admin.utils module, insert my changes, override sys.modules['django.contrib.admin.utils'] = myutils but that's ugly because of maintainability in case of Django updates in the future.
So I decided to override only the display_for_fields method of django.contrib.admin.utils using the following approach to avoid duplication of Django code:
Override display_for_field function in django.contrib.admin.utils in settings.py:
from myapp.contrib.admin import utils
utils.override_method()
In myapp.utils.py:
from django.contrib.admin import utils
from django.contrib.admin.utils import *
def display_for_field_mod(value, field, empty_value_display):
if isinstance(field, models.ForeignKey) and value:
if field.related_model.__name__.lower() != 'user':
link_string = 'admin:myapp_' + field.related_model.__name__.lower() + '_change'
link = reverse(link_string, args=(value.id,))
return format_html('{}', link, value)
else:
return formats.localize(value)
else:
return display_for_field(value, field, empty_value_display)
def override_method():
utils.display_for_field = display_for_field_mod
But the problem is: display_for_field gets imported in django.contrib.admin.helpers using:
from django.contrib.admin.utils import (
display_for_field, [...]
)
So due to the scope of the imported funtion I cannot override this function from outside.
Do I miss some other obvious possibility? Is there a clean method to achieve this or is the only option to duplicate/modify django’s original code?
I came across a similar issue. If anyone's still looking for a solution, you just need to override the display_for_field in helpers, e.g.:
from django.contrib.admin import helpers, utils
def override_method():
helpers.display_for_field = display_for_field_mod
utils.display_for_field = display_for_field_mod
Only overriding helpers is sufficient, but it's a good idea to override utils as well, just in case.
I had forked my shipping app into a folder called forked_apps using oscar_fork_app command and also added in settings.py get_core_apps(['forked_apps.shipping']), I just want to create two shipping methods mentioned, standard and express in the docs in this link:https://django-oscar.readthedocs.io/en/latest/howto/how_to_configure_shipping.html.
In the init.py I have this code pre-existing:
default_app_config = 'forked_apps.shipping.config.ShippingConfig'
In repository.py I have written like this:
from oscar.apps.shipping import repository
from .methods import *
class Repository(repository.Repository):
def get_available_shipping_methods(
self, basket, user=None, shipping_addr=None,
request=None, **kwargs):
methods = (Standard())
print("\n\nFetch availble shipping methods")
if shipping_addr:
# Express is only available in the UK
methods = (Standard(), Express())
return methods
And in the methods.py I had written:
from decimal import Decimal as D
from oscar.apps.shipping import methods
from oscar.core import prices
class Standard(methods.FixedPrice):
code = 'standard'
name = 'Standard shipping'
charge_excl_tax = D('5.00')
class Express(methods.FixedPrice):
code = 'express'
name = 'Express shipping'
charge_excl_tax = D('10.00')
What should happen is, the shipping_methods.html page should show up, but instead, after entering the shipping address it goes to the payment details page directly; this would usually happen only if there are no shipping methods defined, but I have implemented two shipping methods, standard and Express in the above code.I can't figure out how to make this work, even the print statement isn't working.
Is there any other additional code that I must write?
Can someone provide a solution with some code, if you have implemented it?
Remove oscar apps from settings.
For example:
#oscar.apps.checkout
#oscar.apps.shipping
etc.
This section gives me error. I can't fix that.
get_available_shipping_methods(
self, basket, user=None, shipping_addr=None,
request=None, **kwargs):
...
Django ver. > 2.1 | Oscar ver. > Latest
I using it like this;
mkdir customapp
touch customapp/__init__.py
python manage.py oscar_fork_app shipping customapp/
Edit settings.py
from oscar import get_core_apps
INSTALLED_APPS = INSTALLED_APPS + get_core_apps(
['customapp.shipping'])
In our customapp/shipping directory added new file, called (repository.py)
from oscar.apps.shipping import repository
from . import methods
class Repository(repository.Repository):
methods = (methods.Standard(),)
Then add new file in same directory, customapp/shipping, called (methods.py)
from oscar.apps.shipping import methods
from oscar.core import prices
from decimal import Decimal as D
class Standard(methods.Base):
code = 'standard'
name = 'Shipping (Standard)'
def calculate(self, basket):
return prices.Price(
currency=basket.currency,
excl_tax=D('5.00'), incl_tax=D('5.00'))
You can add more methods.
Then run these commands;
python manage.py makemigrations
python manage.py migrate
python manage.py runserver
Hope this helps.
so I have 2 apps running in the same project.
My files are structured as follows:
/project_codebase
/project
__init.py
settings.py
urls.py
wsgi.py
...
/app1
...
/app2
...
manage.py
So, I for some weird reason have a different name for my base directory (that is, it ends with codebase). Hopefully, that is not an issue.
In my settings.py, I have this:
INSTALLED_APPS = [
...
'app1',
'app2',
]
Ok, so in my models.py (from app2), I can easily import models from app1 with from app1.models import *, however, when I use from app2.models import * in my models.py (from app1), I get an ImportError.
Any solutions to this?
This might be due to circular import issues. To avoid this you should load the model dynamically:
For recent versions of django (1.7+) use the application registry:
from django.apps import apps
MyModel1 = apps.get_model('app1', 'MyModel1')
For earlier django versions (<1.7):
from django.db.models.loading import get_model
MyModel1 = get_model('app1', 'MyModel1')
Note 1: If you want to define a ForeignKey relationship, there is no need for a separate import statement. Django has you covered on this:
If app1 is an installed app, you should define the ForeignKey relationship as follows:
# in app2.py
class MyModel2(models.Model):
mymodel1 = models.ForeignKey('app1.MyModel1')
Note 2: The get_model only works if app1 is an installed app and MyModel1 is the model you want to import from app1.
Note 3: Try to avoid wildcard import (from ... import *), as this is bad practice.
It's definitely a circular import.
But i think is what you need is to use models as some sort of RetationFields(ForeignKey, ManyToManyField or OneToOneField) arguments. So you need to skip import and use as so:
# app1/models.py
class Model1(models.Model):
relation_field = models.ForeignKey('app2.Model2')
From docs:
If you need to create a relationship on a model that has not yet been defined, you can use the name of the model, rather than the model object itself
To refer to models defined in another application, you can explicitly specify a model with the full application label
Just put str object as first argument to relation fields that leeds to <app_name>.<Model_name>.
Note: it's better to avoid importing everything from module(from <module_name> import *)
If you want to import only some specific module then do not use import *.
It will take more time load your all library and so can affect the speed of your app also.
If you want to use few modules from your second app then just add module name instead of whole libraries something like this:
from app2.models import Module1, Module2
or it may be circular import issue as other clarify.
Thanks.
i use this code always and it's work :)
from position_app.models import Member
You need to specify the model names you want to import, for ex from app1.models import ModelName1, ModelName2.
Make sure there is no name clash between one of your apps and one of the modules installed in your Python environment. If you use pip, you can run pip freezeto see a list of installed modules.
I had the same error when one of my apps was named 'packaging', and the packaging python module was installed.
I also face this problem when I try to import my model from another app in (django2.2)
But at last I Imported It and Its successfully working.
here is my two app:
INSTALLED_APPS = [
...
'categories',
'videos',
]
and this is the code for how I Imported it into videos/models.py file as a ForeignKey Connectivity
from django.db import models
class Videos(models.Model):
categories = models.ForeignKey('categories.Categories', related_name='categories', on_delete=models.CASCADE)
If want to see my Categories Model from categories/models.py file, you can check this code otherwise neglect it
from django.db import models
class Categories(models.Model):
category_name = models.CharField(max_length=50)
created_at = models.DateTimeField(auto_now_add=True)
updated_at = models.DateTimeField(auto_now=True)
It is a circular import
In my case I needed the imported class not for a relation field, but for use it inside a method instead.
If that's the case, I suggest to import inside the method, otherwise an AppRegistryNotReady("Models aren't loaded yet.") is raised.
class Student(CustomUser):
""" Usuario alumno """
class Meta:
verbose_name = "Alumno"
verbose_name_plural = "Alumnos"
def get_current_school_class(self):
""" Obtiene el curso actual de un alumno """
from school.models import SchoolClass, StudentClass
# proceed with the method...
it's not necessary to import models from others apps
just put the app.models in the foreignkey field and that's work ;)
app 1:
class model1(models.Model):
field=models.field type ...
app 2:
class model2(models.Model):
field=models.ForeignKey('app1.model1', on_delete. ...)
to avoid code correction :D
I couldn't find an answer to the following question, it took me a couple of hours to find out, hence I'm adding it. I'll add my approach of solving it and the answer.
I'm following a YouTube tutorial from This person. For some reason I'm typing the same code, and I checked every single letter. Yet for some reason my cleaning functions aren't called. It's probably something simple, especially since a related question showed something similar. It's probably a framework thing that I get wrong, but I wouldn't know what it is.
Here is the relevant code.
forms.py (complete copy/paste from his Github)
from django import forms
from .models import SignUp
class ContactForm(forms.Form):
full_name = forms.CharField(required=False)
email = forms.EmailField()
message = forms.CharField()
class SignUpForm(forms.ModelForm):
class Meta:
model = SignUp
fields = ['full_name', 'email']
### exclude = ['full_name']
def clean_email(self):
email = self.cleaned_data.get('email')
email_base, provider = email.split("#")
domain, extension = provider.split('.')
# if not domain == 'USC':
# raise forms.ValidationError("Please make sure you use your USC email.")
if not extension == "edu":
raise forms.ValidationError("Please use a valid .EDU email address")
return email
# Final part is ommited, since it's not relevant.
admin.py (typed over from the tutorial)
from django.contrib import admin
# Register your models here.
from .models import SignUp
from .forms import SignUpForm
class SignUpAdmin(admin.ModelAdmin):
list_display = ['__unicode__', 'timestamp', 'updated']
class Meta:
model = SignUp
form = SignUpForm
admin.site.register(SignUp, SignUpAdmin)
After using print statements for a while and reading questions that seemed similar but eventually didn't solve my problem, I decided to look into the source of Django (idea inspired by the most similar question I could find).
Then, I decided to debug the source, since I wanted to know how Django is treating my customized function (thanks to a tutorial + SO answer). In the source I found that the customized functions were called around return super(EmailField, self).clean(value) (line 585, django/forms/fields.py, Django 1.8). When I was stepping through the code I found the critical line if hasattr(self, 'clean_%s' % name): (line 409, django/forms/forms.py, Django 1.8). I checked for the value name which was "email". Yet, the if-statement evaluated as False ((Pdb) p hasattr(self, 'clean_%s' % name)). I didn't understand why, until I figured out that the function name was not registered ((Pdb) pp dir(self)).
I decided to take a look at the whole source code repository and cross-checked every file and then I found that
class Meta:
model = SignUp
form = SignUpForm
means that form / SignUpForm were nested inside the Meta class. At first, I didn't think much of it but slowly I started to realize that it should be outside the Meta class while staying main class (SignUpAdmin).
So form = SignUpForm should have been idented one tab back. For me, as a Django beginner, it still kind of baffles me, because I thought the Meta class was supposed to encapsulate both types of data (models and forms). Apparently it shouldn't, that's what I got wrong.
I am quite new to Django, I'm having few problems with validation
forms in Admin module, more specifically with raising exceptions in the
ModelForm. I can validate and manipulate data in clean methods but
cannot seem to raise any errors. Whenever I include any raise
statement I get this error "'NoneType' object has no attribute
'ValidationError'". When I remove the raise part everything works
fine.
Then if I reimport django.forms (inside clean method) with a different alias (e.g. from django import forms as blahbalh) then I'm able to raise messages using blahblah.ValidateException.
Any tips or suggestions on doing such a thing properly ?
Here's an example of what I'm doing in Admin.py:
admin.py
from django import forms
from proj.models import *
from django.contrib import admin
class FontAdminForm(forms.ModelForm):
class Meta:
model = Font
def clean_name(self):
return self.cleaned_data["name"].upper()
def clean_description(self):
desc = self.cleaned_data['description']
if desc and if len(desc) < 10:
raise forms.ValidationError('Description is too short.')
return desc
class FontAdmin(admin.ModelAdmin):
form = FontAdminForm
list_display = ['name', 'description']
admin.site.register(Font, FontAdmin)
--
Thanks,
A
You problem might be in the * import.
from proj.models import *
if proj.models contains any variable named forms (including some module import like "from django import forms), it could trounce your initial import of:
from django import forms
I would explicitly import from proj.models, e.g.
from proj.models import Font
If that doesn't work, see if there are any other variables name "forms" that could be messing with your scope.
You can use introspection to see what "forms" is. Inside your clean_description method:
print forms.__package__
My guess is it is not going to be "django" (or will return an error, indicating that it is definitely not django.forms).