I am new to Django but not to developing.
I need to make an application in which user can do CRUD operations (Create, Read, Update, Delete). This functionality should apply to all models and the fields for Create & Update will be auto-generated from model attributes.
What I describe is pretty much the functionality that comes with the Admin page. However, I want to use it in my own app instead of using the Admin app.
For example, let's suppose we have Author and Book models:
(models.py)
from django.db import models
class Author(models.Model):
first_name = models.CharField(max_length=60)
last_name = models.CharField(max_length=60)
def __str__(self):
return self.last_name
class Book(models.Model):
title = models.CharField(max_length=60)
author = models.ForeignKey(Author, on_delete=models.CASCADE)
def __str__(self):
return self.title
I suppose that the above information (models) could be enough for CRUD operations, without repeating code of the same logic for each model. I am aiming at a functionality like in admin page where all you have to do is register your model.
I am aware of ModelForm and Generic Views but while they help avoiding hard-coding form fields, I have not found a non-repetitive coding approach. I would like to avoid approaches like the following where same code is being duplicated for each model:
(forms.py)
from django import forms
from todo.models import Author, Book
class AuthorForm(forms.ModelForm):
class Meta:
model = Author
fields = '__all__'
class BookForm(forms.ModelForm):
class Meta:
model = Book
fields = '__all__'
(views.py)
from django.views.generic import CreateView
from.forms import AuthorForm, BookForm
from.models import Author, Book
class AuthorCreateView(CreateView):
model = Author
form_class = AuthorForm
...
class BookCreateView(CreateView):
model = Book
form_class = BookForm
...
So, what is the best approach for a DRY CRUD solution (like in Admin page)? Am I missing any Django features?
Putting my explanation in the comment together, you would get something like this:
from django.views import generic
from myapp.apps import MyAppConfig
from django.forms import modelform_factory
urlpatterns = []
for model in MyAppConfig.get_models():
create_url = path(
f"{model.__class___.__name__.lower()}/create",
generic.CreateView.as_view(
form_class=modelform_factory(model=model, fields='__all__'),
template_name=f"{model._meta.app_label}/create.html",
model=model
),
),
list_url = path(
f"{model.__class__.__name__.lower()}/",
generic.ListView.as_view(
template_name=f"{model._meta.app_label}/list.html",
model=model
),
),
...
urlpatterns.extend([create_url, list_url, read_url, update_url, delete_url])
So the principle is to use the generic view and model form factory, to generate standard crud views, derive path names from model name and use one template per view, in the application (derived from model's app_label).
Using the documentation I linked before, you should be able to piece things together.
Related
Previously I was using my project with sqlite. Then started a new project copied the data from previous project and made some changes, and I'm using this with mysql.
This is my models.py(not full)
from django.db import models
from django.db.models import CheckConstraint, Q, F
class College(models.Model):
CITY_CHOICES=[('BAN','Bangalore')]
id=models.IntegerField(primary_key=True)
name=models.CharField(max_length=50)
city=models.CharField(choices=CITY_CHOICES,default='BAN',max_length=10)
fest_nos=models.IntegerField()
image=models.ImageField(default='default.jpg',upload_to='college_pics')
class Meta():
db_table='college'
def __str__(self):
return self.name
class Organizer(models.Model):
id=models.IntegerField(primary_key=True)
name=models.CharField(max_length=25)
phone=models.IntegerField()
def __str__(self):
return self.name
class Fest(models.Model):
FEST_CHOICES=[
('CUL','Cultural'),
('TEC','Technical'),
('COL','College'),
('SPO','Sports'),
]
id=models.IntegerField(primary_key=True)
name=models.CharField(max_length=50)
clg_id=models.ForeignKey(College,on_delete=models.CASCADE)
fest_type=models.CharField(choices=FEST_CHOICES,default='COL',max_length=10)
fest_desc=models.TextField(default='This is a fest')
#below two field are not showing up in admin page
start_date=models.DateField(auto_now_add=True)
end_date=models.DateField(auto_now_add=True)
event_nos=models.IntegerField()
org_id=models.ManyToManyField(Organizer)
image=models.ImageField(default='default.jpg',upload_to='fest_pics')
class Meta:
constraints = [
CheckConstraint(
check = Q(end_date__gte=F('start_date')),
name = 'check_start_date',
)
]
db_table='fest'
def __str__(self):
return self.name
The start_date and end_date attributes are the new ones added in this project. It was not there in the old one.
My admin.py file
from django.contrib import admin
from .models import College, Event, Fest, Organizer, Participated
admin.site.register(College)
admin.site.register(Organizer)
admin.site.register(Fest)
admin.site.register(Event)
admin.site.register(Participated)
But in my admin dashboard, while adding new fests I'm not getting the option to add start and end date.
I made migrations once again, fake migrated etc. What to do?
Is check constraint under model fest causing this problem?
They fields won't show up on Django Admin because they have auto_now_add=True so, the user shouldn't touch them.
You can make auto_now_add field display in admin by using readonly_fields in the admin class(this only show the data, you still can't edit it because it's auto_now_add)
#register with a class to use
class FestAdmin(admin.ModelAdmin):
readonly_fields = ('start_date', 'end_date')
admin.site.register(Fest, FestAdmin)
I need to display multiple models in django admin change list view. I want to use single search box to filter all of them at once. Is there an easy way to do it?
My idea was to inherit from admin site, add another view to it and iterate over models in modified change_list.html but i can't import models and ModelAdmins because i get django.core.exceptions.AppRegistryNotReady: Apps aren't loaded yet. error so i can't get the same context that django uses to render regular change_list.html.
What's the correct way to do it? Is there simpler approach?
As Ohad suggested, the most robust approach is probably to make formal relationships between the models from which you want the objects to display together. You have a couple of options here. Essentially you will want to make a master class and then subclass your models from it. This makes a lot of sense if your models are ontologically related to a parent concept. For example:
Publication
Book
Magazine issue
Books and magazines are both publications. They both share some fields, like title and publication date. But they differ in that a book usually has a single author and a magazine has volumes and issue dates. Django already provides a couple different approaches to subclassing using Model inheritance. However, after trying these myself I found that the django-polymorphic extension is way better. Here is a code example of a Django 3.0 app using django-polymorphic which has a Book model and a Magazine model with a single listing of all publications that shows all of the books and magazines in the system.
models.py
from django.db import models
from polymorphic.models import PolymorphicModel
class Publication(PolymorphicModel):
title = models.CharField(max_length=256)
publication_year = models.IntegerField()
class Book(Publication):
author_first = models.CharField(max_length=256)
author_last = models.CharField(max_length=256)
class Magazine(Publication):
volume_number = models.IntegerField()
issue_name = models.CharField(max_length=256)
admin.py
from django.contrib import admin
from polymorphic.admin import PolymorphicParentModelAdmin, PolymorphicChildModelAdmin, PolymorphicChildModelFilter
from .models import Publication, Book, Magazine
class PublicationChildAdmin(PolymorphicChildModelAdmin):
""" Base admin class for all child models """
base_model = Publication # Optional, explicitly set here.
#admin.register(Book)
class BookAdmin(PublicationChildAdmin):
base_model = Book # Explicitly set here!
# show_in_index = True # makes child model admin visible in main admin site
list_display = ('title', 'publication_year', 'author_first', 'author_last')
#admin.register(Magazine)
class MagazineAdmin(PublicationChildAdmin):
base_model = Magazine # Explicitly set here!
# show_in_index = True # makes child model admin visible in main admin site
list_display = ('title', 'publication_year', 'issue_name')
#admin.register(Publication)
class PublicationParentAdmin(PolymorphicParentModelAdmin):
""" The parent model admin """
base_model = Publication # Optional, explicitly set here.
child_models = (Book, Magazine)
list_filter = (PolymorphicChildModelFilter,) # This is optional.
list_display = ('title', 'publication_year')
This will of course only display those fields that are common (in the Publication model). If you want to display fields that are particular to each model there are various tricks for this. Here's one quick way to do it:
admin.py
...
#admin.register(Publication)
class PublicationParentAdmin(PolymorphicParentModelAdmin):
""" The parent model admin """
base_model = Publication # Optional, explicitly set here.
child_models = (Book, Magazine)
list_filter = (PolymorphicChildModelFilter,) # This is optional.
list_display = ('title', 'publication_year', 'issue', 'author')
def author(self, obj):
if obj.polymorphic_ctype.model == 'book':
book = Book.objects.get(pk=obj.pk)
return book.author_first + ' ' + book.author_last
return None
def issue(self, obj):
if obj.polymorphic_ctype.model == 'magazine':
return str(Magazine.objects.get(pk=obj.pk).issue_name)
return None
Tada!
From the docs it seems that there is no easy solution.(if there is no relation between the models)
https://docs.djangoproject.com/en/2.2/ref/contrib/admin/#django.contrib.admin.ModelAdmin.search_fields
So if the search is commonly used build a special model/models that combines the data that might be searched
every one,,I am reading a Django practice book,,I saw a code "model = Thing" in admin.py,,,however, when I remove "model = Thing",,,the web program still can run,the admin site looks no difference??,what does this code mean?if without it what gonna happen? my models.py class is Thing
admin.py
from django.contrib import admin
from collection.models import Thing
class ThingAdmin(admin.ModelAdmin):
model = Thing #if I remove this code, the program still can run,,why need this code
list_display = ('name', 'description',)
prepopulated_fields = {'slug': ('name',)}
admin.site.register(Thing, ThingAdmin)
modles.py
from django.db import models
class Thing(models.Model):
name = models.CharField(max_length=255)
description = models.TextField()
slug = models.SlugField(unique=True)
Setting a model attribute on the ModelAdmin class will have no effect. You can safely remove that line from your code.
In the Django admin, you specify the model when you call admin.site.register(), or by using the register decorator. This allows you to use the same model admin class for more than one model.
admin.site.register(Thing, ThingAdmin)
admin.site.register(OtherThing, ThingAdmin)
As Jon pointed out in the comments, you do need to specify the model for InlineModelAdmin objects.
I'm trying relate two resources (models) in an API using Tastypie but I'm getting an error.
I've followed the django tutorial and used:
models.py
from django.db import models
class Poll(models.Model):
question = models.CharField(max_length=200)
pub_date = models.DateTimeField('date published')
class Choice(models.Model):
poll = models.ForeignKey(Poll)
choice_text = models.CharField(max_length=200)
votes = models.IntegerField(default=0)
I tried to create a link between the Poll and Choice based on this stackoverflow answer and wrote the following code:
api.py
class ChoiceResource(ModelResource):
poll = fields.ToOneField('contact.api.PollResource', attribute='poll', related_name='choice')
class Meta:
queryset = Choice.objects.all()
resource_name = 'choice'
class PollResource(ModelResource):
choice = fields.ToOneField(ChoiceResource, 'choice', related_name='poll', full=True)
class Meta:
queryset = Poll.objects.all()
resource_name = 'poll'
When I go to: 127.0.0.1:8088/contact/api/v1/choice/?format=json
Everything works as it should. For example one of my choices links to the right poll:
{
"choice_text": "Nothing",
"id": 1,
"poll": "/contact/api/v1/poll/1/",
"resource_uri": "/contact/api/v1/choice/1/",
"votes": 6
}
When I go to: 127.0.0.1:8088/contact/api/v1/poll/?format=json
I get:
{
"error": "The model '<Poll: What's up?>' has an empty attribute 'choice' and doesn't allow a null value."
}
Do I need to use the fields.ToManyField instead or do I need to change my original model?
Tastypie recommends against creating reverse relationships (what you're trying to do here the relationship is Choice -> Poll and you want Poll -> Choice), but if you still wanted to, you can.
Excerpt from the Tastypie docs:
Unlike Django’s ORM, Tastypie does not automatically create reverse
relations. This is because there is substantial technical complexity
involved, as well as perhaps unintentionally exposing related data in
an incorrect way to the end user of the API.
However, it is still possible to create reverse relations. Instead of
handing the ToOneField or ToManyField a class, pass them a string that
represents the full path to the desired class. Implementing a reverse
relationship looks like so:
# myapp/api/resources.py
from tastypie import fields
from tastypie.resources import ModelResource
from myapp.models import Note, Comment
class NoteResource(ModelResource):
comments = fields.ToManyField('myapp.api.resources.CommentResource', 'comments')
class Meta:
queryset = Note.objects.all()
class CommentResource(ModelResource):
note = fields.ToOneField(NoteResource, 'notes')
class Meta:
queryset = Comment.objects.all()
This is a Model Class
class ModelName(models.Model):
(...)
pasta = TaggableManager(verbose_name=u'Pasta')
and a form template (normal :P )
{{form.as_p}}
I'd like to leave everything very clean and usefull.
But result is a list of TaggedItem Object :( :
[<TaggedItem: id: 2 tagged with general >, <TaggedItem: id: 3 tagged with outer >]
Instead of something like
general, outer
How do it fashionably in Django?
Give a look at the code in: https://github.com/alex/django-taggit/blob/master/taggit/forms.py. You will find the widget used to render the tags. You can use it to render them correctly.
Example:
models.py
from django.db import models
from taggit.managers import TaggableManager
class Example(models.Model):
name = models.CharField(max_length=20)
tags = TaggableManager()
forms.py
.models import Example
from django import forms
from taggit.forms import TagWidget
class ExampleForm(forms.ModelForm):
class Meta:
model = Example
fields = ('name', 'tags',)
widgets = {
'tags': TagWidget(),
}
I'd recommend you to check this answer too.
django - django-taggit form
I would use django-taggit-autosuggest as it offers better UI to the user.