I have the following models, currently connected by ForeignKey:
class Topic(models.Model):
text = models.CharField(max_length=200)
date_added = models.DateTimeField(auto_now_add=True)
def __str__ (self):
return self.text
class Entry(models.Model):
topic = models.ForeignKey(Topic, on_delete=models.CASCADE)
text = models.TextField()
date_added = models.DateTimeField(auto_now_add=True)
class Meta:
verbose_name_plural = 'entries'
def __str__(self):
if len(self.text) > 50:
return self.text[:50] + "..."
else:
return self.text
Is it possible to connect them in a way that in Django admin site I could go to a particular topic and have a list of all the entries displayed? Currently nothing gets displayed and I can only see the connection when I go to a particular entry (there is a dropdown saying to which topic current entry belongs).
EDIT: these are inlines added to the admin.py:
from django.contrib import admin
from javascript.models import Topic, Entry
# Register your models here.
admin.site.register(Topic)
admin.site.register(Entry)
#inlines
class EntryInline(admin.TabularInline):
model = Entry
class TopicAdmin(admin.ModelAdmin):
inlines = [EntryInline]
You can proceed by using inlines(TabularInline or StackedInline)
First of all create inline class for your entries:
class EntryInlines(admin.TabularInline):
model = Entry
# you can customize fields displayed here; make any field as read only if you want
Now include EntryInlines to the model where you want to display data (only if foreign Entry model has foreign key to respective model) as below:
class TopicAdmin(admin.ModelAdmin):
inlines = [EntryInlines] # you can add multiple inline tables too. pass it in the list
# all other customization of your model admin if any
If you have nested foreign keys then you need to add this third-party-package like django-nested-inline
Related
I've got a 'Registration' object in place that users can create on the front end without issue.
It looks like this:
class Registration(models.Model):
person = models.ForeignKey(Person, on_delete=models.PROTECT)
course_detail = models.ForeignKey(CourseDetail, on_delete=models.PROTECT)
camp_shirt = models.ForeignKey(CampShirt, on_delete=models.PROTECT)
comments = models.CharField(max_length=200, blank=True, null=True)
created_at = models.DateTimeField(auto_now_add=True)
updated_at = models.DateTimeField(auto_now=True)
def __str__(self):
return "%s" % (self.course_detail.course.camp)
When I am in the admin and click on a given Registration - it takes a while to load because there are thousands and thousands of Person objects.
For ease of use - there will never be a time when we would need to edit the 'person' associated with a given registration, so I would like to make the 'person' dropdown only show the selected user in the person queryset when editing from the django admin.
So when I go to http://myapp.com/admin/registration/23/change I want the form to only display the currently selected person as the only option in the dropdown.
My admin model looks like this:
class RegistrationAdmin(admin.ModelAdmin):
list_display = ("person", "course_detail")
class Meta:
# I think this is what I do in order to override the default admin form? Not sure.
form = RegistrationAdminForm
My RegistrationAdminForm looks like this:
class RegistrationAdminForm(forms.ModelForm):
# course_detail, person, camp_shirt, comments
person = forms.ModelChoiceField(queryset=Person.objects.filter(
id=registration.person.id)
)
def __init__(self, registration, *args, **kwargs):
super(RegistrationAdminForm, self).__init__(*args, **kwargs)
self.fields['person'].queryset = Person.objects.filter(
id=registration.person.id
)
class Meta:
model = Registration
fields = '__all__'
Main Question : How do I change the admin form so that a specific queryset is returned for one of the fields in the django admin?
If the person field will never be changed you can add the person field to readonly_fields, a select with all Person objects will not be rendered.
class RegistrationAdmin(admin.ModelAdmin):
list_display = ("person", "course_detail")
readonly_fields = ("person", )
Then you do not need your custom form. FYI when you want to add a custom form to a ModelAdmin you do not put it in Meta, you define it on the form itself
class MyModelAdmin(admin.ModelAdmin):
form = MyModelForm
I have two models, which are User and Record. Each has several fields.
from django.db import models
class User(models.Model):
openid = models.CharField(max_length=20)
nickname = models.CharField(max_length=20,null=True)
def __str__(self):
return self.nickname
class Record(models.Model):
expression = models.CharField(max_length=100)
user = models.ForeignKey(User)
time = models.DateTimeField(auto_now_add=True)
def __str__(self):
return self.expression
I register them in admin.py
from django.contrib import admin
from .models import User,Record
class RecordAdmin(admin.ModelAdmin):
list_display = ('expression','user','time')
class UserAdmin(admin.ModelAdmin):
empty_value_display = "çİş"
list_display = ('openid','nickname')
admin.site.register(User,UserAdmin)
admin.site.register(Record,RecordAdmin)
it works well in django admin initially. but one day, the fields of the Record model disppeared. It looks like
.
No field displays. It makes me unable to modify or add the values of the Record model. The other model User works well and all data exists in database. So why?
I think you just have to add on_delete=models.CASCADE in your ForeignKey Field. When you are using this kind of field, you have to specify the comportment when you make an update, a delete or anything else on this field.
So your script should be like this :
class Record(models.Model):
expression = models.CharField(max_length=100)
user = models.ForeignKey(User, on_delete=models.CASCADE)
time = models.DateTimeField(auto_now_add=True)
def __str__(self):
return self.expression
This is the result :
Edit :
You can also modify null=True by default=null
class User(models.Model):
openid = models.CharField(max_length=20)
nickname = models.CharField(max_length=20,default=null)
def __str__(self):
return self.nickname
Say I have this app named Pantry that is to connect to any other app I may come along. To keep the app decoupled, generic relations are used through the model LinkedItem which connects the Ingredients model to apps outside Pantry.
I can make a filter_horizontal show up for the LinkedItem's admin in Django. Now I would like the content on the other end of the generic relation, say an app named Bakery, to be able to do a filter_horizontal with ingredients.
Pantry
models.py
from django.db import models
from django.contrib.contenttypes.models import ContentType
from django.contrib.contenttypes import fields
class Ingredient(models.Model):
'''
Model containing all the ingredients, their slugs, and their descriptions
'''
name = models.CharField(unique=True, max_length=100)
slug = models.SlugField(unique=True, max_length=100)
description = models.CharField(max_length=300)
# method to return the name of the db entry
def __str__(self):
return self.name
class LinkedItem(models.Model):
'''
Model that links ingredients to various other content models
'''
content_type = models.ForeignKey(ContentType)
object_id = models.PositiveIntegerField()
content_object = fields.GenericForeignKey('content_type', 'object_id')
ingredient = models.ManyToManyField(Ingredient)
# method to return the name of the db entry
def __str__(self):
return self.ingredient.name
# defines options for the model itself
class Meta:
unique_together = (('content_type','object_id')) # prevents duplicates
Bakery
admin.py
from django.contrib import admin
from bakery.models import Cake
class CakeAdmin(admin.ModelAdmin):
filter_horizontal = ('') # what to put here so ingredients show up?
Any ideas?
A solution is to create a GenericTabularInline for LinkedItem and putting some restrictions on the display to avoid duplicates like below:
from django.contrib.contenttypes.admin import GenericTabularInline
class LinkedItemAdmin(GenericTabularInline):
model = LinkedItem
# choosing the field and display
field = ['ingredient']
filter_horizontal = ['ingredient']
# These help with removing some potential issues in the admin
extra = 0
min_num = 1
max_num = 1
can_delete = False
Then in the CakeAdmin I can do this to make the ingredients show up.
class CakeAdmin(admin.ModelAdmin):
inlines = [LinkedItemAdmin]
I have a very simple app which at the moment declares two models: one is called "Content" and simply holds content data and the other is "Page" which includes "Content" as a OneToOneField.
The reason I've done this is so that I can have "Page" as an actual concrete class that I use and when other models in other modules I'm planning need page data, they can simply include "Content" as a OneToOneField. I've done it this way to avoid inheritance and use composition instead.
models.py:
from django.db import models
class Content(models.Model):
"""Basic page data which can be used by other modules"""
title = models.CharField(max_length=200)
html_title = models.CharField(max_length=200)
meta_desc = models.CharField(max_length=200)
keywords = models.CharField(max_length=200)
content = models.TextField()
class Page(models.Model):
"""Concrete implementation of a basic page managed by the admin"""
slug = models.SlugField()
content = models.OneToOneField(Content)
def __str__(self):
return self.content.title
admin.py:
from django.contrib import admin
from content.models import Page, Content
class ContentInline(admin.TabularInline):
model = Content
fields = ('title', 'html_title', 'meta_desc', 'keywords', 'content')
class PageAdmin(admin.ModelAdmin):
fields = ('slug',)
inlines = [ContentInline]
On the page admin I get this exception:
Exception at /admin/content/page/add/
<class 'content.models.Content'> has no ForeignKey to <class 'content.models.Page'>
What is says of course is correct, but I cannot seem to find a way of doing what I want, which is to include an inline of the non-defining side of a relationship. I don't want to declare the relationship on "Content" as then I'd have to define every single relationship to it inside that class which would introduce dependencies to other modules, which in my opinion it should know nothing about.
Using Django 1.6 on Python 3.3.
Edit: As indicated in the comments, I've decided to use inheritance. My initial concern about this was that I wanted the flexibility to be able to compose classes from multiple other classes. However, since the Django ORM does support multiple inheritance and if I'd realised that method was called "mixins" (new to Python) I would have got somewhere a lot sooner.
Example mixins with models:
from django.db import models
class Content(models.Model):
"""Basic page data which can be used by other modules"""
title = models.CharField(max_length=200)
html_title = models.CharField(max_length=200)
meta_desc = models.CharField(max_length=200)
keywords = models.CharField(max_length=200)
content = models.TextField()
def __str__(self):
return self.title
class Meta:
abstract = True
class Data(models.Model):
data_name = models.CharField(max_length=200)
class Meta:
abstract = True
class Page(Content, Data):
"""Concrete implementation of a basic page managed by the admin"""
slug = models.SlugField()
And then I can just use it as one model in admin.py.
Another solution is moving the OneToOneField from Content to Page
class Content(models.Model):
"""Basic page data which can be used by other modules"""
title = models.CharField(max_length=200)
html_title = models.CharField(max_length=200)
meta_desc = models.CharField(max_length=200)
keywords = models.CharField(max_length=200)
content = models.TextField()
page = models.OneToOneField(Page, primary_key=True, related_name="content")
class Page(models.Model):
"""Concrete implementation of a basic page managed by the admin"""
slug = models.SlugField()
def __str__(self):
return self.content.title
You can still do page.content and the inline form will work out of the box
EDIT:
One cons of that approach is that it will allow the user to create a page without assigning any content to it (in which case page.content will crash)
Its very easy to overcome this issue by creating custom form
class ContentAdminForm(forms.ModelForm):
def __init__(self, *args, **kwargs):
kwargs["empty_permitted"] = False
super(ContentAdminForm, self).__init__(*args, **kwargs)
Then in the admin page
class ContentInline(admin.TabularInline):
model = Content
form = ContentAdminForm
fields = ('title', 'html_title', 'meta_desc', 'keywords', 'content')
If you don't want to change your models at all, there's a django module to display the non-defining side inline: django_reverse_admin
You'll need to add django_reverse_admin to your requirements.txt:
-e git+https://github.com/anziem/django_reverse_admin.git#egg=django_reverse_admin
Then import it:
admin.py
from django.contrib import admin
from django_reverse_admin import ReverseModelAdmin
from content.models import Page, Content
# don't need to define an inline anymore for Content
class PageAdmin(ReverseModelAdmin):
fields = ('slug',)
inline_reverse = ['content']
inline_type = 'tabular' # or could be 'stacked'
I have scenario in which a user can have multiple books. I can create two different models for user and books and relate them using foreign keys (or one-to-many will be right way ?).
I have created a django forms for User model but when i do like this {{form.as_p}} in templates only user model fields is shown not books field.
I want that with user fields my books model filed also displayed (like book names field more then once because he can have multiple books) , Please let me know if it is possible using django forms/models or I have to user simple html forms with jquery and then save data in models.
Thanks
EDIT:
my models :
class Product(models.Model):
categories = models.CharField(max_length=5, choices = settings.CATEGORIES)
name = models.CharField(max_length=100)
description = models.TextField()
currency = models.CharField(max_length=5, choices = settings.CURRENCY)
status = models.BooleanField(default=True)
def __unicode__(self):
return self.name
class Prices(models.Model):
products = models.ForeignKey(Product)
prices = models.IntegerField()
def __unicode__(self):
return self.id
if you are creating a form for Prices, try putting this in your model form:
products = forms.ModelMultipleChoiceField(queryset=Product.objects.all())
I think you should add required fields in meta class such as
class ThreadForm(ModelForm):
class Meta:
model = Thread
fields = ('Books', 'User')
Please understand the work flow to use foreign keys in model form here.