Is there a way to de-couple django admin inline-models from clustering like models together?
A bit of context: I have a model named Page with two inline-models, TextBlock and GalleryContainer. I would to render TextBlocks and GalleryContainers on a template based on the order they're added in the Page admin editor. The default django-admin display looks like this:
I would like it to display as:
Gallery Container 1
Textblock 1
Gallery Container 2
But I have no idea how to do that. Any suggestions or nudges in the right direction would be a great help. Thanks in advance. (I also hope my question makes sense...)
If you want a relation between a column and a gallery your models should reflect that. So if I understand correctly: A page has name and columns. A column has text, gallery (optional) and ordering.
models.py:
class Page(models.Model):
name = models.CharField(max_length=200)
class Gallery(models.Model):
name = models.CharField(max_length=200)
class Column(models.Model):
page = models.ForeignKey(Page)
text = models.TextField()
gallery = models.ForeignKey(Gallery, null=True, blank=True) # Optional
ordering = models.IntegerField()
class Meta:
ordering = ('ordering', )
This example shows how ordering is done by an IntegerField. If you want to order the columns based on the moment they where added replace
ordering = models.IntegerField()
with
models.datetimeField(auto_now_add=True)
In your admin.py:
class ColumnInline(admin.TabularInline): # or StackedInline
model = Column
class PageAdmin(admin.ModelAdmin):
inlines = [
ColumnInline,
]
Note: I put your 'gallery display name' in Gallery as 'name'. Makes more sense to give the gallery it's name only once. But if a gallery is in more places with different names, than you need a field (e.g. 'gallery_display_name=models.CharField(max_lenght=200)') on the Column model.
Is this the answer to your question? I hope it helps!
Related
I'm new to programming.
In my blog I want to show a list of categories.
If I create a queryset like this:
Category.objects.all()
my django-modeltranslation works perfectly.
But I want to get categories of only published posts. Then my queryset is:
Post.objects.values('category__name').filter(is_published=True)
However, django-modeltranslation doesn't work. I get values from 'name' field instead 'name_en' or 'name_ru' fields.
What is wrong?
Here's my models.py :
class Category(models.Model):
name = models.TextField(max_length=100)
url = models.SlugField(max_length=160, unique=True)
class Post(models.Model):
title = models.TextField('title', max_length=150)
category = models.ManyToManyField(Category, related_name='posts', blank=True)
I think you better query in reverse: with .values(…) you select a specific database column, so this will omit the model logic.
You can retrieve the categories with:
Category.objects.filter(posts__is_published=True).distinct()
I'm learning Django right now and I made this class called Clowns. On the Django Admin page I made two test objects.
I made four attributes(dunno what they're called lol) for clowns. They are title, description, identification, and hobbies. See below:
title = models.CharField(max_length=20)
description = models.TextField()
identification = models.CharField(default='-', max_length=5)
hobbies= models.TextField()
To make the "CLOWN" column you see in the image I added this to the clown class:
def __str__(self):
return self.title
I seem to only be able to do this with one attribute/category, in this case it's title. How do I make another column for another attribute, say identification?
in the admin.py file create custom ModelAdmin for your model clown, and register it
from django.contrib import admin
from .models import Clown
class ClownAdmin(admin.ModelAdmin):
list_display = ('title', 'identification', 'hobbies')
admin.site.register(Clown, ClownAdmin)
I've got a two part question regarding Django Admin.
Firstly, I've got a Django model Classified that has a foreign key field from another table Address. On setting data, I've got no issues with any of the fields and all fields get saved correctly.
However, if I want to edit the foreign field in the entry in Classified, it doesn't display the old/existing data in the fields. Instead it shows empty fields in the popup that opens.
How do I get the fields to display the existing data on clicking the + so that I can edit the correct information?
Secondly, I'm sure I've seen search fields in Django Admin. Am I mistaken? Is there a way for me to implement search in the admin panel? I have over 2 million records which need to be updated deleted from time to time. It's very cumbersome to manually go through all the pages in the admin and delete or edit those.
Adding Model Code:
Classified
class Classified(models.Model):
id = models.AutoField(primary_key=True)
name = models.CharField(max_length=256)
contact_person = models.CharField(max_length=300, blank=True)
email = models.CharField(max_length=100, blank=True)
address = models.ForeignKey(Address)
subcategory = models.ForeignKey(Subcategory)
Address
class Address(models.Model):
id = models.AutoField(primary_key=True)
build_add = models.CharField(max_length=255)
street_add = models.CharField(max_length=255)
area = models.CharField(max_length=255)
city = models.ForeignKey(Cities)
The + means just that - add a new instance of the related object and relate the object you're editing to that. Because you're adding a new object it will be blank to start. If you want to be able to edit existing related objects from another object's admin you need to use inlines.
In your app's admin.py have something like:
from django.contrib import admin
from yourapp.models import Address, Classified
class AddressInline(admin.TabularInline):
model = Address
class ClassifiedAdmin(admin.ModelAdmin):
inlines = [AddressInline,]
admin.site.register(Classified, ClassifiedAdmin)
Adding search from there is really easy.
...
class ClassifiedAdmin(admin.ModelAdmin):
inlines = [AddressInline,]
search_fields = [
'field_you_want_to_search',
'another_field',
'address__field_on_relation',
]
...
Note the double underscore in that last one. That means you can search based on values in related objects' fields.
EDIT: This answer is right in that your foreignkey relationship is the wrong way round to do it this way - with the models shown in your question Classified would be the inline and Address the primary model.
Is it possible to do a reverse relation search on the Django Admin interface?
My Django app database schema consists of the following models:
class Tag(models.Model):
title = models.CharField(max_length=50)
class Publication(models.Model):
title = models.CharField(max_length=200)
tags = models.ManyToManyField(Tag, blank=True, related_name="publications")
I have added a search field for looking up tags by title in my admin.py file by doing:
class TagAdmin(admin.ModelAdmin):
list_display = ('title',)
search_fields = ('title',)
Thus, when I type a tag title into the search field on the django admin interface, a list of matching tag titles comes up. Now I'd like to make it so that if I type a tag title into the search field, matching publications come up.
In other words, I'm imagining something like:
class TagAdmin(admin.ModelAdmin):
list_display = ('title',)
search_fields = ('publications',)
Which of course doesn't work... but that's the idea...
Is this even possible? And/or am I even going about this the right way? If so, could someone suggest a way to do this or a resource? If you are kind enough to do so, please keep in mind that I am very much a beginner. Thanks.
You shouldn't try to do this using an admin class registered to your Tag model. Instead, set up an admin class for Publication and set its search_fields:
class PublicationAdmin(admin.ModelAdmin):
list_display = ('title',)
search_fields = ('tags__title',)
Please have a look:
class Categorie(models.Model):
id = models.AutoField('id', primary_key=True)
title = models.CharField('title', max_length=800)
articles = models.ManyToManyField(Article)
class Article(models.Model):
id = models.AutoField('id', primary_key=True)
title = models.CharField('title', max_length=800)
slug = models.SlugField()
indexPosition = models.IntegerField('indexPosition', unique=True)
class CookRecette(Article):
ingredient = models.CharField('ingredient', max_length=100)
class NewsPaper(Article):
txt = models.CharField('ingredient', max_length=100)
So I created "CookRecette" and "NewsPaper" as "Article".
I Also create a "Categorie" class who link to (manyToMany) "Article".
But in the admin interface, I can't link from "Categorie" to an "CookRecette"or "NewsPaper".
Same from the code.
Any help ?
Cheers,
Martin Magakian
PS: I'm so sorry but actually this code is correct! So everything is working fine, I can see my "CookRecette"or "NewsPaper" from "Categorie"
I'll start by saying that you don't need to define the 'id' field, if you don't define it then Django will add it automatically.
Secondly, the CookRecette and NewsPaper objects are not linked to the Categorie object by any means (ForeignKey, OneToOne, OneToMany, ManyToMany) so they wouldn't be able to be accessed that way anyway.
After you have linked the models together in whichever way you wish, you might want to have a look at http://docs.djangoproject.com/en/1.2/ref/contrib/admin/#django.contrib.admin.InlineModelAdmin which will show you how to quickly edit related objects in the Djano admin console.
NewsPaper has part of it as Article object. If you will create new NewsPaper object, you will see a new object in articles. So in admin interface, when managing Categories, you will be able to select any article, and some of them are NewsPaper.
You can add news paper to a category like this:
category = Categorie(title='Abc')
category.save()
news_paper = NewsPaper(slug='Something new', indexPosition=1, txt='...')
news_paper.save()
category.articles.add(news_paper)
You can retrieve news papers from specific category like this:
specific_category = Categorie.objects.get(title='Abc')
NewsPaper.objects.filter(categorie_set=specific_category)