I have a model OrderPage which is manytomany to Site. In Django admin, I want to restrict the selection of sites(Sites which belong to existing OrderPage can not be selected again). Can I do it with unique_together ? I get an error with following model ManyToManyFields are not supported in unique_together
class OrderPage(models.Model):
description = models.CharField(max_length=255, blank=False)
sites = models.ManyToManyField(Site)
class Meta:
unique_together = (('id', 'sites'),)
class Order(models.Model):
order_page = models.ForeignKey(OrderPage)
class OrderPageAdmin(admin.ModelAdmin):
filter_horizontal = ('sites',)
admin.site.register(OrderPage, OrderPageAdmin)
If an Site can have only one OrderPage, you don't need to worry about unique_together.
Ideally you should subclass Site and use a ForeignKey from that to OrderPage. That would natively give you what you're looking for: each site would be able to have one OrderPage, and each OrderPage multiple Sites. This would be the cleanest but you would have to use your subclass throughout the program in place of the original Site which might be more work than you want right now.
class BetterSite(Site):
order_page = models.ForeignKey('OrderPage')
The dirtier way is to keep your M2M and just set the site as unique, since there should only ever be one entry on each site in the M2M table. You would use a 'through' table so you could set the custom uniqueness value:
class OrderPage(models.Model):
description = models.CharField(max_length=255, blank=False)
sites = models.ManyToManyField(Site, through='OrderPageToSite')
class OrderPageToSite(models.Model):
order_page = models.ForeignKey(OrderPage)
site = models.ForeignKey(Site, unique=True)
(Note that I've left these simple but in your FK fields you should also consider setting on_delete and related_name)
Related
My model is:
class CustomerAccount(models.Model):
name = models.CharField(max_length=50)
class MyUser(AbstractUser):
customer_account = models.ManyToManyField(CustomerAccount, related_name='users', blank=True)
default_customer_account = models.ForeignKey(CustomerAccount, related_name='users_using_default_account', null=True, blank=True)
I want to display in the admin interface of the CustomerAccount this sort of thing:
I don't need to add a MyUser in the CustomerAccount interface.
Most SO questions and docs are related to show an Inline class in the admin, but I don't need it.
How should I do?
The functinality shown above you get by making adding the desired field to the filter_horizontal list (docu).
I have just begun to play around with Django admin views, and to start off, I am trying to do something very simple: showing several fields in the listing of objects using list_display as explained here: https://docs.djangoproject.com/en/dev/ref/contrib/admin/
This is my dead simple code:
class ArticleAdmin(admin.ModelAdmin):
list_display = ('title', 'category')
Unfortunately, the list_display option is causing the columnar view to appear, but only some of the objects (40 out of 85) are now displaying in the listing. I cannot deduce why certain objects are showing over the others - their fields look like they are filled similarly. It's clearly not paginating, because when I tried it on an admin of another model, it showed only 2 objects out of about 70 objects.
What might be going on here?
[UPDATE] Article Model:
class Article(models.Model):
revision = models.ForeignKey('ArticleRevision', related_name="current_revision")
category = models.ForeignKey('meta.Category')
language = models.ForeignKey('meta.Language', default=get_default_language)
created = models.DateTimeField(auto_now_add=True, editable=False)
changed = models.DateTimeField(auto_now=True, editable=False)
title = models.CharField(max_length=256)
resources = models.ManyToManyField('oer.Resource', blank=True)
image = models.ManyToManyField('media.Image', blank=True)
views = models.IntegerField(editable=False, default=0)
license = models.ForeignKey('license.License', default=get_default_license)
slug = models.SlugField(max_length=256)
difficulty = models.PositiveIntegerField(editable=True, default=0)
published = models.NullBooleanField()
citation = models.CharField(max_length=1024, blank=True, null=True)
Before adding list_display:
After adding list_display:
[UPDATE] This behaviour occurs only when ForeignKey fields are included in list_display tuple. Any of them.
[UPDATE] Category model code:
class Category(models.Model):
title = models.CharField(max_length=256)
parent = models.ForeignKey('self')
project = models.NullBooleanField(default=False)
created = models.DateTimeField(auto_now_add=True, editable=False)
slug = models.SlugField(max_length=256, blank=True)
def __unicode__(self):
return self.title
This behavior is caused by a foreign key relation somewhere that is not declared as nullable, but nonetheless has a null value in the database. When you have a ManyToOne relationship in list_display, the change list class will always execute the query using select_related. (See the get_query_set method in django.contrib.admin.views.ChangeList).
select_related by default follows all foreign keys on each object, so any broken foreign key found by this query will cause data to drop out when the query is evaluated. This is not specific to the admin; you can interactively test it by comparing the results of Article.objects.all() to Article.objects.all().select_related().
There's no simple way to control which foreign keys the admin will look up - select_related takes some parameters, but the admin doesn't expose a way to pass them through. In theory you could write your own ChangeList class and override get_query_set, but I don't recommend that.
The real fix is to make sure your foreign key model fields accurately reflect the state of your database in their null settings. Personally, I'd probably do this by commenting out all FKs on Article other than Category, seeing if that helps, then turning them back on one by one until things start breaking. The problem doesn't have to be with a FK on an article itself; if a revision, language or category has a broken FK that will still cause the join to miss rows. Or if something they relate to has a broken FK, etc etc.
I am working with an existing database that I can not modify and having some trouble trying to deal with presenting forms for modifying the database in Django. The structure in question is as follows and all models are unmanaged.
class Persons(models.Model):
personid = models.BigIntegerField(primary_key=True, db_column='PersonID')
....
class Phones(models.Model):
phoneid = models.BigIntegerField(primary_key=True, db_column='PhoneID')
number = models.CharField(max_length=60, db_column='Number', blank=True)
type = models.CharField(max_length=15, db_column='Type', blank=True)
...
class Personsphones(models.Model):
personphoneid = models.BigIntegerField(primary_key=True, db_column='PersonPhoneID')
personid = models.ForeignKey(Persons, db_column='PersonID')
phoneid = models.ForeignKey(Phones, db_column='PhoneID')
...
I want to create a form to display all of the 'Phones' associated with a particular 'Persons' and in addition be able to modify/add/remove 'Phones' belonging to a 'Persons'. Right now the only thing I can think of is to display the 'Phones' in a modelformset and then if one is added or removed manually set the 'Personsphones' relation. Any ideas on how to best deal with this model setup?
For making changes to your models you may want to use django-south http://south.aeracode.org/docs/
As far as displaying your 'Phone' under your forms.py you may want to set up class meta like so. With this any changes made to models will reflect on change
class Meta:
model = Persons
exclude = ('user')
In models you may want to use Foreignkey fore relationships between phones and Persons. Better seen in action here https://docs.djangoproject.com/en/dev/ref/models/fields/#foreignkey
How would I go about prepopulating a slug from a foreign key? Here are how some of my models are setup:
Class Title(models.Model):
title = models.CharField(max_length=256)
slug = models.SlugField()
class Issue(models.Model):
title = models.ForeignKey(Title)
number = models.IntegerField(help_text="Do not include the '#'.")
slug = models.SlugField()
admin.py:
class IssueAdmin (admin.ModelAdmin):
prepopulated_fields = {"slug": ("title",)}
admin.site.register(Issue, IssueAdmin)
What the Issue prepopulates is the ID of the foreign key, but I suppose I would need it to preopulate the slug of the foreign key. How would I go about doing this? I am using Django 1.3. I have checked other threads, but they seem to refer to version of Django a few years older that don't work anymore.
I need the Titles to display the list of issues. So far, it works. And you can click on the link to the issue to see what the issue displays.
I feel as if reworking the Title to abstract classes the way Skidoosh will not allow me to view subsets of objects....
If you check the docs (http://docs.djangoproject.com/en/dev/ref/contrib/admin/#django.contrib.admin.ModelAdmin.prepopulated_fields) it does state that you can't reference a foreign key field.
Looking at you design would this not work better:
class BaseModel(models.Model):
title = models.CharField(max_length=256)
slug = models.SlugField()
class Issue(BaseModel):
number = models.IntegerField(help_text="Do not include the '#'.")
class ComicBookSeries(BaseModel):
issues = models.ForeignKey(Issue)
You need to declare the classes in that order!
Hope that helps!
Django has a unique_for_date property you can set when adding a SlugField to your model. This causes the slug to be unique only for the Date of the field you specify:
class Example(models.Model):
title = models.CharField()
slug = models.SlugField(unique_for_date='publish')
publish = models.DateTimeField()
What would be the best way to achieve the same kind of functionality for a non-DateTime field like a ForeignKey? Ideally, I want to do something like this:
class Example(models.Model):
title = models.CharField()
slug = models.SlugField(unique_for='category')
category = models.ForeignKey(Category)
This way I could create the following urls:
/example/category-one/slug
/example/category-two/slug
/example/category-two/slug <--Rejected as duplicate
My ideas so far:
Add a unique index for the slug and categoryid to the table. This requires code outside of Django. And would the built-in admin handle this correctly when the insert/update fails?
Override the save for the model and add my own validation, throwing an error if a duplicate exists. I know this will work but it doesn't seem very DRY.
Create a new slug field inheriting from the base and add the unique_for functionality there. This seems like the best way but I looked through the core's unique_for_date code and it didn't seem very intuitive to extend it.
Any ideas, suggestions or opinions on the best way to do this?
What about unique_together?
class Example(models.Model):
title = models.CharField()
slug = models.SlugField(db_index=False)
category = models.ForeignKey(Category)
class Meta:
unique_together = (('slug','category'),)
# or also working since Django 1.0:
# unique_together = ('slug','category',)
This creates an index, but it is not outside of Django ;) Or did I miss the point?