I have apps and models similar to these:
In App 'Animals':
class Animal(models.Model):
name = models.CharField()
class Lion(Animal):
roar_loudness_decibel = models.IntegerField()
class Bear(Animal):
salmons_eaten = models.IntegerField()
# And many more that inherit Animal
In Application Zoo:
from Animal.models import Animal
class Zoo(models.Model):
animals = models.ManyToManyField(Animal)
In the Django admin interface, I have registered all app models.
Whenever I create a Zoo and want to add animals in the same form (click the green plus button and get a popup), I get the create form for Animal (not the models which have subclassed it).
Is there any way to be able to select which subclassed model to add for a ManyToManyField?
Or is there a better way to solve this (contenttypes? override the form?)
I'd rather not change the models if I can avoid it.
(I know that it is possible to add Bear or Lion separately and just select those objects when selecting animals for a Zoo, but what I'm asking for seems better usability-wise)
Override add_view() method of the AnimalAdmin and if it is called from popup window then show intermediate page with a list of available species.
Just the proof of concept:
admin.py
class AnimalAdmin(admin.ModelAdmin):
def add_view(self, request, form_url='', extra_context=None):
if request.method == 'GET' and '_popup' in request.GET:
return render(request, 'admin/animals/choose_animal.html')
return super(AnimalAdmin, self).add_view(request, form_url,
extra_context)
templates/admin/animals/choose_animal.html
<body>
Choose an animal:
<ul>
<li>Lion</li>
<li>Bear</li>
</ul>
</body>
Should've spent more time Googling for this.
I found https://django-polymorphic.readthedocs.org which offers selecting the child model in the admin interface and much more:
From the docs:
The add screen gains an additional step where the desired child model
is selected.
The edit screen displays the admin interface of the
child model.
Related
Django newbie here. I keep encountering the exact same design paradigm, which seems like it should be common for everyone, yet can't find out how it's supposed to be resolved.
Picture the following ManyToMany relationships:
An organization could have many members; each person could be a member of many organizations
An organization could manage many objects. An object could be in use by multiple organizations. The same applies to the relationship between people and objects.
An organization, person, or object could have multiple media elements (photos, videos, etc) of it, and a single media element could be tagged with numerous organizations, people, or objects
Nothing unusual. But how does a site user add a new person, organization, or object? It seems that if someone is filling out an "add an organization" form, in addition to choosing from existing people, objects, media, etc there should be a button for "new member", "new object", "new photo", etc, and this should take you to the forms for creating new members, objects, media, etc. And when they're done, it should go back to the previous page - whose form filled-out form entries should persist, and the newly created entry should be listed in its respective ManyToMany field.
The problem is, I don't know how to do this. I don't know how one would add a button in the middle of a form, and can't seem to find anything to clarify how to do it. I assume it would need to be a submit button, with a different name / id or some other way so that views.py can treat it differently, via flagging an "incomplete" record in the database. And the new form will need to be passed information about what page it needs to go back to when it's submitted.
Am I thinking about this correctly? If so, then I think the only knowledge I lack is how to add a second submit button in a form and how to recognize its usage in views.py.
If I'm not thinking about this correctly, however, please suggest an alternative paradigm that you think makes more sense :) This is my first Django project, so I'm learning as I do it.
ED: I'm thinking maybe instead of using {{ form.as_p }} to display it, I need to iterate over fields and use some logic to add the extra submit button in the middle as html: What's the best way to add custom HTML in the middle of a form with many fields?
Then I'll just need to figure out a way to detect which submit button was used and put some logic behind it to handle partially-submitted forms, redirecting to a form to create the relation, and then redirecting back on submit... I can probably figure this out...
The first thing I would recommend is to define your models. Lay them all out with the attributes you require. That'll be the foundation for everything else you want to accomplish. You can do everything you mentioned with Django... it's just a matter of coding it. As far as I know you would need to create each model instance separately, and then you can refer to already created instances in the create form for the Organization model for example. I would look into the docs for generic views that help you create objects easily. Then you can link to other create forms if you wish. I don't know how you can create multiple instances of different models in one form, and I don't think it would be the best way to do things even if you can. Here's an example of a model, a create form, a create view, and corresponding url:
# models.py
class Organization(models.Model):
name = models.CharField(max_length=100, null=True, blank=True)
# forms.py
class OrganizationForm(forms.ModelForm):
class Meta:
model = Organization
fields = ('name',)
def __init__(self, *args, **kwargs):
super(OrganizationForm, self).__init__(*args, **kwargs)
self.fields['name'].required = True
def clean(self):
cleaned_data = super(OrganizationForm, self).clean()
name = cleaned_data.get('name')
# views.py
class OrganizationCreateView(CreateView): # inherits from CreateView
form_class = OrganizationForm
template_name = 'create_org.html'
success_url = 'success'
def form_valid(self, form): # validate the form and save the model instance
org = form.save(commit=False)
org.save()
return redirect(reverse('redirect_url'))
# urls.py
from Project.apps.app_name import views as app_views
app_name = 'app_name'
urlpatterns = [
url(r'^create_org/$', app_views.OrganizationCreateView.as_view(), name='create_org'), # as_view() is used for class based views
# create_org.html
<form method="post">
{% crsf_token %}
{{ form.as_p }}
<a href="{% url 'app_name:create_person' %}>Create person</a> # You can link to other create views, and just style the link as a button.
<input type="submit" value="Submit">
</form>
Hope that helps.
One of my Django admin "edit object" pages started loading very slowly because of a ForeignKey on another object there that has a lot of instances. Is there a way I could tell Django to render the field, but not send any options, because I'm going to pull them via AJAX based on a choice in another SelectBox?
You can set the queryset of that ModelChoiceField to empty in your ModelForm.
class MyAdminForm(forms.ModelForm):
def __init__(self):
self.fields['MY_MODEL_CHOIE_FIELD'].queryset = RelatedModel.objects.empty()
class Meta:
model = MyModel
fields = [...]
I think you can try raw_id_fields
By default, Django’s admin uses a select-box interface () for fields that are ForeignKey. Sometimes you don’t want to incur the overhead of having to select all the related instances to display in the drop-down.
raw_id_fields is a list of fields you would like to change into an Input widget for either a ForeignKey or ManyToManyField
Or you need to create a custom admin form
MY_CHOICES = (
('', '---------'),
)
class MyAdminForm(forms.ModelForm):
my_field = forms.ChoiceField(choices=MY_CHOICES)
class Meta:
model = MyModel
fields = [...]
class MyAdmin(admin.ModelAdmin):
form = MyAdminForm
Neither of the other answers worked for me, so I read Django's internals and tried on my own:
class EmptySelectWidget(Select):
"""
A class that behaves like Select from django.forms.widgets, but doesn't
display any options other than the empty and selected ones. The remaining
ones can be pulled via AJAX in order to perform chaining and save
bandwidth and time on page generation.
To use it, specify the widget as described here in "Overriding the
default fields":
https://docs.djangoproject.com/en/1.9/topics/forms/modelforms/
This class is related to the following StackOverflow problem:
> One of my Django admin "edit object" pages started loading very slowly
> because of a ForeignKey on another object there that has a lot of
> instances. Is there a way I could tell Django to render the field, but
> not send any options, because I'm going to pull them via AJAX based on
> a choice in another SelectBox?
Source: http://stackoverflow.com/q/37327422/1091116
"""
def render_options(self, *args, **kwargs):
# copy the choices so that we don't risk affecting validation by
# references (I hadn't checked if this works without this trick)
choices_copy = self.choices
self.choices = [('', '---------'), ]
ret = super(EmptySelectWidget, self).render_options(*args, **kwargs)
self.choices = choices_copy
return ret
In a Django app, I'm having a model Bet which contains a ManyToMany relation with the User model of Django:
class Bet(models.Model):
...
participants = models.ManyToManyField(User)
User should be able to start new bets using a form. Until now, bets have exactly two participants, one of which is the user who creates the bet himself. That means in the form for the new bet you have to chose exactly one participant. The bet creator is added as participant upon saving of the form data.
I'm using a ModelForm for my NewBetForm:
class NewBetForm(forms.ModelForm):
class Meta:
model = Bet
widgets = {
'participants': forms.Select()
}
def save(self, user):
... # save user as participant
Notice the redefined widget for the participants field which makes sure you can only choose one participant.
However, this gives me a validation error:
Enter a list of values.
I'm not really sure where this comes from. If I look at the POST data in the developer tools, it seems to be exactly the same as if I use the default widget and choose only one participant. However, it seems like the to_python() method of the ManyToManyField has its problems with this data. At least there is no User object created if I enable the Select widget.
I know I could work around this problem by excluding the participants field from the form and define it myself but it would be a lot nicer if the ModelForm's capacities could still be used (after all, it's only a widget change). Maybe I could manipulate the passed data in some way if I knew how.
Can anyone tell me what the problem is exactly and if there is a good way to solve it?
Thanks in advance!
Edit
As suggested in the comments: the (relevant) code of the view.
def new_bet(request):
if request.method == 'POST':
form = NewBetForm(request.POST)
if form.is_valid():
form.save(request.user)
... # success message and redirect
else:
form = NewBetForm()
return render(request, 'bets/new.html', {'form': form})
After digging in the Django code, I can answer my own question.
The problem is that Django's ModelForm maps ManyToManyFields in the model to ModelMultipleChoiceFields of the form. This kind of form field expects the widget object to return a sequence from its value_from_datadict() method. The default widget for ModelMultipleChoiceField (which is SelectMultiple) overrides value_from_datadict() to return a list from the user supplied data. But if I use the Select widget, the default value_from_datadict() method of the superclass is used, which simply returns a string. ModelMultipleChoiceField doesn't like that at all, hence the validation error.
To solutions I could think of:
Overriding the value_from_datadict() of Select either via inheritance or some class decorator.
Handling the m2m field manually by creating a new form field and adjusting the save() method of the ModelForm to save its data in the m2m relation.
The seconds solution seems to be less verbose, so that's what I will be going with.
I don't mean to revive a resolved question but I was working a solution like this and thought I would share my code to help others.
In j0ker's answer he lists two methods to get this to work. I used method 1. In which I borrowed the 'value_from_datadict' method from the SelectMultiple widget.
forms.py
from django.utils.datastructures import MultiValueDict, MergeDict
class M2MSelect(forms.Select):
def value_from_datadict(self, data, files, name):
if isinstance(data, (MultiValueDict, MergeDict)):
return data.getlist(name)
return data.get(name, None)
class WindowsSubnetForm(forms.ModelForm):
port_group = forms.ModelMultipleChoiceField(widget=M2MSelect, required=True, queryset=PortGroup.objects.all())
class Meta:
model = Subnet
The problem is that ManyToMany is the wrong data type for this relationship.
In a sense, the bet itself is the many-to-many relationship. It makes no sense to have the participants as a manytomanyfield. What you need is two ForeignKeys, both to User: one for the creator, one for the other user ('acceptor'?)
You can modify the submitted value before (during) validation in Form.clean_field_name. You could use this method to wrap the select's single value in a list.
class NewBetForm(forms.ModelForm):
class Meta:
model = Bet
widgets = {
'participants': forms.Select()
}
def save(self, user):
... # save user as participant
def clean_participants(self):
data = self.cleaned_data['participants']
return [data]
I'm actually just guessing what the value proivded by the select looks like, so this might need a bit of tweaking, but I think it will work.
Here are the docs.
Inspired by #Ryan Currah I found this to be working out of the box:
class M2MSelect(forms.SelectMultiple):
def render(self, name, value, attrs=None, choices=()):
rendered = super(M2MSelect, self).render(name, value=value, attrs=attrs, choices=choices)
return rendered.replace(u'multiple="multiple"', u'')
The first one of the many to many is displayed and when saved only the selected value is left.
I found an easyer way to do this inspired by #Ryan Currah:
You just have to override "allow_multiple_selected" attribut from SelectMultiple class
class M2MSelect(forms.SelectMultiple):
allow_multiple_selected = False
class NewBetForm(forms.ModelForm):
class Meta:
model = Bet
participants = forms.ModelMultipleChoiceField(widget=M2MSelect, required=True, queryset=User.objects.all())
If a django model contains a foreign key field, and if that field is shown in list mode, then it shows up as text, instead of displaying a link to the foreign object.
Is it possible to automatically display all foreign keys as links instead of flat text?
(of course it is possible to do that on a field by field basis, but is there a general method?)
Example:
class Author(models.Model):
...
class Post(models.Model):
author = models.ForeignKey(Author)
Now I choose a ModelAdmin such that the author shows up in list mode:
class PostAdmin(admin.ModelAdmin):
list_display = [..., 'author',...]
Now in list mode, the author field will just use the __unicode__ method of the Author class to display the author. On the top of that I would like a link pointing to the url of the corresponding author in the admin site. Is that possible?
Manual method:
For the sake of completeness, I add the manual method. It would be to add a method author_link in the PostAdmin class:
def author_link(self, item):
return '%s' % (item.id, unicode(item))
author_link.allow_tags = True
That will work for that particular field but that is not what I want. I want a general method to achieve the same effect. (One of the problems is how to figure out automatically the path to an object in the django admin site.)
I was looking for a solution to the same problem and ran across this question... ended up solving it myself. The OP might not be interested anymore but this could still be useful to someone.
from functools import partial
from django.forms import MediaDefiningClass
class ModelAdminWithForeignKeyLinksMetaclass(MediaDefiningClass):
def __getattr__(cls, name):
def foreign_key_link(instance, field):
target = getattr(instance, field)
return u'%s' % (
target._meta.app_label, target._meta.module_name, target.id, unicode(target))
if name[:8] == 'link_to_':
method = partial(foreign_key_link, field=name[8:])
method.__name__ = name[8:]
method.allow_tags = True
setattr(cls, name, method)
return getattr(cls, name)
raise AttributeError
class Book(models.Model):
title = models.CharField()
author = models.ForeignKey(Author)
class BookAdmin(admin.ModelAdmin):
__metaclass__ = ModelAdminWithForeignKeyLinksMetaclass
list_display = ('title', 'link_to_author')
Replace 'partial' with Django's 'curry' if not using python >= 2.5.
I don't think there is a mechanism to do what you want automatically out of the box.
But as far as determining the path to an admin edit page based on the id of an object, all you need are two pieces of information:
a) self.model._meta.app_label
b) self.model._meta.module_name
Then, for instance, to go to the edit page for that model you would do:
'../%s_%s_change/%d' % (self.model._meta.app_label, self.model._meta.module_name, item.id)
Take a look at django.contrib.admin.options.ModelAdmin.get_urls to see how they do it.
I suppose you could have a callable that takes a model name and an id, creates a model of the specified type just to get the label and name (no need to hit the database) and generates the URL a above.
But are you sure you can't get by using inlines? It would make for a better user interface to have all the related components in one page...
Edit:
Inlines (linked to docs) allow an admin interface to display a parent-child relationship in one page instead of breaking it into two.
In the Post/Author example you provided, using inlines would mean that the page for editing Posts would also display an inline form for adding/editing/removing Authors. Much more natural to the end user.
What you can do in your admin list view is create a callable in the Post model that will render a comma separated list of Authors. So you will have your Post list view showing the proper Authors, and you edit the Authors associated to a Post directly in the Post admin interface.
See https://docs.djangoproject.com/en/stable/ref/contrib/admin/#admin-reverse-urls
Example:
from django.utils.html import format_html
def get_admin_change_link(app_label, model_name, obj_id, name):
url = reverse('admin:%s_%s_change' % (app_label, model_name),
args=(obj_id,))
return format_html('%s' % (
url, unicode(name)
))
Given a model with ForeignKeyField (FKF) or ManyToManyField (MTMF) fields with a foreignkey to 'self' how can I prevent self (recursive) selection within the Django Admin (admin).
In short, it should be possible to prevent self (recursive) selection of a model instance in the admin. This applies when editing existing instances of a model, not creating new instances.
For example, take the following model for an article in a news app;
class Article(models.Model):
title = models.CharField(max_length=100)
slug = models.SlugField()
related_articles = models.ManyToManyField('self')
If there are 3 Article instances (title: a1-3), when editing an existing Article instance via the admin the related_articles field is represented by default by a html (multiple)select box which provides a list of ALL articles (Article.objects.all()). The user should only see and be able to select Article instances other than itself, e.g. When editing Article a1, related_articles available to select = a2, a3.
I can currently see 3 potential to ways to do this, in order of decreasing preference;
Provide a way to set the queryset providing available choices in the admin form field for the related_articles (via an exclude query filter, e.g. Article.objects.filter(~Q(id__iexact=self.id)) to exclude the current instance being edited from the list of related_articles a user can see and select from. Creation/setting of the queryset to use could occur within the constructor (__init__) of a custom Article ModelForm, or, via some kind of dynamic limit_choices_to Model option. This would require a way to grab the instance being edited to use for filtering.
Override the save_model function of the Article Model or ModelAdmin class to check for and remove itself from the related_articles before saving the instance. This still means that admin users can see and select all articles including the instance being edited (for existing articles).
Filter out self references when required for use outside the admin, e.g. templates.
The ideal solution (1) is currently possible to do via custom model forms outside of the admin as it's possible to pass in a filtered queryset variable for the instance being edited to the model form constructor. Question is, can you get at the Article instance, i.e. 'self' being edited the admin before the form is created to do the same thing.
It could be I am going about this the wrong way, but if your allowed to define a FKF / MTMF to the same model then there should be a way to have the admin - do the right thing - and prevent a user from selecting itself by excluding it in the list of available choices.
Note: Solution 2 and 3 are possible to do now and are provided to try and avoid getting these as answers, ideally i'd like to get an answer to solution 1.
Carl is correct, here's a cut and paste code sample that would go in admin.py
I find navigating the Django relationships can be tricky if you don't have a solid grasp, and a living example can be worth 1000 time more than a "go read this" (not that you don't need to understand what is happening).
class MyForm(forms.ModelForm):
class Meta:
model = MyModel
def __init__(self, *args, **kwargs):
super(MyForm, self).__init__(*args, **kwargs)
self.fields['myManyToManyField'].queryset = MyModel.objects.exclude(
id__exact=self.instance.id)
You can use a custom ModelForm in the admin (by setting the "form" attribute of your ModelAdmin subclass). So you do it the same way in the admin as you would anywhere else.
You can also override the get_form method of the ModelAdmin like so:
def get_form(self, request, obj=None, **kwargs):
"""
Modify the fields in the form that are self-referential by
removing self instance from queryset
"""
form = super().get_form(request, obj=None, **kwargs)
# obj won't exist yet for create page
if obj:
# Finds fieldnames of related fields whose model is self
rmself_fields = [f.name for f in self.model._meta.get_fields() if (
f.concrete and f.is_relation and f.related_model is self.model)]
for fieldname in rmself_fields:
form.base_fields[fieldname]._queryset =
form.base_fields[fieldname]._queryset.exclude(id=obj.id)
return form
Note that this is a on-size-fits-all solution that automatically finds self-referencing model fields and removes self from all of them :-)
I like the solution of checking at save() time:
def save(self, *args, **kwargs):
# call full_clean() that in turn will call clean()
self.full_clean()
return super().save(*args, **kwargs)
def clean(self):
obj = self
parents = set()
while obj is not None:
if obj in parents:
raise ValidationError('Loop error', code='infinite_loop')
parents.add(obj)
obj = obj.parent