here is the models page
In this picture, only the title shows up on here, I used:
def __unicode__(self):
return self.title;
here is the each individual objects
How do I show all these fields?
How do I show all the fields in each Model page?
If you want to include all fields without typing all fieldnames, you can use
list_display = BookAdmin._meta.get_all_field_names()
The drawback is, the fields are in sorted order.
Edit:
This method has been deprecated in Django 1.10 See Migrating from old API for reference. Following should work instead for Django >= 1.9 for most cases -
list_display = [field.name for field in Book._meta.get_fields()]
By default, the admin layout only shows what is returned from the object's unicode function. To display something else you need to create a custom admin form in app_dir/admin.py.
See here: https://docs.djangoproject.com/en/dev/ref/contrib/admin/#django.contrib.admin.ModelAdmin.list_display
You need to add an admin form, and setting the list_display field.
In your specific example (admin.py):
class BookAdmin(admin.ModelAdmin):
list_display = ('title', 'author', 'price')
admin.site.register(Book, BookAdmin)
If you want to include all but the ManyToManyField field names, and have them in the same order as in the models.py file, you can use:
list_display = [field.name for field in Book._meta.fields if field.name != "id"]
As you can see, I also excluded the id.
If you find yourself doing this a lot, you could create a subclass of ModelAdmin:
class CustomModelAdmin(admin.ModelAdmin):
def __init__(self, model, admin_site):
self.list_display = [field.name for field in model._meta.fields if field.name != "id"]
super(CustomModelAdmin, self).__init__(model, admin_site)
and then just inherit from that:
class BookAdmin(CustomModelAdmin):
pass
or you can do it as a mixin:
class CustomModelAdminMixin(object):
def __init__(self, model, admin_site):
self.list_display = [field.name for field in model._meta.fields if field.name != "id"]
super(CustomModelAdminMixin, self).__init__(model, admin_site)
class TradeAdmin(CustomModelAdminMixin, admin.ModelAdmin):
pass
The mixin is useful if you want to inherit from something other than admin.ModelAdmin.
I found OBu's answer here to be very useful for me. He mentions:
The drawback is, the fields are in sorted order.
A small adjustment to his method solves this problem as well:
list_display = [f.name for f in Book._meta.fields]
Worked for me.
The problem with most of these answers is that they will break if your model contains ManyToManyField or ForeignKey fields.
For the truly lazy, you can do this in your admin.py:
from django.contrib import admin
from my_app.models import Model1, Model2, Model3
#admin.register(Model1, Model2, Model3)
class UniversalAdmin(admin.ModelAdmin):
def get_list_display(self, request):
return [field.name for field in self.model._meta.concrete_fields]
Many of the answers are broken by Django 1.10. For version 1.10 or above, this should be
list_display = [f.name for f in Book._meta.get_fields()]
Docs
Here is my approach, will work with any model class:
MySpecialAdmin = lambda model: type('SubClass'+model.__name__, (admin.ModelAdmin,), {
'list_display': [x.name for x in model._meta.fields],
'list_select_related': [x.name for x in model._meta.fields if isinstance(x, (ManyToOneRel, ForeignKey, OneToOneField,))]
})
This will do two things:
Add all fields to model admin
Makes sure that there is only a single database call for each related object (instead of one per instance)
Then to register you model:
admin.site.register(MyModel, MySpecialAdmin(MyModel))
Note: if you are using a different default model admin, replace 'admin.ModelAdmin' with your admin base class
Show all fields:
list_display = [field.attname for field in BookModel._meta.fields]
Every solution found here raises an error like this
The value of 'list_display[n]' must not be a ManyToManyField.
If the model contains a Many to Many field.
A possible solution that worked for me is:
list_display = [field.name for field in MyModel._meta.get_fields() if not x.many_to_many]
I like this answer and thought I'd post the complete admin.py code (in this case, I wanted all the User model fields to appear in admin)
from django.contrib import admin
from django.contrib.auth.models import User
from django.db.models import ManyToOneRel, ForeignKey, OneToOneField
MySpecialAdmin = lambda model: type('SubClass'+model.__name__, (admin.ModelAdmin,), {
'list_display': [x.name for x in model._meta.fields],
'list_select_related': [x.name for x in model._meta.fields if isinstance(x, (ManyToOneRel, ForeignKey, OneToOneField,))]
})
admin.site.unregister(User)
admin.site.register(User, MySpecialAdmin(User))
I'm using Django 3.1.4 and here is my solution.
I have a model Qualification
model.py
from django.db import models
TRUE_FALSE_CHOICES = (
(1, 'Yes'),
(0, 'No')
)
class Qualification(models.Model):
qual_key = models.CharField(unique=True, max_length=20)
qual_desc = models.CharField(max_length=255)
is_active = models.IntegerField(choices=TRUE_FALSE_CHOICES)
created_at = models.DateTimeField()
created_by = models.CharField(max_length=255)
updated_at = models.DateTimeField()
updated_by = models.CharField(max_length=255)
class Meta:
managed = False
db_table = 'qualification'
admin.py
from django.contrib import admin
from models import Qualification
#admin.register(Qualification)
class QualificationAdmin(admin.ModelAdmin):
list_display = [field.name for field in Qualification._meta.fields if field.name not in ('id', 'qual_key', 'qual_desc')]
list_display.insert(0, '__str__')
here i am showing all fields in list_display excluding 'id', 'qual_key', 'qual_desc' and inserting '__str__' at the beginning.
This answer is helpful when you have large number of modal fields, though i suggest write all fields one by one for better functionality.
list_display = [field.name for field in Book._meta.get_fields()]
This should work even with python 3.9
happy coding
I needed to show all fields for multiple models, I did using the following style:
#admin.register(
AnalyticsData,
TechnologyData,
TradingData,
)
class CommonAdmin(admin.ModelAdmin):
search_fields = ()
def get_list_display(self, request):
return [
field.name
for field in self.model._meta.concrete_fields
if field.name != "id" and not field.many_to_many
]
But you can also do this by creating a mixin, if your models have different fields.
Related
Say I have the following model in a Wagtail app:
# models.py
from django.db import models
from django.db.models.deletion import CASCADE
from wagtail.admin.edit_handlers import InlinePanel, FieldPanel
from modelcluster.models import ParentalKey, ClusterableModel
class Person(ClusterableModel):
name = models.CharField(max_length=300)
contact_for = ParentalKey(
'self', on_delete=CASCADE, null=True, related_name='contacts'
)
panels = [
FieldPanel('name'),
FieldPanel('contact_for'),
InlinePanel('contacts')
]
And the following hooks:
# wagtail_hooks.py
from wagtail.contrib.modeladmin.options import ModelAdmin, modeladmin_register
from home.models import Person
class PersonAdmin(ModelAdmin):
model = Person
menu_label = 'People'
menu_icon = 'list-ul'
menu_order = 200
add_to_settings_menu = False
exclude_from_explorer = False
list_display = ('name',)
search_fields = ('name',)
list_filter = ('contact_for',)
modeladmin_register(PersonAdmin)
When running the above and navigating to Admin > People > Add Person I will get a RecursionError exception (maximum recursion depth exceeded).
I'm guessing this is a Wagtail related issue, because I can use the class as intended in IPython:
./manage.py shell -i ipython
In [1]: from home.models import Person
In [2]: mike = Person(name='Mike')
In [3]: mike.contacts = [Person(name='Sam'), Person(name='John')]
In [4]: mike.contacts
Out[4]: <modelcluster.fields.create_deferring_foreign_related_manager.<locals>.DeferringRelatedManager at 0x7f36e9548af0>
In [5]: mike.save()
In [6]: [person.name for person in Person.objects.all()]
Out[6]: ['Mike', 'Sam', 'John']
In [7]: Person.objects.all()[1].contact_for.name
Out[7]: 'Mike'
So, is there a way I can make use of a recursive ParentalKey in Wagtail? What am I doing wrong/missing?
Edit: I just found this answer. So I'm wondering if I should even be trying to use ParentalKey (and ParentalManyToManyField) for non Page models.
Edit 2: For anyone interested, I ended up splitting my Model to avoid a recursive key. InlinePanels do work with ModelAdmin classes, it's just Wagtail doesn't seem to support recursive keys. I have also opted to use Snippets and SnippetChooserPanel where it felt appropriate.
A ParentalKey means that the child model is notionally treated as 'part of' the parent model and doesn't exist as an independent entity - for example, an image gallery being part of a page - for purposes such as versioning, and moderation workflow. (In Wagtail, non-page models handled through snippets or ModelAdmin don't have these features, but in order for them to share the same InlinePanel mechanism as pages, they also use ParentalKey.)
In this case, a Person is not part of another Person, and needs to be editable independently of the 'parent', so a ParentalKey isn't appropriate here. Instead, you should use a ForeignKey, which just indicates some relation between the models. This does mean that you can't use an InlinePanel to manage multiple Person records within the same view - you'll have to edit them separately, and use something like SnippetChooserPanel or a simple dropdown to set up the relations between them.
I am using subclassed model of Wagtail Page.
In below code you can see that PhoenixPage is base page which subclasses Wagtail Page model.
PhoenixArticlePage & PhoenixMealPrepPage subclasses PhoenixPage
PhoenixArticleIndexPage subclasses PhoenixBaseIndexPage which in turn subclasses PhoenixPage
Idea is to use PhoenixArticleIndexPage for all other article pages.
Problem is even after using the specific() method on queryset i am unable to use filter or any other operation on the queryset.
i tried using order_by() as well as filter()
Can someone share some insights here ? what might be wrong ?
Here is a model example:
class PhoenixPage(Page):
"""
General use page with caching, templating, and SEO functionality.
All pages should inherit from this.
"""
class Meta:
verbose_name = _("Phoenix Page")
# Do not allow this page type to be created in wagtail admin
is_creatable = False
tags = ClusterTaggableManager(
through=PhoenixBaseTag,
verbose_name="Tags",
blank=True,
related_name="phoenixpage_tags",
)
class PhoenixBaseIndexPage(PaginatedListPageMixin, PhoenixPage):
class meta:
verbose_name = "Phoenix Base Index Page"
app_label = "v1"
index_show_subpages_default = True
is_creatable = False
class PhoenixArticleIndexPage(PhoenixBaseIndexPage):
class Meta:
verbose_name = "Phoenix Article Index Page"
app_label = "v1"
class PhoenixArticlePage(PhoenixPage):
class Meta:
verbose_name = "Phoenix Article Page"
app_label = "v1"
subpage_types = []
parent_page_types = ["v1.PhoenixArticleIndexPage"]
class PhoenixMealPrepPage(PhoenixPage):
class Meta:
verbose_name = "Phoenix Meal Prep Page"
app_label = "v1"
subpage_types = []
parent_page_types = ["v1.PhoenixArticleIndexPage"]
Here are shell queries i tried.
Index page
In [4]: a = PhoenixArticleIndexPage.objects.all()[0]
In [5]: a
Out[5]: <PhoenixArticleIndexPage: articles>
As expected, get_children returning all instances of Wagtail Page.
In [6]: a.get_children()
Out[6]: <PageQuerySet [<Page: article title>, <Page: article title2>, <Page: Our 30-Day Reset Recipes Are So Easy AND Delicious>]>
Getting specific children from the Index page.
In [7]: a.get_children().specific()
Out[7]: <PageQuerySet [<PhoenixArticlePage: article title>, <PhoenixArticlePage: article title2>, <PhoenixMealPrepPage: Our 30-Day Reset Recipes Are So Easy AND Delicious>]>
Get Tag and try to filter the queryset
In [8]: q = a.get_children().specific()
In [12]: m = PhoenixTag.objects.get(slug='meal')
In [16]: k={"tags":m}
In [19]: q.filter(**k)
***FieldError: Cannot resolve keyword 'tags' into field. Choices are ...***
But if i go to particular entry in queryset then i can see tags field on it.
In [15]: q[2]
Out[15]: <PhoenixMealPrepPage: Our 30-Day Reset Recipes Are So Easy AND Delicious>
In [16]: q[2].tags
Out[16]: <modelcluster.contrib.taggit._ClusterTaggableManager at 0x1060832b0>
Could be different question all together but for reference adding it here.
Found the corner case of using difference() and specific() method on a queryset.
In [87]: q = PhoenixPage.objects.child_of(a).live()
In [89]: f = q.filter(featured=True)[:3]
In [91]: l = q.difference(f)
In [93]: l.order_by(a.index_order_by).specific() . <-- does not work
DatabaseError: ORDER BY term does not match any column in the result set.
The specific() method on PageQuerySet works by running the initial query on the basic Page model as normal, then running additional queries - one for each distinct page type found in the results - to retrieve the information from the specific page models. This means it's not possible to use fields from the specific model in filter or order_by clauses, because those have to be part of the initial query, and at that point Django has no way to know which page models are involved.
However, if you know that your query should only ever return pages of one particular type (PhoenixPage in this case) containing the field you want to filter/order on, you can reorganise your query expression so that the query happens on that model instead:
PhoenixPage.objects.child_of(a).filter(tags=m).specific()
I have a Django-tastypie resource that represents a banner and has a field called impression that I increment whenever the banner appears on the site.
class BannerResource(ModelResource):
owner = fields.ForeignKey('advertisment.api.AdvertiserResource', 'owner', full=True)
class Meta:
queryset = Banner.objects.all()
resource_name = 'banner'
authorization = Authorization()
I would like to get the banner that has the minimum impression, in the official documentation there is nothing like
filtering = {'impressions': ('min',)}
I'm using BackboneJS in the front end and I could get all the banners with Backbone collection and do the filtering with JavaScript but I'm looking for a quicker way to do it.
Any ideas?
Thanks
If you'd like to retrieve banners with number of impressions greater than X you need to things. For one you need to define possible filtering operations on your resource like so (given your model has impressions field):
class BannerResource(ModelResource):
owner = fields.ForeignKey('advertisment.api.AdvertiserResource', 'owner', full=True)
class Meta:
queryset = Banner.objects.all()
resource_name = 'banner'
authorization = Authorization()
filtering = { 'impressions' : ALL }
for available options take a look at Tastypie's documentation on filtering.
Then if you made the following request:
GET http://<your_host>/v1/banners?impressions__gte=X
you should get what you need.
I am trying to achieve a category model where name has unique=True,
but practically I can still add same category name with different cases.
i.e. I have a category called Food
I am still able to add food, FOOD, fOod, FOOd
Is their any philosophy behind this? or it is a work in progress.
Cause in real world if I think of Category Food, it will always be food, no matter what case it has used to mention itself.
Thank you in advance to look at this.
To answer my own question:
I have found I can have clean method on my model. So I added
class Category(models.Model):
name = models.CharField(max_length=200, unique=True)
def clean(self):
self.name = self.name.capitalize()
It is capitalising the first letter, which is then handled by the save method, which calls the validate_unique method to raise error.
You can use Postgre specific model field called Citext fields (case insensitive fields).
There are three option at the moment:
class CICharField(**options), class CIEmailField(**options) and class CITextField(**options)
Example:
from django.db import models
from django.contrib.postgres.fields import CICharField
class Category(models.Model):
name = CICharField(verbose_name="Name", max_length=255)
But don't forget to create an extension for the citext fields.
See here.
Basically, you have to add the extension class in the migration file, inside the operations array, before the first CreateModel operation.
# migration file
operations = [
CITextExtension(), # <------ here
migrations.CreateModel(
...
),
...,
]
Setting the column to case-insensitive collation should fix this. You may need to do it at the SQL level.
i'm experimenting with django and the builtin admin interface.
I basically want to have a field that is a drop down in the admin UI. The drop down choices should be all the directories available in a specified directory.
If i define a field like this:
test_folder_list = models.FilePathField(path=/some/file/path)
it shows me all the files in the directory, but not the directories.
Does anyone know how i can display the folders?
also i tried doing
test_folder_list = models.charField(max_length=100, choices=SOME_LIST)
where SOME_LIST is a list i populate using some custom code to read the folders in a directory. This works but it doesn't refresh. i.e. the choice list is limited to a snapshot of whatever was there when running the app for the first time.
thanks in advance.
update:
after some thinking and research i discovered what i want may be to either
1. create my own widget that is based on forms.ChoiceField
or
2. pass my list of folders to the choice list when it is rendered to the client
for 1. i tried a custom widget.
my model looks like
class Test1(models.Model):
test_folder_ddl = models.CharField(max_length=100)
then this is my custom widget:
class FolderListDropDown(forms.Select):
def __init__(self, attrs=None, target_path):
target_folder = '/some/file/path'
dir_contents = os.listdir(target_folder)
directories = []
for item in dir_contents:
if os.path.isdir(''.join((target_folder,item,))):
directories.append((item, item),)
folder_list = tuple(directories)
super(FolderListDropDown, self).__init__(attrs=attrs, choices=folder_list)
then i did this in my modelForm
class test1Form(ModelForm):
test_folder_ddl = forms.CharField(widget=FolderListDropDown())
and it didn't seem to work.What i mean by that is django didn't want to use my widget and instead rendered the default textinput you get when you use a CharField.
for 2. I tried this in my ModelForm
class test1Form(ModelForm):
test_folder_ddl = forms.CharField(widget=FolderListDropDown())
test_folder_ddl.choices = {some list}
I also tried
class test1Form(ModelForm):
test_folder_ddl = forms.ChoiceField(choices={some list})
and it would still render the default char field widget.
Anyone know what i'm doing wrong?
Yay solved. after beating my head all day and going through all sorts of examples by people i got this to work.
basically i had the right idea with #2. The steps are
- Create a ModelForm of our model
- override the default form field user for a models.CharField. i.e. we want to explcitly say use a choiceField.
- Then we have to override how the form is instantiated so that we call the thing we want to use to generate our dynamic list of choices
- then in our ModelAdmin make sure we explicitly tell the admin to use our ModelForm
class Test1(models.Model):
test_folder_ddl = models.CharField(max_length=100)
class Test1Form(ModelForm):
test_folder_ddl = forms.choiceField()
def __init__(self, *args, **kwargs):
super(Test1Form, self).__init__(*args, **kwargs)
self.fields['test_folder_ddl'].choices = utility.get_folder_list()
class Test1Admin(admin.ModelAdmin):
form = Test1Form
I use a generator:
see git://gist.github.com/1118279.git
import pysvn
class SVNChoices(DynamicChoice):
"""
Generate a choice from somes files in a svn repo
""""
SVNPATH = 'http://xxxxx.com/svn/project/trunk/choices/'
def generate(self):
def get_login( realm, username, may_save ):
return True, 'XXX', 'xxxxx', True
client = pysvn.Client()
client.callback_get_login = get_login
return [os.path.basename(sql[0].repos_path) for sql in client.list(self.SVNPATH)[1:]]