I'm new in django. I need a google map in my model.
1. I tried to use django-google-maps.my models.py is:
from django.db import models
from django_google_maps import fields as map_fields
class Rental(models.Model):
address = map_fields.AddressField(max_length=200)
geolocation = map_fields.GeoLocationField(max_length=200)
and my admin.py:
from django.contrib import admin
from django_google_maps import widgets as map_widgets
from django_google_maps import fields as map_fields
from .models import Rental
class RentalAdmin(admin.ModelAdmin):
formfield_overrides = {
map_fields.AddressField: {'widget': map_widgets.GoogleMapsAddressWidget},
}
admin.site.register(Rental, RentalAdmin)
but when i click on the map in admin interface that fields are still empty .
I used this guide to install it https://github.com/madisona/django-google-maps
2.I tried to use easy-map by this guide http://pypi.python.org/pypi/django-easy-maps but I can't see address preview in admin interface.
my admin.py is:
from django import forms
from django.contrib import admin
from easy_maps.widgets import AddressWithMapWidget
from .models import Address
class AddressAdmin(admin.ModelAdmin):
class form(forms.ModelForm):
class Meta:
widgets = {
'address': AddressWithMapWidget({'class': 'vTextField'})
}
admin.site.register(Address, AddressAdmin)
You might be missing Google API Key in settings.py
GOOGLE_MAPS_API_KEY = 'your key here'
generate your Google Maps Javascript API from https://code.google.com/apis/console.
Related
I am trying to intregrate django-summernote in my blog post admin area.
when I set summernote_fields = ('description',) and register to admin area, I am getting error like below :
for model in model_or_iterable:
TypeError: 'MediaDefiningClass' object is not iterable
My admin.py is given below :
from django.contrib import admin
from django_summernote.admin import SummernoteModelAdmin
from .models import Blog
class BlogPostSummerNote(SummernoteModelAdmin):
summernote_fields = ('description',)
#admin.register(Blog)
class BlogAdmin(admin.ModelAdmin):
list_display = ('title','slug','author','publish','status')
prepopulated_fields = {'slug':('title',)}
admin.site.register(BlogPostSummerNote)
I can't able to figure this out . Can any one tell me why this is happening ??
############# SOLUTION #################
I Solved this problem by doing as below in admin.py:
from django.contrib import admin
from django_summernote.admin import SummernoteModelAdmin
from .models import Blog
#admin.register(Blog)
class BlogPostSummerNote(SummernoteModelAdmin):
list_display = ('title','slug','author','publish','status')
summernote_fields = ('description',)
prepopulated_fields = {'slug':('title',)}
I have this code:
from django.contrib import admin
from test_app.models import Master, Detail
class DetailInline(admin.TabularInline):
model = Detail
class MasterAdmin(admin.ModelAdmin):
inlines = [DetailInline]
admin.register(Master, MasterAdmin)
But somehow it does not get displayed in the django admin index page.
Other apps are listed, and their admin.py file looks similar.
What's wrong with this code?
Use admin.site.register(Master, MasterAdmin)
Method-1
from django.contrib import admin
from test_app.models import Master, Detail
class DetailInline(admin.TabularInline):
model = Detail
class MasterAdmin(admin.ModelAdmin):
inlines = [DetailInline]
admin.site.register(Master, MasterAdmin)
Method-2 : using #register decorator
from django.contrib import admin
from test_app.models import Master, Detail
class DetailInline(admin.TabularInline):
model = Detail
#admin.register(Master)
class MasterAdmin(admin.ModelAdmin):
inlines = [DetailInline]
Ref : ModelAdmin objects
I'm trying to achive the Django TinyMCE widget with Django Flatpages (into Admin).
I already read this Embedding tinyMCE in django flatpage
I did the sabe but is not working. Here is my admin.py code:
from django.contrib.flatpages.admin import FlatpageForm, FlatPageAdmin
from django.contrib.flatpages.models import FlatPage
from tinymce.widgets import TinyMCE
from django.contrib import admin
class PageForm(FlatpageForm):
class Meta:
model = FlatPage
widgets = {
'content': TinyMCE(attrs={'cols': 100, 'rows': 15}),
}
class PageAdmin(FlatPageAdmin):
"""
Page Admin
"""
form = PageForm
admin.site.unregister(FlatPage)
admin.site.register(FlatPage, PageAdmin)
My site is running well, without erros, is possible to edit or create a new Django Flatpage but the widget doesn't appear. Any ideas?
The accepted answer wasn't working for me (Django 1.10) so this is what worked for me.
I first created a new app called content in my project to keep organized.
manage.py startapp content
I added this to my admin.py
from django.contrib import admin
from django.contrib.flatpages.admin import FlatPageAdmin
from django.contrib.flatpages.models import FlatPage
from django.db import models
from tinymce.widgets import TinyMCE
# Register your models here.
class PageAdmin(FlatPageAdmin):
"""
Page Admin
"""
formfield_overrides = {
models.TextField: {'widget': TinyMCE(attrs={'cols': 100, 'rows': 15})},
}
admin.site.unregister(FlatPage)
admin.site.register(FlatPage, PageAdmin)
Finally run manage.py collectstatic, restart your webserver and refresh.
I did something different, and I was missing the "fields" property in meta Class, following the new code, tested and running:
from django.contrib.flatpages.admin import FlatpageForm, FlatPageAdmin
from django.contrib import admin
from django import forms
from django.contrib.flatpages.models import FlatPage
from tinymce.widgets import TinyMCE
class FlatPageForm(forms.ModelForm):
content = forms.CharField(widget=TinyMCE(attrs={'cols': 80, 'rows': 30}))
class Meta:
model = FlatPage
fields = '__all__'
class PageAdmin(FlatPageAdmin):
"""
Page Admin
"""
form = FlatPageForm
admin.site.unregister(FlatPage)
admin.site.register(FlatPage, PageAdmin)
If I want to register my models with the admin I have to do this like this:
#admin.py
admin.site.register(models.About)
But with multiple models you can't do something like this:
models = (models.Project, models.Client, models.About)
for m in models:
admin.site.register(m)
First of all: why not!? Secondly: imagine one has a lot of models which all should be accessible from the admin interface. How do you do that in a generic way?
admin.site.register has this definition in the library:
def register(self, model_or_iterable, admin_class=None, **options):
so models to be registered can be a single model or iterable object so just use this:
myModels = [models.Project, models.Client, models.About] # iterable list
admin.site.register(myModels)
I tested this in my site and works perfectly fine.
# File: admin.py
from django.contrib import admin
from .models import Project, Client, About
admin.register(Project, Client, About)(admin.ModelAdmin)
With respect to the recent release of Django 1.7, you can use the django.contrib.admin.register decorator to register multiple models that using the same admin class.
from django.contrib import admin
from .models import Project, Client, About
#admin.register(Project, Client, About)
class DefaultAdmin(admin.ModelAdmin):
pass
Update
Consider making a simple call instead of declaring a dummy class
Based on the snippet here, what I usually do is have the following code in my admin.py
from django.db.models import get_models, get_app
from django.contrib import admin
from django.contrib.admin.sites import AlreadyRegistered
def autoregister(*app_list):
for app_name in app_list:
app_models = get_app(app_name)
for model in get_models(app_models):
try:
admin.site.register(model)
except AlreadyRegistered:
pass
autoregister('myapp')
With Django 3.1.3
in admin.py
from .models import Course, Category
admin.site.register(Course)
admin.site.register(Category)
This is better way to regestering multiple models
from django.contrib import admin
from myapp.models import Course, Category
admin.site.register(Course)
admin.site.register(Category)
from django.contrib import admin
from .models import *
#admin.register(Project, Client, About)
class AppAdmin(admin.ModelAdmin):
pass
Here is a wrokaround that attempts to register a custom models to the admin UI if existing else the default one, for that to work, you must follow the convention of naming your custom admin classes as "MyModelAdmin" by appending "Admin" to the end.
for model_name, model in apps.get_app_config('myapp').models.items() :
if '_' not in model_name :
if globals().get(model.__name__+'Admin') :
admin.site.register(model, globals().get(model.__name__+'Admin'))
else :
admin.site.register(model)
-> if you have just one model:
from django.contrib import admin
from .models import MyModel1
class myModel1Admin(admin.ModelAdmin):
list_display = ("name", "address")
admin.site.register(MyModel1, myModel1Admin)
-> but if you have more than one model or ModelAdmin, you can register them separately:
e.g:
from django.contrib import admin
from .models import MyModel1, MyModel2
class myModel1Admin(admin.ModelAdmin):
list_display = ("name", "address")
class myModel2Admin(admin.ModelAdmin):
list_display = ("name", "photo")
admin.site.register(MyModel1, myModel1Admin)
admin.site.register(MyModel2, myModel2Admin)
I have an app named doors and my models.py for the app has 10 tables/class. Under my admin.py, how do I register every model in the file models.py?
For example, currently I have to hardcode it:
from django.contrib import admin
from doors.models import *
admin.site.register(Group)
admin.site.register(Item)
admin.site.register(ItemType)
admin.site.register(Location)
admin.site.register(Log)
admin.site.register(Order)
admin.site.register(Property)
admin.site.register(User)
admin.site.register(Vendor)
Is there a way I perhaps find every class in models.py and loop through and register each class? Or is there some kind of wildcard I can use with Django?
Seems get_models and get_app are no longer available in django 1.8.
The following can be used:
from django.contrib import admin
from django.apps import apps
app = apps.get_app_config('dashboard')
for model_name, model in app.models.items():
admin.site.register(model)
EXTENSION: If you would like to show all or select fields of the model as a grid instead of a single column unicode representation of the model objects you may use this:
app = apps.get_app_config('your_app_name')
for model_name, model in app.models.items():
model_admin = type(model_name + "Admin", (admin.ModelAdmin,), {})
model_admin.list_display = model.admin_list_display if hasattr(model, 'admin_list_display') else tuple([field.name for field in model._meta.fields])
model_admin.list_filter = model.admin_list_filter if hasattr(model, 'admin_list_filter') else model_admin.list_display
model_admin.list_display_links = model.admin_list_display_links if hasattr(model, 'admin_list_display_links') else ()
model_admin.list_editable = model.admin_list_editable if hasattr(model, 'admin_list_editable') else ()
model_admin.search_fields = model.admin_search_fields if hasattr(model, 'admin_search_fields') else ()
admin.site.register(model, model_admin)
What this does is, it extends ModelAdmin class on the fly and sets the list_display field which is required for showing model data in grid representation in the admin. If you list your desired fields in your model as admin_list_display it takes that one, or generates a tuple of all fields available in the model, otherwise.
Other optional fields can similarly be set, such as list_filter.
See django documentation for more info on list_display.
I figured it out with #arie's link (for django < 1.8):
from django.contrib import admin
from django.db.models import get_models, get_app
for model in get_models(get_app('doors')):
admin.site.register(model)
But I wonder if I can do this without get_app... Couldn't the code be smart enough to know the name of its own app?
From Django 1.7 on, you can use this code in your admin.py:
from django.apps import apps
from django.contrib import admin
from django.contrib.admin.sites import AlreadyRegistered
app_models = apps.get_app_config('my_app').get_models()
for model in app_models:
try:
admin.site.register(model)
except AlreadyRegistered:
pass
From Django 1.8, to fix the error message
RemovedInDjango19Warning: django.db.models.get_app is deprecated.
We can use this approach in 2 lines
from django.contrib import admin
from my_app.models import *
from django.apps import apps
for model in apps.get_app_config('my_app').models.values():
admin.site.register(model)
from django.apps import apps
from django.contrib.admin.sites import AlreadyRegistered
app_models = apps.get_app_config('app-name').get_models()
for model in app_models:
try:
admin.site.register(model)
except AlreadyRegistered:
pass
from django.contrib import admin
from .models import Projects, ProjectsUsers, Comments, ProjectsDescription
Models = (Projects, ProjectsUsers, Comments, ProjectsDescription)
admin.site.register(Models)
From Django3.0,you can try add the following code in admin.py
from . import models
class ListAdminMixin(object):
def __init__(self, model, admin_site):
self.list_display = [field.name for field in model._meta.fields if field.name != "id"]
super(ListAdminMixin, self).__init__(model, admin_site)
for m in [your_model_name]:
mod = getattr(models, m)
admin_class = type('AdminClass', (ListAdminMixin, admin.ModelAdmin), {})
try:
admin.site.register(mod, admin_class)
except admin.sites.AlreadyRegistered:
pass