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  • Django Admin Actions missing

    - by Andrew C
    One of my Django sites is missing the Django Admin Action bar shown here: http://docs.djangoproject.com/en/dev/ref/contrib/admin/actions/#ref-contrib-admin-actions There is no checkbox next to each row and no Action select box near the top of the page. This is happening on every model. I have several sites running Django 1.1, and they all show the Admin Actions, so it feels like a local configuration issue. Anyone seen this before?

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  • /users/tags should contain scores

    - by Sean Patrick Floyd
    I am implementing some simple JavaScript/bookmarklet based apps that show some reputation info, including the score in the User's top tags (roughly based on this previous bookmarklet of mine). Now I can get a user's top tags (using the API), and I can also get the per-tag score if the user is logged in, by dynamically parsing the tag's top users page. But it costs me one AJAX request per tag and I have to download 10+k to extract a single numeric value. It would save a lot of traffic if the tags in <api>/users/<userid>/tags had a score field. The data seems to be there, after all the top users pages use it, so it would just be a question of exposing the data. Suggested structure: "tags": [ { "name": { "description": "name of the tag", "values": "string", "optional": false, "suggested_buffer_size": 25 }, "score": { "description": "tag score, sum of up votes for answers on non-wiki questions", "values": "32-bit signed integer", "optional": false }, "count": { "description": "tag count, exact meaning depends on context", "values": "32-bit signed integer", "optional": false }, "restricted_to": { "description": "user types that can make use of this tag, lack of this field indicates it is useable by all", "values": "one of anonymous, unregistered, registered, or moderator", "optional": true }, "fulfills_required": { "description": "indicates whether this tag is one of those that is required to be on a post", "values": "boolean", "optional": false }, "user_id": { "description": "user associated with this tag, depends on context", "values": "32-bit signed integer", "optional": true } } ]

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  • C++ template-function -> passing a template-class as template-argument

    - by SeMa
    Hello, i try to make intensive use of templates to wrap a factory class: The wrapping class (i.e. classA) gets the wrapped class (i.e. classB) via an template-argument to provide 'pluggability'. Additionally i have to provide an inner-class (innerA) that inherits from the wrapped inner-class (innerB). The problem is the following error-message of the g++ "gcc version 4.4.3 (Ubuntu 4.4.3-4ubuntu5)": sebastian@tecuhtli:~/Development/cppExercises/functionTemplate$ g++ -o test test.cpp test.cpp: In static member function ‘static classA<A>::innerA<iB>* classA<A>::createInnerAs(iB&) [with iB = int, A = classB]’: test.cpp:39: instantiated from here test.cpp:32: error: dependent-name ‘classA::innerA<>’ is parsed as a non-type, but instantiation yields a type test.cpp:32: note: say ‘typename classA::innerA<>’ if a type is meant As you can see in the definition of method createInnerBs, i intend to pass a non-type argument. So the use of typename is wrong! The code of test.cpp is below: class classB{ public: template < class iB> class innerB{ iB& ib; innerB(iB& b) :ib(b){} }; template<template <class> class classShell, class iB> static classShell<iB>* createInnerBs(iB& b){ // this function creates instances of innerB and its subclasses, // because B holds a certain allocator return new classShell<iB>(b); } }; template<class A> class classA{ // intention of this class is meant to be a pluggable interface // using templates for compile-time checking public: template <class iB> class innerA: A::template innerB<iB>{ innerA(iB& b) :A::template innerB<iB>(b){} }; template<class iB> static inline innerA<iB>* createInnerAs(iB& b){ return A::createInnerBs<classA<A>::template innerA<> >(b); // line 32: error occurs here } }; typedef classA<classB> usable; int main (int argc, char* argv[]){ int a = 5; usable::innerA<int>* myVar = usable::createInnerAs(a); return 0; } Please help me, i have been faced to this problem for several days. Is it just impossible, what i'm trying to do? Or did i forgot something? Thanks, Sebastian

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  • Insert django form into template dynamically using javascript??

    - by qulzam
    I want to add same django form instance on same template. i already add one before and other add dynamically using javascript. for example 'form" is a django form: newcell.innerHTML = {{ form.firstname }}; The problem is that when i submit the form, in view the request object has only one value (that is not add using javascript). how can i get the values of other form elements values that is added dynamically runtime.

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  • Django authentication in django nonrel on GAE

    - by tooba
    I'm using the Django nonrel project on a google app engine project running locally in development. I've created my own models and these are fine when they are saved and retrieved in the datastore. I'm hoping to use django.contrib.auth to provide the user functionality. I can use the shell to create users and these get assigned an ID. When I create one of my own models which references User I have to pass in a user ID as it quite rightly fails otherwise. However, checking via the gae admin interface I can't see the User model in the datastore for the users I've created via the shell. Nor can I retreive the user details from one of my models which references them. Calls to mymodel.user.username return nothing. Nor can I log into admin using the username and password I've set up. I can see saved versions of the models I've made in the gae admin app. I get the impression that users are being created somewhere other than the datastore. Is there something else I need to do to use the standard contrib.auth users with django-nonrel and gae?

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  • Website Title Tags For SEO - Creating Excellent Title Tags

    What are title tags? One of the most basic, but important things you can do for your website (in terms of search engine optimisation - SEO), is to ensure that each and every page has its own unique, keyword rich title tag. The title tag can be found at the top of your source code within the and tags (on a web page, click on View Source Code to view the page code).

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  • Django: How do I add arbitrary html attributes to input fields on a form?

    - by User
    I have an input field that is rendered with a template like so: <div class="field"> {{ form.city }} </div> Which is rendered as: <div class="field"> <input id="id_city" type="text" name="city" maxlength="100" /> </div> Now suppose I want to add an autocomplete="off" attribute to the input element that is rendered, how would I do that? Or onclick="xyz()" or class="my-special-css-class"?

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  • Django - provide additional information in template

    - by Ninefingers
    Hi all, I am building an app to learn Django and have started with a Contact system that currently stores Contacts and Addresses. C's are a many to many relationship with A's, but rather than use Django's models.ManyToManyField() I've created my own link-table providing additional information about the link, such as what the address type is to the that contact (home, work etc). What I'm trying to do is pass this information out to a view, so in my full view of a contact I can do this: def contact_view_full(request, contact_id): c = get_object_or_404(Contact, id=contact_id) a = [] links = ContactAddressLink.objects.filter(ContactID=c.id) for link in links: b = Address.objects.get(id=link.AddressID_id) a.append(b) return render_to_response('contact_full.html', {'contact_item': c, 'addresses' : a }, context_instance=RequestContext(request)) And so I can do the equivalent of c.Addresses.all() or however the ManyToManyField works. What I'm interested to know is how can I pass out information about the link in the link object with the 'addresses' : a information, so that when my template does this: {% for address in addresses %} <!-- ... --> {% endfor %} and properly associate the correct link object data with the address. So what's the best way to achieve this? I'm thinking a union of two objects might be an idea but I haven't enough experience with Django to know if that's considered the best way of doing it. Suggestions? Thanks in advance. Nf

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  • PyDev and Django: PyDev breaking Django shell?

    - by Rosarch
    I've set up a new project, and populated it with simple models. (Essentially I'm following the tut.) When I run python manage.py shell on the command line, it works fine: >python manage.py shell Python 2.6.4 (r264:75708, Oct 26 2009, 08:23:19) [MSC v.1500 32 bit (Intel)] on win32 Type "help", "copyright", "credits" or "license" for more information. (InteractiveConsole) >>> from mysite.myapp.models import School >>> School.objects.all() [] Works great. Then, I try to do the same thing in Eclipse (using a Django project that is composed of the same files.) Right click on mysite project Django Shell with Django environment This is the output from the PyDev Console: >>> import sys; print('%s %s' % (sys.executable or sys.platform, sys.version)) C:\Python26\python.exe 2.6.4 (r264:75708, Oct 26 2009, 08:23:19) [MSC v.1500 32 bit (Intel)] >>> >>> from django.core import management;import mysite.settings as settings;management.setup_environ(settings) 'path\\to\\mysite' >>> from mysite.myapp.models import School >>> School.objects.all() Traceback (most recent call last): File "<console>", line 1, in <module> File "C:\Python26\lib\site-packages\django\db\models\query.py", line 68, in __repr__ data = list(self[:REPR_OUTPUT_SIZE + 1]) File "C:\Python26\lib\site-packages\django\db\models\query.py", line 83, in __len__ self._result_cache.extend(list(self._iter)) File "C:\Python26\lib\site-packages\django\db\models\query.py", line 238, in iterator for row in self.query.results_iter(): File "C:\Python26\lib\site-packages\django\db\models\sql\query.py", line 287, in results_iter for rows in self.execute_sql(MULTI): File "C:\Python26\lib\site-packages\django\db\models\sql\query.py", line 2368, in execute_sql cursor = self.connection.cursor() File "C:\Python26\lib\site-packages\django\db\backends\__init__.py", line 81, in cursor cursor = self._cursor() File "C:\Python26\lib\site-packages\django\db\backends\sqlite3\base.py", line 170, in _cursor self.connection = Database.connect(**kwargs) OperationalError: unable to open database file What am I doing wrong here?

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  • Custom template for Django's comments application does not display fields

    - by Jannis
    Hi, I want to use django.contrib.comments in a blogging application and customize the way the form is displayed. My problem is that I can't get the fields to display although displaying the hidden fields works just fine. I had a look at the docs and compared it with the regular way of displaying forms but honestly I don't know why the following doesn't work out: {% get_comment_form for comments_object as form %} <form action="{% comment_form_target %}" method="POST"> […] {% for hidden in form.hidden_fields %} {{ hidden }} {% endfor %} {% for field in form.fields %} {{field}} {% endfor %} […] </form> The output looks like this: <form action="/comments/post/" method="POST"> <input type="hidden" name="content_type" value="flatpages.flatpage" id="id_content_type" /> <input type="hidden" name="object_pk" value="1" id="id_object_pk" /> <input type="hidden" name="timestamp" value="1269522506" id="id_timestamp" /> <input type="hidden" name="security_hash" value="ec4…0fd" id="id_security_hash" /> content_type object_pk timestamp security_hash name email url comment honeypot […] </form> </div> Can you tell me what I'm doing wrong? Thanks in advance

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  • Django: UserProfile with Unique Foreign Key in Django Admin

    - by lazerscience
    Hi, I have extended Django's User Model using a custom user profile called UserExtension. It is related to User through a unique ForeignKey Relationship, which enables me to edit it in the admin in an inline form! I'm using a signal to create a new profile for every new user: def create_user_profile(sender, instance, created, **kwargs): if created: try: profile, created = UserExtension.objects.get_or_create(user=instance) except: pass post_save.connect(create_user_profile, sender=User) (as described here for example: http://stackoverflow.com/questions/44109/extending-the-user-model-with-custom-fields-in-django) The problem is, that, if I create a new user through the admin, I get an IntegritiyError on saving "column user_id is not unique". It doesnt seem that the signal is called twice, but i guess the admin is trying to save the profile AFTERWARDS? But I need the creation through signal if I create a new user in other parts of the system!

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  • Django templates onchange data

    - by Hulk
    In the following code, i have a drop down box and a multi select box. My question is that using javascript and django .how will i changes the designation with changes in names from drop down box. <tr><td> name:</td><td><select id="name" name="name">{% for name in names %} <option value="{{name.id}}" {% for selected_id in names %}{% ifequal name.id selected_id %} {{ selected }} {% endifequal %} {% endfor %}>{{name.name}}</option>{% endfor %} </select> </td></tr> {% for desg in designation %} <tr><td><p>Topics:</td><td> <select id="desg" name="desg" multiple="multiple"> <option value="{{desg.id}}" >{{desg.desg}}</option> </select></p></td></tr> {% endfor %} Thanks..

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  • How to open python scripts directly by typing in their name in terminal (Mac OS X)

    - by Haffi112
    I'm working on installing django and running it on my system. I have a problem though, in this tutorial creating a project is explained by running the command django-admin.py startproject mysite My issue is that this doesn't work. I changed to the directory where django-admin.py is located and ran the command chmod +x django-admin.py with no results. I tried adding the directory with the file to my path without results. I ended up fixing my problem with this command python /location/of/django-admin.py startproject mysite which yielded the outcome I expected. My problem is: What do I need to change/configure such that command django-admin.py startproject mysite would be sufficient? Here are some experiments: 21:09~/Desktop/HI/NSN/Polls > django-admin.py startproject mysite -bash: django-admin.py: command not found 21:09~/Desktop/HI/NSN/Polls > ./django-admin.py startproject mysite -bash: ./django-admin.py: No such file or directory 21:09~/Desktop/HI/NSN/Polls > python django-admin.py startproject mysite python: can't open file 'django-admin.py': [Errno 2] No such file or directory 21:09~/Desktop/HI/NSN/Polls > /opt/local/lib/python2.4/site-packages/django/bin/django-admin.py startproject prufa1 -bash: /opt/local/lib/python2.4/site-packages/django/bin/django-admin.py: /opt/local/bin: bad interpreter: Permission denied 21:09~/Desktop/HI/NSN/Polls > sudo /opt/local/lib/python2.4/site-packages/django/bin/django-admin.py startproject prufa1Password: sudo: unable to execute /opt/local/lib/python2.4/site-packages/django/bin/django-admin.py: Permission denied 21:09~/Desktop/HI/NSN/Polls > sudo /opt/local/lib/python2.4/site-packages/django/bin/django-admin.py startproject prufa1sudo: unable to execute /opt/local/lib/python2.4/site-packages/django/bin/django-admin.py: Permission denied 21:09~/Desktop/HI/NSN/Polls > python /opt/local/lib/python2.4/site-packages/django/bin/django-admin.py startproject prufa1 21:09~/Desktop/HI/NSN/Polls > ls mysite prufa1 Final edit: The problem is solved, see Ian C's answer for the right solution. Thank you everyone for helping my out, this was very fast!

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  • Django: Validation error in Admin

    - by tomwolber
    NEWBIE ALERT! background: For the first time, I am writing a model that needs to be validated. I cannot have two Items that have overlapping "date ranges". I have everything working, except when I raise forms.ValidationError, I get the yellow screen of death (debug=true) or a 500 page (debug=false). My question: How can I have an error message show up in the Admin (like when you leave a required filed blank)? Sorry for my inexperience, please let me know if I can clarify the question better. Models.py from django.db import models from django import forms from django.forms import ModelForm from django.db.models import Q class Item(models.Model): name = models.CharField(max_length=500) slug = models.SlugField(unique=True) startDate = models.DateField("Start Date", unique="true") endDate = models.DateField("End Date") def save(self, *args, **kwargs): try: Item.objects.get(Q(startDate__range=(self.startDate,self.endDate))|Q(endDate__range=(self.startDate,self.endDate))|Q(startDate__lt=self.startDate,endDate__gt=self.endDate)) #check for validation, which may raise an Item.DoesNotExist error, excepted below #if the validation fails, raise this error: raise forms.ValidationError('Someone has already got that date, or somesuch error message') except Item.DoesNotExist: super(Item,self).save(*args,**kwargs) def __unicode__(self): return self.name def get_absolute_url(self): return "/adtest/%s/" % self.slug

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  • Django - Override admin site's login form

    - by TrojanCentaur
    I'm currently trying to override the default form used in Django 1.4 when logging in to the admin site (my site uses an additional 'token' field required for users who opt in to Two Factor Authentication, and is mandatory for site staff). Django's default form does not support what I need. Currently, I've got a file in my templates/ directory called templates/admin/login.html, which seems to be correctly overriding the template used with the one I use throughout the rest of my site. The contents of the file are simply as below: # admin/login.html: {% extends "login.html" %} The actual login form is as below: # login.html: {% load url from future %}<!DOCTYPE html> <html> <head> <title>Please log in</title> </head> <body> <div id="loginform"> <form method="post" action="{% url 'id.views.auth' %}"> {% csrf_token %} <input type="hidden" name="next" value="{{ next }}" /> {{ form.username.label_tag }}<br/> {{ form.username }}<br/> {{ form.password.label_tag }}<br/> {{ form.password }}<br/> {{ form.token.label_tag }}<br/> {{ form.token }}<br/> <input type="submit" value="Log In" /> </form> </div> </body> </html> My issue is that the form provided works perfectly fine when accessed using my normal login URLs because I supply my own AuthenticationForm as the form to display, but through the Django Admin login route, Django likes to supply it's own form to this template and thus only the username and password fields render. Is there any way I can make this work, or is this something I am just better off 'hard coding' the HTML fields into the form for?

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  • Django tests failing on invalid keyword argument

    - by Darwin Tech
    I have a models.py like so: from django.db import models from django.contrib.auth.models import User from datetime import datetime class UserProfile(models.Model): user = models.OneToOneField(User) def __unicode__(self): return self.user.username class Project(models.Model): user = models.ForeignKey(UserProfile) created = models.DateTimeField(auto_now_add=True) updated = models.DateTimeField(auto_now=True) product = models.ForeignKey('tool.product') module = models.ForeignKey('tool.module') model = models.ForeignKey('tool.model') zipcode = models.IntegerField(max_length=5) def __unicode__(self): return unicode(self.id) And my tests.py: from django.test import TestCase, Client # --- import app models from django.contrib.auth.models import User from tool.models import Module, Model, Product from user_profile.models import Project, UserProfile # --- unit tests --- # class UserProjectTests(TestCase): fixtures = ['admin_user.json'] def setUp(self): self.product1 = Product.objects.create( name='bar', ) self.module1 = Module.objects.create( name='foo', enable=True ) self.model1 = Model.objects.create( module=self.module1, name='baz', enable=True ) self.user1 = User.objects.get(pk=1) ... def test_can_create_project(self): self.project1 = Model.objects.create( user=self.user1, product=self.product1, module=self.module1, model=self.model1, zipcode=90210 ) self.assertEquals(self.project1.zipcode, 90210) But I get a TypeError: 'product' is an invalid keyword argument for this function error. I'm not sure what is failing but I'm guessing something to do with the FK relationships... Any help would be much appreciated.

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  • Django: Overriding the save() method: how do I call the delete() method of a child class

    - by Patti
    The setup = I have this class, Transcript: class Transcript(models.Model): body = models.TextField('Body') doPagination = models.BooleanField('Paginate') numPages = models.PositiveIntegerField('Number of Pages') and this class, TranscriptPages(models.Model): class TranscriptPages(models.Model): transcript = models.ForeignKey(Transcript) order = models.PositiveIntegerField('Order') content = models.TextField('Page Content', null=True, blank=True) The Admin behavior I’m trying to create is to let a user populate Transcript.body with the entire contents of a long document and, if they set Transcript.doPagination = True and save the Transcript admin, I will automatically split the body into n Transcript pages. In the admin, TranscriptPages is a StackedInline of the Transcript Admin. To do this I’m overridding Transcript’s save method: def save(self): if self.doPagination: #do stuff super(Transcript, self).save() else: super(Transcript, self).save() The problem = When Transcript.doPagination is True, I want to manually delete all of the TranscriptPages that reference this Transcript so I can then create them again from scratch. So, I thought this would work: #do stuff TranscriptPages.objects.filter(transcript__id=self.id).delete() super(Transcript, self).save() but when I try I get this error: Exception Type: ValidationError Exception Value: [u'Select a valid choice. That choice is not one of the available choices.'] ... and this is the last thing in the stack trace before the exception is raised: .../django/forms/models.py in save_existing_objects pk_value = form.fields[pk_name].clean(raw_pk_value) Other attempts to fix: t = self.transcriptpages_set.all().delete() (where self = Transcript from the save() method) looping over t (above) and deleting each item individually making a post_save signal on TranscriptPages that calls the delete method Any ideas? How does the Admin do it? UPDATE: Every once in a while as I'm playing around with the code I can get a different error (below), but then it just goes away and I can't replicate it again... until the next random time. Exception Type: MultiValueDictKeyError Exception Value: "Key 'transcriptpages_set-0-id' not found in " Exception Location: .../django/utils/datastructures.py in getitem, line 203 and the last lines from the trace: .../django/forms/models.py in _construct_form form = super(BaseInlineFormSet, self)._construct_form(i, **kwargs) .../django/utils/datastructures.py in getitem pk = self.data[pk_key]

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  • Configuration problems with django and mod_wsgi

    - by Jimbo
    Hi, I've got problems on getting django to work on apache 2.2 with mod_wsgi. Django is installed and mod_wsgi too. I can even see a 404 page when accessing the path and I can login to django admin. But if I want to install the tagging module I get the following error: Traceback (most recent call last): File "setup.py", line 49, in <module> version_tuple = __import__('tagging').VERSION File "/home/jim/django-tagging/tagging/__init__.py", line 3, in <module> from tagging.managers import ModelTaggedItemManager, TagDescriptor File "/home/jim/django-tagging/tagging/managers.py", line 5, in <module> from django.contrib.contenttypes.models import ContentType File "/usr/lib/python2.5/site-packages/django/contrib/contenttypes/models.py", line 1, in <module> from django.db import models File "/usr/lib/python2.5/site-packages/django/db/__init__.py", line 10, in <module> if not settings.DATABASE_ENGINE: File "/usr/lib/python2.5/site-packages/django/utils/functional.py", line 269, in __getattr__ self._setup() File "/usr/lib/python2.5/site-packages/django/conf/__init__.py", line 40, in _setup self._wrapped = Settings(settings_module) File "/usr/lib/python2.5/site-packages/django/conf/__init__.py", line 75, in __init__ raise ImportError, "Could not import settings '%s' (Is it on sys.path? Does it have syntax errors?): %s" % (self.SETTINGS_MODULE, e) ImportError: Could not import settings 'mysite.settings' (Is it on sys.path? Does it have syntax errors?): No module named mysite.settings My httpd.conf: Alias /media/ /home/jim/django/mysite/media/ <Directory /home/jim/django/mysite/media> Order deny,allow Allow from all </Directory> Alias /admin/media/ "/usr/lib/python2.5/site-packages/django/contrib/admin/media/" <Directory "/usr/lib/python2.5/site-packages/django/contrib/admin/media/"> Order allow,deny Allow from all </Directory> WSGIScriptAlias /dj /home/jim/django/mysite/apache/django.wsgi <Directory /home/jim/django/mysite/apache> Order deny,allow Allow from all </Directory> My django.wsgi: import sys, os sys.path.append('/home/jim/django') sys.path.append('/home/jim/django/mysite') os.chdir('/home/jim/django/mysite') os.environ['DJANGO_SETTINGS_MODULE'] = 'mysite.settings' import django.core.handlers.wsgi application = django.core.handlers.wsgi.WSGIHandler() I try to get this to work since a few days and have read several blogs and answers here on so but nothing worked.

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  • Django Error: NameError name 'current_datetime' is not defined

    - by Diego
    I'm working through the book "The Definitive Guide to Django" and am stuck on a piece of code. This is the code in my settings.py: ROOT_URLCONF = 'mysite.urls' I have the following code in my urls.py from django.conf.urls.defaults import * from mysite.views import hello, my_homepage_view urlpatterns = patterns('', ('^hello/$', hello), ) urlpatterns = patterns('', ('^time/$', current_datetime), ) And the following is the code in my views.py file: from django.http import HttpResponse import datetime def hello(request): return HttpResponse("Hello World") def current_datetime(request): now = datetime.datetime.now() html = "<html><body>It is now %s.</body></html>" % now return HttpResponse(html) Yet, I get the following error when I test the code in the development server. NameError at /time/ name 'current_datetime' is not defined Can someone help me out here? This really is just a copy-paste from the book. I don't see any mistyping.

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  • Project name inserted automatically in url when using django template url tag

    - by thebossman
    I am applying the 'url' template tag to all links in my current Django project. I have my urls named like so... url(r'^login/$', 'login', name='site_login'), This allows me to access /login at my site's root. I have my template tag defined like so... <a href="{% url site_login %}"> It works fine, except that Django automatically resolves that url as /myprojectname/login, not /login. Both urls are accessible. Why? Is there an option to remove the projectname? This occurs for all url tags, not just this one.

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  • Load django template from the database

    - by Björn Lindqvist
    Hello, Im trying to render a django template from a database outside of djangos normal request-response structure. But it appears to be non-trivial due to the way django templates are compiled. I want to do something like this: >>> s = Template.objects.get(pk = 123).content >>> some_method_to_render(s, {'a' : 123, 'b' : 456}) >>> ... the rendered output here ... How do you do this?

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  • Django Admin drop down combobox and assigned values

    - by Daniel Garcia
    I have several question for the Django Admin feature. Im kind of new in Django so im not sure how to do it. Basically what Im looking to do is when Im adding information on the model. Some of the fields i want them to be drop-downs and maybe combo-boxes with AutoCompleteMode. Also looking for some fields to have the same information, for example if i have a datatime field I want that information to feed the fields day, month and year from hoti.hotiapp.models import Occurrence from django.contrib import admin class MyModelAdmin(admin.ModelAdmin): exclude = ['reference',] admin.site.register(Occurrence, MyModelAdmin) Anything helps Thanks in advance

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