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  • Django models query

    - by Hulk
    Code: class criteria(models.Model): details = models.CharField(max_length = 512) Headerid = models.ForeignKey(Header) def __unicode__(self): return self.id() the details corresponds to a textarea in the UI and a validation is done for 512 characters but when this is saved. /home/project/django/django/core/handlers/base.py in get_response, line 109 Is this any thing related with schema or number of characters entered from UI

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  • Tying in to Django Admin's Model History

    - by akdom
    The Setup: I'm working on a Django application which allows users to create an object in the database and then go back and edit it as much as they desire. Django's admin site keeps a history of the changes made to objects through the admin site. The Question: How do I hook my application in to the admin site's change history so that I can see the history of changes users make to their "content" ?

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  • Django: Generating a queryset from a GET request

    - by Nimmy Lebby
    I have a Django form setup using GET method. Each value corresponds to attributes of a Django model. What would be the most elegant way to generate the query? Currently this is what I do in the view: def search_items(request): if 'search_name' in request.GET: query_attributes = {} query_attributes['color'] = request.GET.get('color', '') if not query_attributes['color']: del query_attributes['color'] query_attributes['shape'] = request.GET.get('shape', '') if not query_attributes['shape']: del query_attributes['shape'] items = Items.objects.filter(**query_attributes) But I'm pretty sure there's a better way to go about it.

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  • Django attribute error: 'module' object has no attribute 'is_usable'

    - by Robert A Henru
    Hi, I got the following error when calling the url in Django. It's working before, I guess it's related with some accidental changes I made, but I have no idea what they are. Thanks before for the help, Robert Environment: Request Method: GET Request URL: http://localhost:8000/time/ Django Version: 1.2 Python Version: 2.6.1 Installed Applications: ['django.contrib.auth', 'django.contrib.contenttypes', 'django.contrib.sessions', 'django.contrib.sites', 'django.contrib.messages', 'django.contrib.admin', 'djlearn.books'] Installed Middleware: ('django.middleware.common.CommonMiddleware', 'django.contrib.sessions.middleware.SessionMiddleware', 'django.middleware.csrf.CsrfViewMiddleware', 'django.contrib.auth.middleware.AuthenticationMiddleware', 'django.contrib.messages.middleware.MessageMiddleware') Traceback: File "/Library/Python/2.6/site-packages/django/core/handlers/base.py" in get_response 100. response = callback(request, *callback_args, **callback_kwargs) File "/Users/rhenru/Workspace/django/djlearn/src/djlearn/../djlearn/views.py" in current_datetime 16. return render_to_response('current_datetime.html',{'current_date':now,}) File "/Library/Python/2.6/site-packages/django/shortcuts/__init__.py" in render_to_response 20. return HttpResponse(loader.render_to_string(*args, **kwargs), **httpresponse_kwargs) File "/Library/Python/2.6/site-packages/django/template/loader.py" in render_to_string 181. t = get_template(template_name) File "/Library/Python/2.6/site-packages/django/template/loader.py" in get_template 157. template, origin = find_template(template_name) File "/Library/Python/2.6/site-packages/django/template/loader.py" in find_template 128. loader = find_template_loader(loader_name) File "/Library/Python/2.6/site-packages/django/template/loader.py" in find_template_loader 111. if not func.is_usable: Exception Type: AttributeError at /time/ Exception Value: 'module' object has no attribute 'is_usable'

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  • django crispy-forms inline forms

    - by abolotnov
    I'm trying to adopt crispy-forms and bootstrap and use as much of their functionality as possible instead of inventing something over and over again. Is there a way to have inline forms functionality with crispy-forms/bootstrap like django-admin forms have? Here is an example: class NewProjectForm(forms.Form): name = forms.CharField(required=True, label=_(u'???????? ???????'), widget=forms.TextInput(attrs={'class':'input-block-level'})) group = forms.ModelChoiceField(required=False, queryset=Group.objects.all(), label=_(u'?????? ????????'), widget=forms.Select(attrs={'class':'input-block-level'})) description = forms.CharField(required=False, label=_(u'???????? ???????'), widget=forms.Textarea(attrs={'class':'input-block-level'})) class Meta: model = Project fields = ('name','description','group') def __init__(self, *args, **kwargs): self.helper = FormHelper() self.helper.form_class = 'horizontal-form' self.helper.form_action = 'submit_new_project' self.helper.layout = Layout( Field('name', css_class='input-block-level'), Field('group', css_class='input-block-level'), Field('description',css_class='input-block-level'), ) self.helper.add_input(Submit('submit',_(u'??????? ??????'))) self.helper.add_input(Submit('cancel',_(u'? ?????????'))) super(NewProjectForm, self).__init__(*args, **kwargs) it will display a decent form: How do I go about adding a form that basically represents this model: class Link(models.Model): name = models.CharField(max_length=255, blank=False, null=False, verbose_name=_(u'????????')) url = models.URLField(blank=False, null=False, verbose_name=_(u'??????')) project = models.ForeignKey('Project') So there will be a project and name/url links and way to add many, like same thing is done in django-admin where you are able to add extra 'rows' with data related to your main model. On the sreenshot below you are able to fill out data for 'Question' object and below that you are able to add data for QuestionOption objects -you are able to click the '+' icon to add as many QuestionOptions as you want. I'm not looking for a way to get the forms auto-generated from models (that's nice but not the most important) - is there a way to construct a form that will let you add 'rows' of data like django-admin does?

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  • ProgrammingError when aggregating over an annotated & grouped Django ORM query

    - by ento
    I'm trying to construct a query to get the "average, maximum, minimum number of items purchased by a single user". The data source is this simple sales record table: class SalesRecord(models.Model): id = models.IntegerField(primary_key=True) user_id = models.IntegerField() product_code = models.CharField() price = models.IntegerField() created_at = models.DateTimeField() A new record is inserted into this table for every item purchased by a user. Here's my attempt at building the query: q = SalesRecord.objects.all() q = q.values('user_id').annotate( # group by user and count the # of records count=Count('id'), # (= # of items) ).order_by() result = q.aggregate(Max('count'), Min('count'), Avg('count')) When I try to execute the code, a ProgrammingError is raised at the last line: (1064, "You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'FROM (SELECT sales_records.user_id AS user_id, COUNT(sales_records.`' at line 1") Django's error screen shows that the SQL is SELECT FROM (SELECT `sales_records`.`player_id` AS `player_id`, COUNT(`sales_records`.`id`) AS `count` FROM `sales_records` WHERE (`sales_records`.`created_at` >= %s AND `sales_records`.`created_at` <= %s ) GROUP BY `sales_records`.`player_id` ORDER BY NULL) subquery It's not selecting anything! Can someone please show me the right way to do this? Hacking Django I've found that clearing the cache of selected fields in django.db.models.sql.BaseQuery.get_aggregation() seems to solve the problem. Though I'm not really sure this is a fix or a workaround. @@ -327,10 +327,13 @@ # Remove any aggregates marked for reduction from the subquery # and move them to the outer AggregateQuery. + self._aggregate_select_cache = None + self.aggregate_select_mask = None for alias, aggregate in self.aggregate_select.items(): if aggregate.is_summary: query.aggregate_select[alias] = aggregate - del obj.aggregate_select[alias] + if alias in obj.aggregate_select: + del obj.aggregate_select[alias] ... yields result: {'count__max': 267, 'count__avg': 26.2563, 'count__min': 1}

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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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  • What is causing this OverflowError in Django?

    - by orokusaki
    I'm using a normal ModelForm.save() to create an object, and this exception comes up. It worked fine before until I added commit_manually, transaction.rollback() and transaction.commit() to my view. Has anyone else ran into this? Is this because of sqlite3? OverflowError: long too big to convert C:\Python26\Lib\site-packages\django-trunk\django\db\backends\sqlite3\base.py in execute, line 197 params: (203866156270872165269663274649746494334L,) query: u'SELECT (1) AS "a", "auth_user"."id", "auth_user"."username", "auth_user"."first_name", "auth_user"."last_name", "auth_user"."email", "auth_user"."password", "auth_user"."is_staff", "auth_user"."is_active", "auth_user"."is_superuser", "auth_user"."last_login", "auth_user"."date_joined" FROM "auth_user" WHERE "auth_user"."id" = ? LIMIT 1' self <django.db.backends.sqlite3.base.SQLiteCursorWrapper object at 0x015D5A98> Why would that L param be passed in, and

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  • Django model operating on a queryset

    - by jmoz
    I'm new to Django and somewhat to Python as well. I'm trying to find the idiomatic way to loop over a queryset and set a variable on each model. Basically my model depends on a value from an api, and a model method must multiply one of it's attribs by this api value to get an up-to-date correct value. At the moment I am doing it in the view and it works, but I'm not sure it's the correct way to achieve what I want. I have to replicate this looping elsewhere. Is there a way I can encapsulate the looping logic into a queryset method so it can be used in multiple places? I have this atm (I am using django-rest-framework): class FooViewSet(viewsets.ModelViewSet): model = Foo serializer_class = FooSerializer bar = # some call to an api def get_queryset(self): # Dynamically set the bar variable on each instance! foos = Foo.objects.filter(baz__pk=1).order_by('date') for item in foos: item.needs_bar = self.bar return items I would think something like so would be better: def get_queryset(self): bar = # some call to an api # Dynamically set the bar variable on each instance! return Foo.objects.filter(baz__pk=1).order_by('date').set_bar(bar) I'm thinking the api hit should be in the controller and then injected to instances of the model, but I'm not sure how you do this. I've been looking around querysets and managers but still can't figure it out nor decided if it's the best method to achieve what I want. Can anyone suggest the correct way to model this with django? Thanks.

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  • Asynchronous daemon processing / ORM interaction with Django

    - by perrierism
    I'm looking for a way to do asynchronous data processing with a daemon that uses Django ORM. However, the ORM isn't thread-safe; it's not thread-safe to try to retrieve / modify django objects from within threads. So I'm wondering what the correct way to achieve asynchrony is? Basically what I need to accomplish is taking a list of users in the db, querying a third party api and then making updates to user-profile rows for those users. As a daemon or background process. Doing this in series per user is easy, but it takes too long to be at all scalable. If the daemon is retrieving and updating the users through the ORM, how do I achieve processing 10-20 users at a time? I would use a standard threading / queue system for this but you can't thread interactions like models.User.objects.get(id=foo) ... Django itself is an asynchronous processing system which makes asynchronous ORM calls(?) for each request, so there should be a way to do it? I haven't found anything in the documentation so far. Cheers

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  • In Django-pagination Paginate does not working...

    - by mosg
    Hello. Python 2.6.2 django-pagination 1.0.5 Question: How to force pagination work correctly? The problem is that {% paginate %} does not work, but other {% load pagination_tags %} and {% autopaginate object_list 10 %} works! Error message appeared, when I add {% paginate %} into html page: TemplateSyntaxError at /logging Caught an exception while rendering: pagination/pagination.html What I have done: Install django-pagination without any problems. When I do in python import pagination, it's work well. Added pagination to INSTALLED_APP in settings.py: INSTALLED_APPS = ( # ..., 'pagination', ) Added in settings.py: TEMPLATE_CONTEXT_PROCESSORS = ( "django.core.context_processors.auth", "django.core.context_processors.debug", "django.core.context_processors.i18n", "django.core.context_processors.media", "django.core.context_processors.request" ) Also add to settings.py middleware: MIDDLEWARE_CLASSES = ( # ... 'pagination.middleware.PaginationMiddleware', ) Add to top in views.py: from django.template import RequestContext And finally add to my HTML template page lines: {% load pagination_tags %} ... {% autopaginate item_list 50 %} {% for item in item_list %} ... {% endfor %} {% paginate %} Thanks. PS: some edits required, because I can't django code style work well here :)

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  • Login URL using authentication information in Django

    - by fuSi0N
    I'm working on a platform for online labs registration for my university. Login View [project views.py] from django.http import HttpResponse, HttpResponseRedirect, Http404 from django.shortcuts import render_to_response from django.template import RequestContext from django.contrib import auth def index(request): return render_to_response('index.html', {}, context_instance = RequestContext(request)) def login(request): if request.method == "POST": post = request.POST.copy() if post.has_key('username') and post.has_key('password'): usr = post['username'] pwd = post['password'] user = auth.authenticate(username=usr, password=pwd) if user is not None and user.is_active: auth.login(request, user) if user.get_profile().is_teacher: return HttpResponseRedirect('/teachers/'+user.username+'/') else: return HttpResponseRedirect('/students/'+user.username+'/') else: return render_to_response('index.html', {'msg': 'You don\'t belong here.'}, context_instance = RequestContext(request) return render_to_response('login.html', {}, context_instance = RequestContext(request)) def logout(request): auth.logout(request) return render_to_response('index.html', {}, context_instance = RequestContext(request)) URLS #========== PROJECT URLS ==========# urlpatterns = patterns('', (r'^media/(?P<path>.*)$', 'django.views.static.serve', {'document_root': settings.MEDIA_ROOT }), (r'^admin/', include(admin.site.urls)), (r'^teachers/', include('diogenis.teachers.urls')), (r'^students/', include('diogenis.students.urls')), (r'^login/', login), (r'^logout/', logout), (r'^$', index), ) #========== TEACHERS APP URLS ==========# urlpatterns = patterns('', (r'^(?P<username>\w{0,50})/', labs), ) The login view basically checks whether the logged in user is_teacher [UserProfile attribute via get_profile()] and redirects the user to his profile. Labs View [teachers app views.py] from django.http import HttpResponse, HttpResponseRedirect, Http404 from django.shortcuts import render_to_response from django.template import RequestContext from django.contrib.auth.decorators import user_passes_test from django.contrib.auth.models import User from accounts.models import * from labs.models import * def user_is_teacher(user): return user.is_authenticated() and user.get_profile().is_teacher @user_passes_test(user_is_teacher, login_url="/login/") def labs(request, username): q1 = User.objects.get(username=username) q2 = u'%s %s' % (q1.last_name, q1.first_name) q2 = Teacher.objects.get(name=q2) results = TeacherToLab.objects.filter(teacher=q2) return render_to_response('teachers/labs.html', {'results': results}, context_instance = RequestContext(request)) I'm using @user_passes_test decorator for checking whether the authenticated user has the permission to use this view [labs view]. The problem I'm having with the current logic is that once Django authenticates a teacher user he has access to all teachers profiles basically by typing the teachers username in the url. Once a teacher finds a co-worker's username he has direct access to his data. Any suggestions would be much appreciated.

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  • CodeIgniter & Datamapper as frontend, Django Admin as backend, database tables inconsistent

    - by Rasiel
    I created a database for a site i'm doing using Django as the admin backend. However because the server where the site is hosted on, won't be able to support Python, I find myself needing to do the front end in PHP and as such i've decided to use CodeIgniter along with Datamapper to map the models/relationship. However DataMapper requires the tables to be in a specific format for it to work, and Django maps its tables differently, using the App name as the prefix in the table. I've tried using the prefix & join_prefix vars in datamapper but still doesn't map them correctly. Has anyone used a combination of this? and if so how have the fixed the issue of db table names being inconsistent? Is there anything out there that i can use to make them work together?

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  • Django: Summing values of records grouped by foreign key

    - by Dan0
    Hi there In django, given the following models (slightly simplified), I'm struggling to work out how I would get a view including sums of groups class Client(models.Model): api_key = models.CharField(unique=True, max_length=250, primary_key=True) name = models.CharField(unique=True, max_length=250) class Purchase(models.Model): purchase_date = models.DateTimeField() client = models.ForeignKey(SavedClient, to_field='api_key') amount_to_invoice = models.FloatField(null=True) For a given month, I'd like to see e.g. April 2010 For Each Client that purchased this month: * CLient: Name * Total amount of Purchases for this month * Total cost of purchases for this month For each Purchase made by client: * Date * Amount * Etc I've been looking into django annotation, but can't get my head around how to sum values of a field for a particular group over a particular month and send the information to a template as a variable/tag. Any info would be appreciated

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  • Django: Country drop down list?

    - by User
    I have a form for address information. One of the fields is for the address country. Currently this is just a textbox. I would like a drop down list (of ISO 3166 countries) for this. I'm a django newbie so I haven't even used a Django Select widget yet. What is a good way to do this? Hard-code the choices in a file somewhere? Put them in the database? In the template?

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  • django cross-site reverse a url

    - by tutuca
    I have a similar question than django cross-site reverse. But i think I can't apply the same solution. I'm creating an app that lets the users create their own site. After completing the signup form the user should be redirected to his site's new post form. Something along this lines: new_post_url = 'http://%s.domain:9292/manage/new_post %site.domain' logged_user = authenticate(username=user.username, password=user.password) if logged_user is not None: login(request, logged_user) return redirect(new_product_url) Now, I know that "new_post_url" is awful and makes babies cry so I need to reverse it in some way. I thought in using django.core.urlresolvers.reverse to solve this but that only returns urls on my domain, and not in the user's newly created site, so it doesn't works for me. So, do you know a better/smarter way to solve this?

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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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  • How to customize a many-to-many inline model in django admin

    - by Jonathan
    I'm using the admin interface to view invoices and products. To make things easy, I've set the products as inline to invoices, so I will see the related products in the invoice's form. As you can see I'm using a many-to-many relationship. In models.py: class Product(models.Model): name = models.TextField() price = models.DecimalField(max_digits=10,decimal_places=2) class Invoice(models.Model): company = models.ForeignKey(Company) customer = models.ForeignKey(Customer) products = models.ManyToManyField(Product) In admin.py: class ProductInline(admin.StackedInline): model = Invoice.products.through class InvoiceAdmin(admin.ModelAdmin): inlines = [FilteredApartmentInline,] admin.site.register(Product, ProductAdmin) The problem is that django presents the products as a table of drop down menus (one per associated product). Each drop down contains all the products listed. So if I have 5000 products and 300 are associated with a certain invoice, django actually loads 300x5000 product names. Also the table is not aesthetic. How can I change it so that it'll just display the product's name in the inline table? Which form should I override, and how?

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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: reverse lookup URL of feeds?

    - by Santa
    I am having trouble doing a reverse URL lookup for Django-generated feeds. I have the following setup in urls.py: feeds = { 'latest': LatestEntries, } urlpatterns = patterns('', # ... # enable feeds (RSS) url(r'^feeds/(?P<url>.*)/$', 'django.contrib.syndication.views.feed', {'feed_dict': feeds}, name='feeds_view'), ) I have tried using the following template tag: <a href="{% url feeds_view latest %}">RSS feeds</a> But the resulting link is not what want (http://my.domain.com/feeds//). It should be http://my.domain.com/feeds/latest/. For now, I am using a hack to generate the URL for the template: <a href="http://{{ request.META.HTTP_HOST }}/feeds/latest">RSS feeds</a> But, as you can see, it clearly is not DRY. Is there something I am missing?

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  • django auth_views.login and redirects

    - by Zayatzz
    Hello I could not understand why after logging in from address: http://localhost/en/accounts/login/?next=/en/test/ I get refirected to http://localhost/accounts/profile/ So i ran search in django files and found that this address is the default LOGIN_REDIRECT_URL for django. What i did not understand is why it gets redirected to there. I guessed, that my login form's post address should be : /accounts/login/?next=/en/test/ instead of /accounts/login/ I wrote it into template and it worked. But since the redirect url changes dynamically, how can i make this login post forms address change dynamically too? is there a templatetag for that or something? Alan

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  • How to use multiple flatpages models in a django app?

    - by the_drow
    I have multiple models that can be converted to flatpages but have to have some extra information (For example I have an about us page but I also have a blog). However I understand that there must be only one flatpages model since the middleware only returns the flatpages instance and does not resolve the child models. What do I have to do? EDIT: It seems I need to change the views. Here's the current code: from django.contrib.flatpages.models import FlatPage from django.template import loader, RequestContext from django.shortcuts import get_object_or_404 from django.http import HttpResponse, HttpResponseRedirect from django.conf import settings from django.core.xheaders import populate_xheaders from django.utils.safestring import mark_safe from django.views.decorators.csrf import csrf_protect DEFAULT_TEMPLATE = 'flatpages/default.html' # This view is called from FlatpageFallbackMiddleware.process_response # when a 404 is raised, which often means CsrfViewMiddleware.process_view # has not been called even if CsrfViewMiddleware is installed. So we need # to use @csrf_protect, in case the template needs {% csrf_token %}. # However, we can't just wrap this view; if no matching flatpage exists, # or a redirect is required for authentication, the 404 needs to be returned # without any CSRF checks. Therefore, we only # CSRF protect the internal implementation. def flatpage(request, url): """ Public interface to the flat page view. Models: `flatpages.flatpages` Templates: Uses the template defined by the ``template_name`` field, or `flatpages/default.html` if template_name is not defined. Context: flatpage `flatpages.flatpages` object """ if not url.endswith('/') and settings.APPEND_SLASH: return HttpResponseRedirect("%s/" % request.path) if not url.startswith('/'): url = "/" + url # Here instead of getting the flat page it needs to find if it has a page with a child model. f = get_object_or_404(FlatPage, url__exact=url, sites__id__exact=settings.SITE_ID) return render_flatpage(request, f) @csrf_protect def render_flatpage(request, f): """ Internal interface to the flat page view. """ # If registration is required for accessing this page, and the user isn't # logged in, redirect to the login page. if f.registration_required and not request.user.is_authenticated(): from django.contrib.auth.views import redirect_to_login return redirect_to_login(request.path) if f.template_name: t = loader.select_template((f.template_name, DEFAULT_TEMPLATE)) else: t = loader.get_template(DEFAULT_TEMPLATE) # To avoid having to always use the "|safe" filter in flatpage templates, # mark the title and content as already safe (since they are raw HTML # content in the first place). f.title = mark_safe(f.title) f.content = mark_safe(f.content) # Here I need to be able to configure what I am passing in the context c = RequestContext(request, { 'flatpage': f, }) response = HttpResponse(t.render(c)) populate_xheaders(request, response, FlatPage, f.id) return response

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  • Create Django formset wihtout multiple queries

    - by Martin
    I need to display multiple forms (up to 10) of a model on a page. This is the code I use for to accomplish this. TheFormSet = formset_factory(SomeForm, extra=10) ... formset = TheFormSet(prefix='party') return render_to_response('template.html', { 'formset' : formset, }) The problem is, that it seems to me that Django queries the database for each of the forms in the formset, even though the data displayed in them is the same. Is this the way Formsets work or am I doing something wrong? Is there a way around it inside django or would I have to use JavaScript for a workaround?

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