Search Results

Search found 11188 results on 448 pages for 'django manage py'.

Page 16/448 | < Previous Page | 12 13 14 15 16 17 18 19 20 21 22 23  | Next Page >

  • Django internationalization for admin pages - translate model name and attributes

    - by geekQ
    Django's internationalization is very nice (gettext based, LocaleMiddleware), but what is the proper way to translate the model name and the attributes for admin pages? I did not find anything about this in the documentation: http://docs.djangoproject.com/en/dev/topics/i18n/internationalization/ http://www.djangobook.com/en/2.0/chapter19/ I would like to have "???????? ????? ??? ?????????" instead of "???????? order ??? ?????????". Note, the 'order' is not translated. First, I defined a model, activated USE_I18N = True in settings.py, run django-admin makemessages -l ru. No entries are created by default for model names and attributes. Grepping in the Django source code I found: $ ack "Select %s to change" contrib/admin/views/main.py 70: self.title = (self.is_popup and ugettext('Select %s') % force_unicode(self.opts.verbose_name) or ugettext('Select %s to change') % force_unicode(self.opts.verbose_name)) So the verbose_name meta property seems to play some role here. Tried to use it: class Order(models.Model): subject = models.CharField(max_length=150) description = models.TextField() class Meta: verbose_name = _('order') Now the updated po file contains msgid 'order' that can be translated. So I put the translation in. Unfortunately running the admin pages show the same mix of "???????? order ??? ?????????". I'm currently using Django 1.1.1. Could somebody point me to the relevant documentation? Because google can not. ;-) In the mean time I'll dig deeper into the django source code...

    Read the article

  • django forms- register user script

    - by itsandy
    Hi all, I want to make something like http://www.djangosnippets.org/accounts/register/ using django..the register form. I am new to django. i have made a simple view form using django forms but unable o understand how to connect my form to a database. Im using postgresql. is there an easy way to use some snippet or script to achieve this. Please Help

    Read the article

  • Django admin interface upload failing on request data read error

    - by Jake
    Hi All, This is an updated version of an old question I asked. I've now done a lot more testing, plus the old question got hijacked. I'm getting a request data read error when trying to upload files to the Django admin interface. Files under about 150k work, but bigger files always fail and almost always at around 192k (that's 3 chunks) completed, sometimes at around 160k. The Exception I get is below. File "/usr/lib/python2.4/site-packages/django/http/multipartparser.py", line 405, in read return self._file.read(num_bytes) IOError: request data read error I've tried Chrome and Firefox on Windows and Firefox on Mac - Same results. I can upload to other sites so I don't think it's my connection. I'm running python 2.4, django 1.1, mod_wsgi, on CentOS (a media temple DV server) Locally it's fine (Django development server) Everything I've found on this issue says it's a mod_python issue and that changing to mod_wsgi will fix it, but I am running mod_wsgi. Can anyone help?

    Read the article

  • Django admin fails when using includes in urlpatterns

    - by zenWeasel
    I am trying to refactor out my application a little bit to keep it from getting too unwieldily. So I started to move some of the urlpatterns out to sub files as the documentation proposes. Besides that fact that it just doesn't seem to be working (the items are not being rerouted) but when I go to the admin, it says that 'urlpatterns has not been defined'. The urls.py I have at the root of my application is: if settings.ENABLE_SSL: urlpatterns = patterns('', (r'^checkout/orderform/onepage/(\w*)/$','checkout.views.one_page_orderform',{'SSL':True},'commerce.checkout.views.single_product_orderform'), ) else: urlpatterns = patterns('', (r'^checkout/orderform/onepage/(\w*)/$','commerce.checkout.views.single_product_orderform'), ) urlpatterns+= patterns('', (r'^$', 'alchemysites.views.route_to_home'), (r'^%s/' % settings.DAJAXICE_MEDIA_PREFIX, include('dajaxice.urls')), (r'^/checkout/', include('commerce.urls')), (r'^/offers',include('commerce.urls')), (r'^/order/',include('commerce.urls')), (r'^admin/', include(admin.site.urls)), (r'^accounts/login/$', login), (r'^accounts/logout/$', logout), (r'^(?P<path>.*)/$','alchemysites.views.get_path'), (r'^static/(?P<path>.*)$', 'django.views.static.serve', {'document_root':settings.MEDIA_ROOT}), The urls I have moved out so far are the checkout/offers/order which are all subapps of 'commerce' where the urls.py for the apps are so to be clear. /urls.py in questions (included here) /commerce/urls.py where the urls.py I want to include is: order_info = { 'queryset': Order.objects.all(), } urlpatterns+= patterns('', (r'^offers/$','offers.views.start_offers'), (r'^offers/([a-zA-Z0-9-]*)/order/(\d*)/add/([a-zA-Z0-9-]*)/(\w*)/next/([a-zA-Z0-9-)/$','offers.views.show_offer'), (r'^reports/orders/$', list_detail.object_list,order_info), ) and the applications offers lies under commerce. And so the additional problem is that admin will not work at all, so I'm thinking because I killed it somewhere with my includes. Things I have checked for: Is the urlpatterns variable accidentally getting reset somewhere (i.e. urlpatterns = patterns, instead of urlpatterns+= patterns) Are the patterns in commerce.urls valid (yes, when moved back to root they work). So from there I am stumped. I can move everything back into the root, but was trying to get a little decoupled, not just for theoretical reason but for some short terms ones. Lastly if I enter www.domainname/checkout/orderform/onepage/xxxjsd I get the correct page. However, entering www.domainname/checkout/ gets handled by the alchemysites.views.get_path. If not the answer (because this is pretty darn specific), then is there a good way for troubleshoot urls.py? It seems to just be trial and error. Seems there should be some sort of parser that will tell you what your urlpatterns will do.

    Read the article

  • Writing good tests for Django applications

    - by Ludwik Trammer
    I've never written any tests in my life, but I'd like to start writing tests for my Django projects. I've read some articles about tests and decided to try to write some tests for an extremely simple Django app or a start. The app has two views (a list view, and a detail view) and a model with four fields: class News(models.Model): title = models.CharField(max_length=250) content = models.TextField() pub_date = models.DateTimeField(default=datetime.datetime.now) slug = models.SlugField(unique=True) I would like to show you my tests.py file and ask: Does it make sense? Am I even testing for the right things? Are there best practices I'm not following, and you could point me to? my tests.py (it contains 11 tests): # -*- coding: utf-8 -*- from django.test import TestCase from django.test.client import Client from django.core.urlresolvers import reverse import datetime from someproject.myapp.models import News class viewTest(TestCase): def setUp(self): self.test_title = u'Test title: bareksc' self.test_content = u'This is a content 156' self.test_slug = u'test-title-bareksc' self.test_pub_date = datetime.datetime.today() self.test_item = News.objects.create( title=self.test_title, content=self.test_content, slug=self.test_slug, pub_date=self.test_pub_date, ) client = Client() self.response_detail = client.get(self.test_item.get_absolute_url()) self.response_index = client.get(reverse('the-list-view')) def test_detail_status_code(self): """ HTTP status code for the detail view """ self.failUnlessEqual(self.response_detail.status_code, 200) def test_list_status_code(self): """ HTTP status code for the list view """ self.failUnlessEqual(self.response_index.status_code, 200) def test_list_numer_of_items(self): self.failUnlessEqual(len(self.response_index.context['object_list']), 1) def test_detail_title(self): self.failUnlessEqual(self.response_detail.context['object'].title, self.test_title) def test_list_title(self): self.failUnlessEqual(self.response_index.context['object_list'][0].title, self.test_title) def test_detail_content(self): self.failUnlessEqual(self.response_detail.context['object'].content, self.test_content) def test_list_content(self): self.failUnlessEqual(self.response_index.context['object_list'][0].content, self.test_content) def test_detail_slug(self): self.failUnlessEqual(self.response_detail.context['object'].slug, self.test_slug) def test_list_slug(self): self.failUnlessEqual(self.response_index.context['object_list'][0].slug, self.test_slug) def test_detail_template(self): self.assertContains(self.response_detail, self.test_title) self.assertContains(self.response_detail, self.test_content) def test_list_template(self): self.assertContains(self.response_index, self.test_title)

    Read the article

  • AddThis Social SignIn and Django

    - by piokuc
    I am developing a Django website. I've been using django-registration for user registration so far but I would really like to allow users to login to my site using their Facebook, Twitter, Google, etc accounts. I am using addthis sharing buttons. I just noticed they introduced a social sign in solution. The idea seems great, you integrate your authentication system with their service once, and your users can login via all of the popular social networking sites. Has anybody integrated addthis social signin plugin with a django website? How can you use it along side django-registration? Are there any similar, alternative solutions?

    Read the article

  • Django admin output extra HTML in ModelSite

    - by VoteyDisciple
    Ultimately, I want to add an <iframe> to the display of a particular model on Django's admin page. Django is already rendering the form for this model correctly, but I want to add this <iframe> in addition to Django's form. The src attribute needs to involve the primary key for the currently-displayed record. I've learned how to properly override the change_form.html template through Django's documentation, and I can add markup to the right block, but I can't figure out how to access the primary key value. (No amount of determined Googling has helped at all.) Alternatively, is there a direct way to specify that I want to produce extra output in my ModelSite definition?

    Read the article

  • Problem with anchor tags in Django after using lighttpd + fastcgi

    - by Drew A
    I just started using lighttpd and fastcgi for my django site, but I've noticed my anchor links are no longer working. I used the anchor links for sorting links on the page, for example I use an anchor to sort links by the number of points (or votes) they have received. For example: the code in the html template: ... {% load sorting_tags %} ... {% ifequal sort_order "points" %} {% trans "total points" %} {% trans "or" %} {% anchor "date" "date posted" %} {% order_by_votes links request.direction %} {% else %} {% anchor "points" "total points" %} {% trans "or" %} {% trans "date posted" %} ... The anchor link on "www.mysite.com/my_app/" for total points will be directed to "my_app/?sort=points" But the correct URL should be "www.mysite.com/my_app/?sort=points" All my other links work, the problem is specific to anchor links. The {% anchor %} tag is taken from django-sorting, the code can be found at http://github.com/directeur/django-sorting Specifically in django-sorting/templatetags/sorting_tags.py Thanks in advance.

    Read the article

  • csrf error in django

    - by niklasfi
    Hello, I want to realize a login for my site. I basically copied and pasted the following bits from the Django Book together. However I still get an error (CSRF verification failed. Request aborted.), when submitting my registration form. Can somebody tell my what raised this error and how to fix it? Here is my code: views.py: # Create your views here. from django import forms from django.contrib.auth.forms import UserCreationForm from django.http import HttpResponseRedirect from django.shortcuts import render_to_response def register(request): if request.method == 'POST': form = UserCreationForm(request.POST) if form.is_valid(): new_user = form.save() return HttpResponseRedirect("/books/") else: form = UserCreationForm() return render_to_response("registration/register.html", { 'form': form, }) register.html: <html> <body> {% block title %}Create an account{% endblock %} {% block content %} <h1>Create an account</h1> <form action="" method="post">{% csrf_token %} {{ form.as_p }} <input type="submit" value="Create the account"> </form> {% endblock %} </body> </html>

    Read the article

  • Django URL Conf Returns Incorrect "Current URL"

    - by natnit
    I have a django app that is mostly done, and the URLs work perfectly when I run it with the manage.py runserver command. However, I've recently tried to get it running via lighttpd, and many links have stopped working. For example: http://mysite.com/races/32 should work, but instead throws this error message. Page not found (404) Request Method: GET Request URL: http://mysite.com/races/32 Using the URLconf defined in racetrack.urls, Django tried these URL patterns, in this order: ^admin/ ^create/$ ^races/$ ^races/(?P<race_id>\d+)/$ ^races/(?P<race_id>\d+)/manage/$ ^races/(?P<text>\w+)/$ ^user/(?P<kol_id>\d+)/$ ^$ ^login/$ ^logout/$ The current URL, 32, didn't match any of these. The request URL is accurate, but the last line (which displays the current URL) is giving 32 instead of races/32 as expected. Here is my urlconf: from django.conf.urls.defaults import * from django.contrib import admin admin.autodiscover() urlpatterns = patterns('racetrack.races.views', (r'^admin/', include(admin.site.urls)), (r'^create/$', 'create'), (r'^races/$', 'index'), (r'^races/(?P<race_id>\d+)/$', 'detail'), (r'^races/(?P<race_id>\d+)/manage/$', 'manage'), (r'^races/(?P<text>\w+)/$', 'index'), (r'^user/(?P<kol_id>\d+)/$', 'user'), # temporary for index page replace with welcome page (r'^$', 'index'), ) urlpatterns += patterns('django.contrib.auth.views', (r'^login/$', 'login', {'template_name': 'races/login.html'}), (r'^logout/$', 'logout', {'next_page': '/'}), ) Thank you.

    Read the article

  • Doing a count over a filter query efficiently in django

    - by apple_pie
    Hello, Django newbie here, I need to do a count over a certain filter in a django model. If I do it like so: my_model.objects.filter(...).count() I'm guessing it does the SQL query that retrieves all the rows and only afterwards does the count. To my knowledge it's much more efficient to do the count without retrieving those rows like so "SELECT COUNT(*) FROM ...". Is there a way to do so in django?

    Read the article

  • customizing Django look and feel in Python

    - by user248237
    I am learning Django and got it to work with wsgi. I'm following the tutorial here: http://docs.djangoproject.com/en/1.1/intro/tutorial01/ My question is: how can I customize the look and feel of Django? Is there a repository of templates that "look good", kind of like there are for Wordpress, that I can start from? I find the tutorial counterintuitive in that it goes immediately toward customizing the admin page of Django, rather than the main pages visible to users of the site. Is there an example of a "typical" Django site, with a decent template, that I can look at and built on/modify? The polls application is again not very representative since it's so specialized. any references on this would be greatly appreciated. thanks.

    Read the article

  • Crossed import in django

    - by Kuhtraphalji
    On example, i have 2 apps: alpha and beta in alpha/models.py import of model from beta.models and in beta/models.py import of model from alpha.models manage.py validate says that ImportError: cannot import name ModelName how to solve this problem?

    Read the article

  • Getting a KeyError in DB backend of django-digest

    - by rtmie
    I have just started to integrate django_digest into my app. As a start I have added the @httpdigest decorator to one of my views. If I try to connect to it I get a KeyError exception thrown in django_digest/backend/db.py . Depending on which db I configure I get a different KeyError in a different location. I am using Django 1.2.1, with MySql (also tested with sqlite). I am using the default values for all the settings options. As far as I can see I have followed all instructions but am struggling all day with this. I am using the repository versions of django-digest and python-digest. Any steer would be greatly appreciated. Tracebacks for sqlite and mysql below: with sqlite: Traceback (most recent call last): File "/home/robm/projects/gcs/server/gcs2.5/lib/python2.5/site-packages/django/core/servers/basehttp.py", line 674, in __call__ return self.application(environ, start_response) File "/home/robm/projects/gcs/server/gcs2.5/lib/python2.5/site-packages/django/core/handlers/wsgi.py", line 248, in __call__ signals.request_finished.send(sender=self.__class__) File "/home/robm/projects/gcs/server/gcs2.5/lib/python2.5/site-packages/django/dispatch/dispatcher.py", line 162, in send response = receiver(signal=self, sender=sender, **named) File "/home/robm/projects/gcs/server/gcs2.5/lib/python2.5/site-packages/django_digest-1.8-py2.5.egg/django_digest/backend/db.py", line 16, in close_connection _connection.close() File "/home/robm/projects/gcs/server/gcs2.5/lib/python2.5/site-packages/django/db/backends/sqlite3/base.py", line 186, in close if self.settings_dict['NAME'] != ":memory:": KeyError: 'NAME' with mysql: Traceback (most recent call last): File "/home/robm/projects/gcs/server/gcs2.5/lib/python2.5/site-packages/django/core/servers/basehttp.py", line 674, in __call__ return self.application(environ, start_response) File "/home/robm/projects/gcs/server/gcs2.5/lib/python2.5/site-packages/django/core/handlers/wsgi.py", line 241, in __call__ response = self.get_response(request) File "/home/robm/projects/gcs/server/gcs2.5/lib/python2.5/site-packages/django/core/handlers/base.py", line 142, in get_response return self.handle_uncaught_exception(request, resolver, exc_info) File "/home/robm/projects/gcs/server/gcs2.5/lib/python2.5/site-packages/django/core/handlers/base.py", line 166, in handle_uncaught_exception return debug.technical_500_response(request, *exc_info) File "/home/robm/projects/gcs/server/gcs2.5/lib/python2.5/site-packages/django/core/handlers/base.py", line 80, in get_response response = middleware_method(request) File "/home/robm/projects/gcs/server/gcs2.5/lib/python2.5/site-packages/django_digest-1.8-py2.5.egg/django_digest/middleware.py", line 13, in process_request if (not self._authenticator.authenticate(request) and File "/home/robm/projects/gcs/server/gcs2.5/lib/python2.5/site-packages/django_digest-1.8-py2.5.egg/django_digest/__init__.py", line 86, in authenticate partial_digest = self._account_storage.get_partial_digest(digest_response.username) File "/home/robm/projects/gcs/server/gcs2.5/lib/python2.5/site-packages/django_digest-1.8-py2.5.egg/django_digest/backend/db.py", line 97, in get_partial_digest cursor = get_connection().cursor() File "/home/robm/projects/gcs/server/gcs2.5/lib/python2.5/site-packages/django/db/backends/__init__.py", line 75, in cursor cursor = self._cursor() File "/home/robm/projects/gcs/server/gcs2.5/lib/python2.5/site-packages/django/db/backends/mysql/base.py", line 281, in _cursor if settings_dict['USER']: KeyError: 'USER'

    Read the article

  • Can django's auth_user.username be varchar(75)? How could that be done?

    - by perrierism
    Is there anything wrong with running alter table on auth_user to make username be varchar(75) so it can fit an email? What does that break if anything? If you were to change auth_user.username to be varchar(75) where would you need to modify django? Is it simply a matter of changing 30 to 75 in the source code?: username = models.CharField(_('username'), max_length=30, unique=True, help_text=_("Required. 30 characters or fewer. Letters, numbers and @/./+/-/_ characters")) Or is there other validation on this field that would have to be changed or any other repercussions to doing so? See comment discussion with bartek below regarding the reason for doing it.

    Read the article

  • Adding links to full change forms for inline items in django admin?

    - by David Eyk
    I have a standard admin change form for an object, with the usual StackedInline forms for a ForeignKey relationship. I would like to be able to link each inline item to its corresponding full-sized change form, as the inline item has inlined items of its own, and I can't nest them. I've tried everything from custom widgets to custom templates, and can't make anything work. So far, the "solutions" I've seen in the form of snippets just plain don't seem to work for inlines. I'm getting ready to try some DOM hacking with jQuery just to get it working and move on. I hope I must be missing something very simple, as this seems like such a simple task! Using Django 1.2.

    Read the article

  • How do I flag only one of the formsets in django admin?

    - by azuer88
    I have these (simplified) models: class Question(models.Model): question = models.CharField(max_length=60) class Choices(models.Model): question = models.ForeignKey(Question) text = models.CharField(max_length=60) is_correct = models.BooleanField(default=False) I've made Choices as an inline of Question (in admin). Is there a way to make sure that only one Choice will have is_correct = True? Ideally, is_correct will be displayed as a radio button when it is displayed in the admin formset (TabularInline). my admin.py has: from django.contrib import admin class OptionInline(admin.TabularInline): model = Option extra = 5 max_num = 5 class QuestionAdmin(admin.ModelAdmin): inlines = [OptionInline, ] admin.site.register(QType) admin.site.register(Question, QuestionAdmin)

    Read the article

  • How can I call model methods or properties from Django Admin?

    - by kg
    Is there a natural way to display model methods or properties in the Django admin site? In my case I have base statistics for a character that are part of the model, but other things such as status effects which affect the total calculation for that statistic: class Character(models.Model): base_dexterity = models.IntegerField(default=0) @property def dexterity(stat_name): total = self.base_dexterity total += sum(s.dexterity for s in self.status.all()]) return total It would be nice if I could display the total calculated statistic alongside the field to change the base statistic in the Change Character admin page, but it is not clear to me how to incorporate that information into the page.

    Read the article

  • Django 1.5 custom User model error. "Manager isn't available; User has been swapped"

    - by bpetit
    I extend the django user model as described in the dev doc. I wan't to keep most of the original User model features so I extend the AbstractUser class. I've defined in settings.py: AUTH_USER_MODEL = 'myapp.CustomUser' My user class: class CustomUser(AbstractUser): custom_field = models.ForeignKey('OtherModel') objects = UserManager() Everything seems to work fine but when I try to make it managed by the admin site: admin.site.register(CustomUser, UserAdmin) I get this error on the admin CustomUser creation page (after validation of the password confirmation form): AttributeError: Manager isn't available; User has been swapped for 'myapp.CustomUser' The point is that I need this model managed by the admin site in order to have the same creation process as with the original User model (two step process with password validation). Thanks for any reply

    Read the article

  • Specifying default value for django hidden form field - bone DRY?

    - by jMyles
    So let's say at the last minute (in the view) I decide I want to specify a default for a field and make it hidden, like so: form.fields['coconut'] = forms.ModelChoiceField(label="", widget=forms.HiddenInput(), queryset=swallow.coconuts.all(), initial=some_particular_coconut) My question is this: Do I really need to specify queryset here? I mean, I already know, from initial, exactly which coconut I'm talking about. Why do I also need to specify that the universe of available coconuts is the set of coconuts which this particular swallow carried (by the husk)? Is there a way I can refrain from specifying queryset? Simply omitting causes django to raise TypeError. If indeed it is required, isn't this a bit damp?

    Read the article

  • Installing Django on Windows

    - by Pranav
    Ever needed to install Django in a Microsoft Windows environment, here is a quick start guide to make that happen: Read through the official Django installation documentation, it might just save you a world of hut down the road. Download Python for your version of Windows. Install Python, my preference here is to put it into the Program Files folder under a folder named Python<Version> Add your chosen Python installation path into your Windows path environment variable. This is an optional step, however it allows you to just type python in the command line and have it fire up the Python interpreter. An easy way of adding it is going into Control Panel, System and into the Environment Variables section. Download Django, you can either download a compressed file or if you’re comfortable with using version control – check it out from the Django Subversion repository. Create a folder named django under your <Python installation directory>\Lib\site-packages\ folder. Using my example above that would have been C:\Program Files\Python25\Lib\site-packages\. If you chose to download the compressed file, open it and extract the contents of the django folder into your newly created folder. If you’d prefer to check it out from Subversion, the normal check out points are http://code.djangoproject.com/svn/django/trunk/ for the latest development copy or a named release which you’ll find under http://code.djangoproject.com/svn/django/tags/releases/. Done, you now have a working Django installation on Windows. At this point, it’d be pertinent to confirm that everything is working properly, which you can do by following the first Django tutorial. The tutorial will make mention of django-admin.py, which is a utility which offers some basic functionality to get you off the ground. The file is located in the bin folder under your Django installation directory. When you need to use it, you can either type in the full path to it or simply add that file path into your environment variables as well. Hope this helps!

    Read the article

  • Django + Apache wsgi = paths problem

    - by Shamanu4
    Hello. I have this view which generates interface language options menu def lang_menu(request,language): lang_choices = [] import os.path for lang in settings.LANGUAGES: if os.path.isfile("gui/%s.py" % lang) or os.path.isfile("gui/%s.pyc" % lang): langimport = "from gui.%s import menu" % lang try: exec(langimport) except ImportError: lang_choices.append({'error':'invalid language file'}) else: lang_choices.append(menu) else: lang_choices.append({'error':'lang file not found'}) t = loader.get_template('gui/blocks/lang_menu_options.html') data = '' for lang in lang_choices: if not 'error' in lang: data = "%s\n%s" % (data,t.render(Context(lang))) if not data: data = "Error! No languages configured or incorrect language files!" return Context({'content':data}) When I'am using development server (python manage.py runserver ...) it works fine. But when I ported my app to apache wsgi server I've got error "No languages configured or incorrect language files!" Here is my Apache config <VirtualHost *:9999> WSGIScriptAlias / "/usr/local/etc/django/terminal/django.wsgi" <Directory "/usr/local/etc/django/terminal"> Options +ExecCGI Allow From All </Directory> Alias /media/ "/usr/local/lib/python2.5/site-packages/django/contrib/admin/media/" <Location /media/> SetHandler None </Location> <Directory "/usr/local/lib/python2.5/site-packages/django/contrib/admin/media/> Allow from all </Directory> Alias /static/ "/usr/local/etc/django/terminal/media/" <Location /static/> SetHandler None </Location> ServerName ******* ServerAlias ******* ErrorLog /var/log/django.error.log TransferLog /var/log/django.access.log </VirtualHost> django.wsgi: import os, sys sys.path.append('/usr/local/etc/django') sys.path.append('/usr/local/etc/django/terminal') os.environ['DJANGO_SETTINGS_MODULE'] = 'terminal.settings' import django.core.handlers.wsgi application = django.core.handlers.wsgi.WSGIHandler() It's look like as problem with path configuration but I'm stuck here ...

    Read the article

< Previous Page | 12 13 14 15 16 17 18 19 20 21 22 23  | Next Page >