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  • Right way to have a thread in parallel to django project on wsgi.

    - by Enrico Carlesso
    Hi guys. I'm writing a django project, and I need to have a parallel thread which performs certain tasks. The project will be deployed in Apache2.2 with mod_wsgi. Actually my implementation consists on a thread with a while True - Sleep which is called from my django.wsgi file. Is this implementation correct? Two problems raises: does django.wsgi get called only once? Will I have just that instance of the thread running? And second, I need to "manually" visit at least a page to have the Thread run. Is there a workaround? Does anyone has some hints on better solutions? Thanks in advance.

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  • Django Querysets -- need a less expensive way to do this..

    - by rh0dium
    Hi all, I have a problem with some code and I believe it is because of the expense of the queryset. I am looking for a much less expensive (in terms of time) way to to this.. log.info("Getting Users") employees = Employee.objects.filter(is_active = True) log.info("Have Users") if opt.supervisor: if opt.hierarchical: people = getSubs(employees, " ".join(args)) else: people = employees.filter(supervisor__name__icontains = " ".join(args)) else: log.info("Filtering Users") people = employees.filter(name__icontains = " ".join(args)) | \ employees.filter(unix_accounts__username__icontains = " ".join(args)) log.info("Filtered Users") log.info("Processing data") np = [] for person in people: unix, p4, bugz = "No", "No", "No" if len(person.unix_accounts.all()): unix = "Yes" if len(person.perforce_accounts.all()): p4 = "Yes" if len(person.bugzilla_accounts.all()): bugz = "Yes" if person.cell_phone != "": exphone = fixphone(person.cell_phone) elif person.other_phone != "": exphone = fixphone(person.other_phone) else: exphone = "" np.append({ 'name':person.name, 'office_phone': fixphone(person.office_phone), 'position': person.position, 'location': person.location.description, 'email': person.email, 'functional_area': person.functional_area.name, 'department': person.department.name, 'supervisor': person.supervisor.name, 'unix': unix, 'perforce': p4, 'bugzilla':bugz, 'cell_phone': fixphone(exphone), 'fax': fixphone(person.fax), 'last_update': person.last_update.ctime() }) log.info("Have data") Now this results in a log which looks like this.. 19:00:55 INFO phone phone Getting Users 19:00:57 INFO phone phone Have Users 19:00:57 INFO phone phone Processing data 19:01:30 INFO phone phone Have data As you can see it's taking over 30 seconds to simply iterate over the data. That is way too expensive. Can someone clue me into a more efficient way to do this. I thought that if I did the first filter that would make things easier but seems to have no effect. I'm at a loss on this one. Thanks To be clear this is about 1500 employees -- Not too many!!

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  • How to provide an inline model field with a queryset choices without losing field value for inline r

    - by Judith Boonstra
    The code displayed below is providing the choices I need for the app field, and the choices I need for the attr field when using Admin. I am having a problem with the attr field on the inline form for already saved records. The attr selected for these saved does show in small print above the field, but not within the field itself. # MODELS: Class Vocab(models.Model): entity = models.Charfield, max_length = 40, unique = True) Class App(models.Model): name = models.ForeignKey(Vocab, related_name = 'vocab_appname', unique = True) app = SelfForeignKey('self, verbose_name = 'parent', blank = True, null = True) attr = models.ManyToManyField(Vocab, related_name = 'vocab_appattr', through ='AppAttr' def parqs(self): a method that provides a queryset consisting of available apps from vocab, excluding self and any apps within the current app's dependent line. def attrqs(self): a method that provides a queryset consisting of available attr from vocab excluding those already selected by current app, 2) those already selected by any apps within the current app's parent line, and 3) those selected by any apps within the current app's dependent line. Class AppAttr(models.Model): app = models.ForeignKey(App) attr = models.ForeignKey(Vocab) # FORMS: from models import AppAttr def appattr_form_callback(instance, field, *args, **kwargs) if field.name = 'attr': if instance: return field.formfield(queryset = instance.attrqs(), *kwargs) return field.formfield(*kwargs) # ADMIN: necessary imports class AppAttrInline(admin.TabularInline): model = AppAttr def get_formset(self, request, obj = None, **kwargs): kwargs['formfield_callback'] = curry(appattr_form_callback, obj) return super(AppAttrInline, self).get_formset(request, obj, **kwargs) class AppForm(forms.ModelForm): class Meta: model = App def __init__(self, *args, **kwargs): super(AppForm, self).__init__(*args, **kwargs) if self.instance.id is None: working = App.objects.all() else: thisrec = App.objects.get(id = self.instance.id) working = thisrec.parqs() self.fields['par'].queryset = working class AppAdmin(admin.ModelAdmin): form = AppForm inlines = [AppAttrInline,] fieldsets = .......... necessary register statements

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  • Overwrite queryset which builds filter sidebar

    - by cw
    Hi, I'm writing a hockey database/manager. So I have the following models: class Team(models.Model): name = models.CharField(max_length=60) class Game(models.Model): home_team = models.ForeignKey(Team,related_name='home_team') away_team = models.ForeignKey(Team,related_name='away_team') class SeasonStats(models.Model): team = models.ForeignKey(Team) Ok, so my problem is the following. There are a lot of teams, but Stats are just managed for my Club. So if I use "list_display" in the admin backend, I'd like to modify/overwrite the queryset which builds the sidebar for filtering, to just display our home teams as a filter option. Is this somehow possible in Django? I already made a custom form like this class SeasonPlayerStatsAdminForm(forms.ModelForm): team = forms.ModelChoiceField(Team.objects.filter(club__home=True)) So now just the filtering is missing. Any ideas?

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  • Is there a way to simplify this Django query?

    - by Mark
    accepted_bids = Bid.objects.filter(shipment__user=u, status='acc').select_related('shipment') completed_shipments = [] for b in accepted_bids: completed_shipments.append(b.shipment) vehicles_shipped = [] for s in completed_shipments: vehicles_shipped.extend(s.items.all()) In the end, I want a list of shipped vehicles. A vehicle is shipped if it's part of a shipment that's completed. A shipment is completed if it has an accepted bid. I'd prefer not to iterate over the querysets thereby forcing a hit to the DB before its necessary... isn't there a way to get all the associated shipments from a list of bids, for example?

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  • How to get all objects with their children using django orm?

    - by kender
    Hi, I got very simple hierarchical structure: every object can have 0 or 1 parent. There's no limit on how many children each object can have. So in my application I got such a model: class O(Model): name = CharField(max_length = 20) parent = ForeignKey('O', related_name = 'children') Now I would like to be able to fetch all objects who have a particular one Object1 in their parent-tree (as in their parent or parent of their parents, etc). Should I use mptt or is there a simpler approach?

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  • multiple models in Rails with a shared interface

    - by dfondente
    I'm not sure of the best structure for a particular situation in Rails. We have several types of workshops. The administration of the workshops is the same regardless of workshop type, so the data for the workshops is in a single model. We collect feedback from participants about the workshops, and the questionnaire is different for each type of workshop. I want to access the feedback about the workshop from the workshop model, but the class of the associated model will depend on the type of workshop. If I was doing this in something other than Rails, I would set up an abstract class for WorkshopFeedback, and then have subclasses for each type of workshop: WorkshopFeedbackOne, WorkshopFeedbackTwo, WorkshopFeedbackThree. I'm unsure how to best handle this with Rails. I currently have: class Workshop < ActiveRecord::Base has_many :workshop_feedbacks end class Feedback < ActiveRecord::Base belongs_to :workshop has_many :feedback_ones has_many :feedback_twos has_many :feedback_threes end class FeedbackOne < ActiveRecord::Base belongs_to :feedback end class FeedbackTwo < ActiveRecord::Base belongs_to :feedback end class FeedbackThree < ActiveRecord::Base belongs_to :feedback end This doesn't seem like to the cleanest way to access the feedback from the workshop model, as accessing the correct feedback will require logic investigating the Workshop type and then choosing, for instance, @workshop.feedback.feedback_one. Is there a better way to handle this situation? Would it be better to use a polymorphic association for feedback? Or maybe using a Module or Mixin for the shared Feedback interface? Note: I am avoiding using Single Table Inheritance here because the FeedbackOne, FeedbackTwo, FeedbackThree models do not share much common data, so I would end up with a large sparsely populated table with STI.

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  • Django - The included urlconf doesn't have any patterns in it

    - by unsorted
    My website, which was working before, suddenly started breaking with the error "ImproperlyConfigured at / The included urlconf resume.urls doesn't have any patterns in it" The project base is called resume. In settings.py I have set ROOT_URLCONF = 'resume.urls' Here's my resume.urls, which sits in the project root directory. from django.conf.urls.defaults import * # Uncomment the next two lines to enable the admin: from django.contrib import admin admin.autodiscover() urlpatterns = patterns('', # Example: # (r'^resume/', include('resume.foo.urls')), # Uncomment the admin/doc line below and add 'django.contrib.admindocs' # to INSTALLED_APPS to enable admin documentation: (r'^admin/doc/', include('django.contrib.admindocs.urls')), # Uncomment the next line to enable the admin: (r'^admin/', include(admin.site.urls)), (r'^accounts/login/$', 'django.contrib.auth.views.login'), #(r'^employer/', include(students.urls)), (r'^ajax/', include('urls.ajax')), (r'^student/', include('students.urls')), (r'^club/(?P<object_id>\d+)/$', 'resume.students.views.club_detail'), (r'^company/(?P<object_id>\d+)/$', 'resume.students.views.company_detail'), (r'^program/(?P<object_id>\d+)/$', 'resume.students.views.program_detail'), (r'^course/(?P<object_id>\d+)/$', 'resume.students.views.course_detail'), (r'^career/(?P<object_id>\d+)/$', 'resume.students.views.career_detail'), (r'^site_media/(?P<path>.*)$', 'django.views.static.serve', {'document_root': 'C:/code/django/resume/media'}), ) Anyone know what's wrong? This is driving me crazy. Thanks,

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  • Settings module not found deploying django on a shared server

    - by mcanes
    I'm trying to deploy my django project on a shared hosting as describe here I have my project on /home/user/www/testa I'm using this script #!/usr/bin/python import sys, os sys.path.append("/home/user/bin/python") sys.path.append('/home/user/www/testa') os.chdir("/home/user/www/testa") os.environ['DJANGO_SETTINGS_MODULE'] = "settings.py" from django.core.servers.fastcgi import runfastcgi runfastcgi(method="threaded", daemonize="false") And here's the error I get when trying to run it from shell: WSGIServer: missing FastCGI param REQUEST_METHOD required by WSGI! WSGIServer: missing FastCGI param SERVER_NAME required by WSGI! WSGIServer: missing FastCGI param SERVER_PORT required by WSGI! WSGIServer: missing FastCGI param SERVER_PROTOCOL required by WSGI! Traceback (most recent call last): File "build/bdist.linux-i686/egg/flup/server/fcgi_base.py", line 558, in run File "build/bdist.linux-i686/egg/flup/server/fcgi_base.py", line 1118, in handler File "/home/user/lib/python2.4/site-packages/django/core/handlers/wsgi.py", line 230, in __call__ self.load_middleware() File "/home/user/lib/python2.4/site-packages/django/core/handlers/base.py", line 33, in load_middleware for middleware_path in settings.MIDDLEWARE_CLASSES: File "/home/user/lib/python2.4/site-packages/django/utils/functional.py", line 269, in __getattr__ self._setup() File "/home/usr/lib/python2.4/site-packages/django/conf/__init__.py", line 40, in _setup self._wrapped = Settings(settings_module) File "/home/user/lib/python2.4/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 'settings.py' (Is it on sys.path? Does it have syntax errors?): No module named settings.py Content-Type: text/html Unhandled Exception Unhandled Exception An unhandled exception was thrown by the application. What am I doing wrong? Running the script from the browser just gives me an internal server error.

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  • Django Image Upload: IOErrno2 Could not find path -- and yet it's saving the image there anyway?

    - by Rob
    I have an issue where the local version of django is handling image upload as expected but my server is not. Note: I am using a Django Container on MediaTemple.net (grid server) Here is my code. def view_settings(request): <snip> if request.POST: success_msgs = () mForm = MainProfileForm(request.POST, request.FILES, instance = mProfile) pForm = ChangePasswordForm(request.POST) eForm = ChangeEmailForm(request.POST) if mForm.is_valid(): m = mForm.save(commit = False) if mForm.cleaned_data['avatar']: m.avatar = upload_photo(request.FILES['avatar'], settings.AVATAR_SAVE_LOCATION) m.save() success_msgs += ('profile pictured updated',) <snip> def upload_photo(data,saveLocation): savePath = os.path.join(settings.MEDIA_ROOT, saveLocation, data.name) destination = open(savePath, 'wb+') for chunk in data.chunks(): destination.write(chunk) destination.close() return os.path.join(saveLocation, data.name) Here's where it gets whacky and I was hoping someone could shed a light on this error, because either a) it's the wrong error code, or b) something is happening with the file before it's completely handled. To recap, the file was actually uploaded to the server in the intended directory - and yet this err msg continues to persist. IOError at /user/settings [Errno 2] No such file or directory: u'/home/user66666/domains/example.com/html/media/images/avatars/DSC03852.JPG' Environment: Request Method: POST Request URL: http://111.111.111.111:2011/user/settings Django Version: 1.0.2 final Python Version: 2.4.4 Installed Applications: ['django.contrib.auth', 'django.contrib.contenttypes', 'django.contrib.sessions', 'django.contrib.sites', 'ctrlme', 'usertools', 'easy_thumbnails'] Installed Middleware: ('django.middleware.common.CommonMiddleware', 'django.contrib.sessions.middleware.SessionMiddleware', 'django.contrib.auth.middleware.AuthenticationMiddleware') Traceback: File "/home/user6666/containers/django/leonidas/usertools/views.py" in view_settings m.avatar = upload_photo(request.FILES['avatar'], settings.AVATAR_SAVE_LOCATION) File "/home/user666666/containers/django/leonidas/usertools/functions.py" in upload_photo destination = open(savePath, 'wb+')

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  • Installing django on dreamhost (help a newb out)

    - by augustfirst
    I'm trying to get django running on my dreahost account. I've been trying to sort of use two tutorials at once: the one on the dreamhost wiki and the one in the django book. I installed django using the script on the wiki page, but I ran into trouble immediately while trying to work through the django book. It says: To start the server, change into your project directory (cd mysite), if you haven’t already, and run this command: python manage.py runserver This launches the server locally, on port 8000, accessible only to connections from your own computer. Now that it’s running, visit 127.0.0.1:8000 with your Web browser. You’ll see a “Welcome to Django” page shaded in a pleasant pastel blue. It worked! Those instructions seem to assume that you're developing locally, not on a shared server. Where the heck am I supposed to look for the "Welcome to Django" page after starting the server? In my webroot? No dice. Anyway, I tried to blunder ahead through the django book to its hello world tutorial (chapter 3). But once I've edited the view file and the URLconf, I don't get a nice clean "hello world" text. Instead (as you can see) I get an "import error". Any help would be greatly appreciated.

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  • Need help with Django tutorial

    - by Nai
    I'm doing the Django tutorial here: http://docs.djangoproject.com/en/1.2/intro/tutorial03/ My TEMPLATE_DIRS in the settings.py looks like this: TEMPLATE_DIRS = ( "/webapp2/templates/" "/webapp2/templates/polls" # Put strings here, like "/home/html/django_templates" or "C:/www/django/templates". # Always use forward slashes, even on Windows. # Don't forget to use absolute paths, not relative paths. ) My urls.py looks like this: from django.conf.urls.defaults import * from django.contrib import admin admin.autodiscover() urlpatterns = patterns('', (r'^polls/$', 'polls.views.index'), (r'^polls/(?P<poll_id>\d+)/$', 'polls.views.detail'), (r'^polls/(?P<poll_id>\d+)/results/$', 'polls.views.results'), (r'^polls/(?P<poll_id>\d+)/vote/$', 'polls.views.vote'), (r'^admin/', include(admin.site.urls)), ) My views.py looks like this: from django.template import Context, loader from polls.models import Poll from django.http import HttpResponse def index(request): latest_poll_list = Poll.objects.all().order_by('-pub_date')[:5] t = loader.get_template('c:/webapp2/templates/polls/index.html') c = Context({ 'latest_poll_list': latest_poll_list, }) return HttpResponse(t.render(c)) I think I am getting the path of my template wrong because when I simplify the views.py code to something like this, I am able to load the page. from django.http import HttpResponse def index(request): return HttpResponse("Hello, world. You're at the poll index.") My index template file is located at C:/webapp2/templates/polls/index.html. What am I doing wrong?

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  • Django: getting the list of related records for a list of objects

    - by Silver Light
    Hello! I have two models related one-to many: class Person(models.Model): name = models.CharField(max_length=255); surname = models.CharField(max_length=255); age = models.IntegerField(); class Dog(models.Model): name = models.CharField(max_length=255); owner = models.ForeignKey('Person'); I want to output a list of persons below each person a list of dogs he has. Here's how I can do it: in view: persons = Person.objects.all()[0:100]; in template: {% for p in persons %} {{ p.name }} has dogs:<br /> {% for d in persons.dog_set.all %} - {{ d.name }}<br /> {% endfor %} {% endfor %} But if I do it like that, Django will execute 101 SQL queries which is very inefficient. I tried to make a custom manager, which will get all the persons, then all the dogs and links them in python, but then I can't use paginator (my another question: http://stackoverflow.com/questions/2532475/django-paginator-raw-sql-query ) and it looks quite ugly. Is there a more graceful way doing this?

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  • Ruby on Rails: create records for multiple models with one form and one submit

    - by notblakeshelton
    I have a 3 models: quote, customer, and item. Each quote has one customer and one item. I would like to create a new quote, a new customer, and a new item in their respective tables when I press the submit button. I have looked at other questions and railscasts and either they don't work for my situation or I don't know how to implement them. I also want my index page to be the page where I can create everything. quote.rb class Quote < ActiveRecord::Base attr_accessible :quote_number has_one :customer has_one :item end customer.rb class Customer < ActiveRecord::Base #unsure of what to put here #a customer can have multiple quotes, so would i use: has_many :quotes #<----? end item.rb class Item < ActiveRecord::Base #also unsure about this #each item can also be in multiple quotes quotes_controller.rb class QuotesController < ApplicationController def index @quote = Quote.new @customer = Customer.new @item = item.new end def create @quote = Quote.new(params[:quote]) @quote.save @customer = Customer.new(params[:customer]) @customer.save @item = Item.new(params[:item]) @item.save end end items_controller.rb class ItemsController < ApplicationController def index end def new @item = Item.new end def create @item = Item.new(params[:item]) @item.save end end customers_controller.rb class CustomersController < ApplicationController def index end def new @customer = Customer.new end def create @customer = Customer.new(params[:customer]) @customer.save end end quotes/index.html.erb <%= form_for @quote do |f| %> <%= f.fields_for @customer do |builder| %> <%= label_tag :firstname %> <%= builder.text_field :firstname %> <%= label_tag :lastname %> <%= builder.text_field :lastname %> <% end %> <%= f.fields_for @item do |builder| %> <%= label_tag :name %> <%= builder.text_field :name %> <%= label_tag :description %> <%= builder.text_field :description %> <% end %> <%= label_tag :quote_number %> <%= f.text_field :quote_number %> <%= f.submit %> <% end %> When I try submitting that I get an error: Can't mass-assign protected attributes: item, customer So to try and fix it I updated the attr_accessible in quote.rb to include :item, :customer but then I get this error: Item(#) expected, got ActiveSupport::HashWithIndifferentAccess(#) Any help would be greatly appreciated.

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  • django wsgi multiple projects different url same apache server

    - by Thomas Schultz
    Hello, I'm trying to get 2 separate django projects running on the same apache server with mod_wsgi that are also under the same domain but different urls. Like www.example.com/site1/ and www.example.com/site2 What I'm trying to do is something like... <VirtualHost *:80> ServerName www.example.com <location "/site1/"> DocumentRoot "/var/www/html/site1" WSGIScriptAlias / /var/www/html/site1/django.wsgi </location> <location "/site2/"> DocumentRoot "/var/www/html/site2" WSGIScriptAlias / /var/www/html/site2/django.wsgi </location> </VirtualHost> The closes thing I've seen is this http://docs.djangoproject.com/en/dev/howto/deployment/modpython/ but "mysite" is different for both of these cases and they're using mod_python instead of mod_wsgi. Any help with this would be great thanks!

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  • Django HttpResponseRedirect acting as proxy rather than 302

    - by Trevor Burnham
    I have a Django method that's returning return HttpResponseRedirect("/redirect-target") When running the server locally, if I visit the page that returns that redirect, I get the log output [17/Oct/2013 15:26:02] "GET /redirecter HTTP/1.1" 302 0 [17/Oct/2013 15:26:02] "GET /redirect-target HTTP/1.1" 404 0 as expected. But, when I visit that page in Chrome, the Network tab shows the request to /redirecter with the response from /redirect-target, rather than showing the 302. cURL does the same: $ curl -I -X GET http://localhost/redirecter HTTP/1.1 404 Not Found date: Thu, 17 Oct 2013 19:32:30 GMT connection: keep-alive transfer-encoding: chunked In production, the same Django code does show a 302 redirect in Chrome and cURL. What could be going on here? Is there some kind of Django setting that might be causing it to proxy the target rather than send a redirect when HttpResponseRedirect is used (but lie about it in the log)? Or is there a quirk on my system (OS X) that might cause localhost redirects to behave this way?

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  • Django fails to find static files served by nginx

    - by Simon
    I know this is a really noobish question but I can't find any solution despite finding the problem trivial. I have a django application deployed with gunicorn. The static files are served by the nginx server with the following url : myserver.com/static/admin/css/base.css. However, my django application keep looking for the static files at myserver.com:8001/static/admin/css/base.css and is obviously failing (404). I don't know how to fix this. Is it a django or an nginx problem ? Here is my nginx configuration file : server { server_name myserver.com; access_log off; location /static/ { alias /home/myproject/static/; } location / { proxy_pass http://127.0.0.1:8001; proxy_set_header X-Forwarded-Host $server_name; proxy_set_header X-Real-IP $remote_addr; add_header P3P 'CP="ALL DSP COR PSAa PSDa OUR NOR ONL UNI COM NAV"'; } } Thanks for the help !

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  • gevent with Django as daemon

    - by jonathonmorgan
    I've been developing an app using django_socketio (a python port of the Node equivalent), which relies on gevent. It ships with a Django management command that runs gevent's pywsgi server, but that of course stops when I close my terminal window, just like Django's dev server. This is a proof of concept, and there's no expectation that it would hold up in a production environment, but I'd like to have the server at least "permanently" process HTTP requests, so I don't need to manually start the dev server in order to demo. I'm assuming I need to run this as a daemon process, but prior to this I've only used apache and mod_wsgi, so unsure of where to begin, or even how I would go about starting a daemon. I found gevent-spawn, which looks promising, but it's unclear to me how that code is executed. Basically, how would I use gevent to serve a Django app in a setting without manually starting/stopping the server?

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  • Deploy multiple django instances on one Host [migrated]

    - by tvn
    I am trying to setup multiple Django instances on one Host with lighttpd. My problem is to get Djangos FCGI working on subdirectories served by my Webserver. So my aim is the following: www.myhost.org/django0 - django1.fcgi on localhost:3000 www.myhost.org/django1 - django2.fcgi on localhost:3001 www.myhost.org/django2 - django3.fcgi on localhost:3002 Unfortunately the following configuration doesn't even work for one: $HTTP["url"] =~ "^/django0/static($|/)" { server.document-root = "/home/django0/django/static/" } $HTTP["url"] =~ "^/django0/media($|/)" { server.document-root = "/usr/lib/python2.7/dist-packages/django/contrib/admin/media/" } $HTTP["url"] =~ "^/django0($|/)" { proxy.server = ( "" => ( ( "host" => "127.0.0.1", "port" => "3001", "check-local" => "disable", ) ) ) } The only response I get is an 404 and even this takes a long time till I get this. I found nothing suspicious neither in the access.log nor in the error.log.

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  • Use localeURL middleware with apache prefix

    - by Olivier R.
    Good morning everyone, I Got a question about localeURL usage. Everything works great for me with url like this : www.mysite.com/ If I type www.mysite.com/ in adress bar, it turns correctly in www.mysite.com/en/ for example. If I use the view change_locale, it's also all right (ie change www.mysite.com/en/ in www.mysite.com/fr/). But my application use apache as server, and use a prefix for the site, that gives url like this : www.mysite.com/prefix/ If I type www.mysite.com/prefix/ in the adress bar, the adress turns into www.mysite.com/en/ without prefix (so 404) I change code of view to manage our settings.SERVER_PREFIX value : def change_locale(request) : """ Redirect to a given url while changing the locale in the path The url and the locale code need to be specified in the request parameters. O. Rochaix; Taken from localeURL view, and tuned to manage : - SERVER_PREFIX from settings.py """ next = request.REQUEST.get('next', None) if not next: next = request.META.get('HTTP_REFERER', None) if not next: next = settings.SERVER_PREFIX + '/' next = urlsplit(next).path prefix = False if settings.SERVER_PREFIX!="" and next.startswith(settings.SERVER_PREFIX) : prefix = True next = "/" + next.lstrip(settings.SERVER_PREFIX) _, path = utils.strip_path (next) if request.method == 'POST': locale = request.POST.get('locale', None) if locale and check_for_language(locale): path = utils.locale_path(path, locale) if prefix : path = settings.SERVER_PREFIX + path response = http.HttpResponseRedirect(path) return response with this customized view, i'm able to correctly change language, but i'm not sure that's the right way of doing stuff. Is there any option on localeURL to manage prefix of apache ?

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  • How to return number of rows in the template

    - by xRobot
    In my view I return all posts of one blog: posts = Post.objects.filter(blog=blog) and pass it to context. But.. How can I get the number of posts in the template ? This is my template: <h1>Number of posts: {{ ??? }} </h1> {% for post in posts %} {{ post.title }} {{ post.body }} {% endfor %}

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  • Returning modified data to a template

    - by Duncan
    I need to amend QuerySet data when i return it to a template. for example, model.objects.all() returns a date (with other fields), but i also want to return the number of days since that date has passed. This is so that in the template, i can say "you last logged in 4 days ago". What is the best way to do this?

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