django-views

I'm new to Django and am trying to create the back end code for a music application on my website. I have created the correct view in my views.py file (in the correct directory) as shown below: def detail(request, album_id): return HttpResponse("<h1>Details for Album ID:" + str(album_id) + "</h1>") however, when creating the url or path for this (shown below) #/music/71/ (pk) path(r'^(?P<album_id>[0-9])/$', views.detail, name='detail'), I am experiencing a warning on my terminal stating: ?: (2_0.W001) Your URL pattern '^(?P<album_id>[0-9])/$' [name='detail'] has a route that contains '(?P<',

Tried to run command: from django.urls import path Getting error: Traceback (most recent call last): File "< stdin >", line 1, in ImportError: cannot import name 'path' I am using django version 1.11 The reason you cannot import path is because it is new in Django 2.0 as is mentioned here: https://docs.djangoproject.com/en/2.0/ref/urls/#path . On that page in the bottom right hand corner you can change the documentation version to the version that you have installed. If you do this you will see that there is no entry for path on the 1.11 docs. You need Django version 2 pip install --upgrade

I'm new to Django. I read the Documentation and I am literally confused with media root files and static root files and where I should be placing my css files along with my javascript files. Can you guys point me in the correct direction? Project setup: Project -myapp -model,view,and test -template - base.html - index.html - view1 - view2 - media - css - style.css - ie.css - js - img -setting.py -url.py What should my static_url, static_root, media_url , media_root, STATICFILES_DIRS and STATICFILES_FINDERS look like? How I currently load my css file is through the base.html with the following:

I'm getting a ValueError when running python manage.py test . My project is named fellow_go , and I'm currently working on an App called pickup . Please note that this error is added in a relatively recent commit to Django: Fixed #24452 -- Fixed HashedFilesMixin correctness with nested paths. . ====================================================================== ERROR: test_view_url_exists_at_desired_location (pickup.tests.test_view.HomePageViewTest) ---------------------------------------------------------------------- Traceback (most recent call last): File "/Users/sunqingyao

I'm using Class based generic views, can anybody suggest me how can i set the initial values to update form? I tried using get_initial() method but didn't got any success. Following is the code which i tried class IncidentUpdateView(UpdateView): form_class = IncidentForm form_class.initial = {"badge_number": '88888'} model = Incident template_name = 'hse/incident/incident_update.html' def get_initial(self, form_class): initials = { "badge_number": '88888' } form = form_class(initial=initials) return form def get_success_url(self): return reverse_lazy('hse-incident', args=[self.object.id]) You

This question already has an answer here: How to update an object from edit form in Django? 2 answers I need to have a form that allows the creation or addition of sessions on a planning Model class Session(models.Model): tutor = models.ForeignKey(User) start_time = models.DateTimeField() end_time = models.DateTimeField() Form class SessionForm(forms.ModelForm): class Meta: model = Session exclude = ['tutor'] View to render the form def editor(request): if request.method == 'GET': if request.GET['id'] != '0': # The user has selected a session session = Session.objects.get(id=request.GET['id'])

Can I have a Form in my template which is not backed by a model. I do not need to store the data just need that data to generate a POST request of my own in the view. Template - The form with text fields. View - get data from form, and generate another request. Flow --> Form submit takes to a url which calls the view " def form_handle(request): if request.method=='POST' form = request.POST #blah blah encode parameters for a url blah blah #and make another post request but this puts only the csrf token into the form variable. Is there some way I can access those text fields of the template in