django-urls

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<',

I am currently trying out django. I use the namespace argument in one of my include() s in urls.py. When I run the server and try to browse, I get this error. File "C:\Users\User\AppData\Local\Programs\Python\Python36-32\lib\site-packages\django\urls\conf.py", line 39, in include 'Specifying a namespace in include() without providing an app_name ' django.core.exceptions.ImproperlyConfigured: Specifying a namespace in include() without providing an app_name is not supported. Set the app_name attribute in the included module, or pass a 2-tuple containing the list of patterns and app_name instead

Very basic question, but I'm having trouble tracking down the answer on the web. I have a template, which I want to link to the django admin site (i.e. localhost:8000/admin). What is the code for this? I'm imagining something like <a href="{% url admin.site.root %}">link to admin panel</a> However, when I try the above snippet I get: Caught an exception while rendering: Reverse for 'project_name.django.contrib.admin.sites.root' with arguments '()' and keyword arguments '{}' not found. Help? Romain Try what Oggy is suggesting but then use ':' instead of '_' with the current Django: <a href="{%

I'm doing a slideshow and each slide has a url format like this: articles/1234#slide=5 . I want to retrive the slide=5 part from the url in my url.py file and then pass it to the corresponding view function and finally, pass it to the template and render the right slide. The url settings is as follows: url(r'^(?P<article_id>\d+)#slide=(?P<current_slide>\d{1,2})$', 'articles.views.show_article') But it seems that it cannot get the current_slide variable from the url. I guess it has something to do with the anchor part cause it's not transferred to the server. But if I ignore the anchor part in

In Django what is the url pattern I need to use to handle urlencode characters such as %20 I am using (?P<name>[\w]+) but this only handles alphanumeric characters so % is causing an error I was able to make it work using the configuration given below. Check if it will suit your needs. (?P<name>[\w|\W]+) If you only want to allow space: (?P<name>[\w\ ]+) The best way to do that and allow others chars is using '\s' that is any spaces, tabs and new lines (?P<name>[\w\s]+) 来源： https://stackoverflow.com/questions/3675368/django-url-pattern-for-20

I have a view that validates and saves a form. After the form is saved, I'd like redirect back to a list_object view with a success message "form for customer xyz was successfully updated..." HttpResponseRedirect doesn't seem like it would work, because it only has an argument for the url, no way to pass dictionary with it. I've tried modifying my wrapper for object_list to take a dict as a parameter that has the necessary context. I the return a call to this wrapper from inside the view that saves the form. However, when the page is rendered, the url is '/customer_form/' rather than '/list