Django - Redirect uppercase urls to lowercase - django

In my Django apps, I have many urls including /(?P<project_name>[_\w]+)/.
The project_name is defined by users and it is an attribute of the Project model.
I've added a validator on the project_name to check if it's lowercase.
So new names are all lowercase but some old names include uppercase characters.
I would like to change all the names stored to make them lowercase but at the same time, I don't want users to get an error when trying to access to one of the url with the old project name including uppercase characters.
As I have many urls and many views, I don't want to update each to manually .lower() the project_name.
Is there a way to redirect all urls including /<project_NAME>/ to /<project_name>/?

Hacky way with decorators
You could create a decorator for all your views that use a project_name:
def project_lowercase(_func=None):
def checkLowercase(func):
#functools.wraps(func)
def wrapper(*args, **kwargs):
request = args[0]
project_name = args[1]
if not project_name.islower():
return HttpResponseRedirect(reverse(your_url_name, kwargs{'project_name' = project_name.lower()))
return func(*args, **kwargs)
return wrapper
return checkLowercase(_func)
Replace your_url_name with whatever you named your url route, then import the function and add the decorator above each view function:
#project_lowercase
def view_project(request, project_name):
# ...
return "Hello World!"
Slugs would be better
Alternitively as suggested in the comments and a better solution you should use a slug to store urls for your projects, have a look here to see how to add them to your models and generate slugs.
To move your existing data to use slugs or just to update project names you can create a data migration that will alter existing data in your database.

The first option I would suggest is to avoid regular expressions, switch to path for your urlpatterns and use custom url converter for your project names.
In case you want to keep using regular expressions, you could restrict your pattern to only accept lower case project names /(?P<project_name>[_a-z0-9]+)/. After that add url pattern, which is (effectively) the same as your current one /(?P<project_name>\w+)/ (note missing _ - \w already includes that) to match all project names, including the legacy ones, the view for that pattern would redirect to your first view with the lower case project_name.

Related

uuid as keyword argument <pk> in django urls.py

I am using a uuid4 as id in some django models. Thats why I cannot use <int:pk> in the urls.py because it wont match. Is there another prefix I can use to match these uuids like a2182835-4518-cb95-8eaf-0d5a34105cb2 ?
You can use the uuid path converter instead of int. From the docs
uuid - Matches a formatted UUID. To prevent multiple URLs from mapping to the same page, dashes must be included and letters must be lowercase. For example, 075194d3-6885-417e-a8a8-6c931e272f00. Returns a UUID instance.
So for your purposes changing <int:pk> to <uuid:pk> should do the trick

Django syndication framework: prevent appending SITE_ID to the links

According to the documentation here: https://djangobook.com/syndication-feed-framework/
If link doesn’t return the domain, the syndication framework will
insert the domain of the current site, according to your SITE_ID
setting
However, I'm trying to generate a feed of magnet: links. The framework doesn't recognize this and attempts to append the SITE_ID, such that the links end up like this (on localhost):
<link>http://localhost:8000magnet:?xt=...</link>
Is there a way to bypass this?
Here's a way to do it with monkey patching, much cleaner.
I like to create a separate folder "django_patches" for these kinds of things:
myproject/django_patches/__init__.py
from django.contrib.syndication import views
from django.contrib.syndication.views import add_domain
def add_domain_if_we_should(domain, url, secure=False):
if url.startswith('magnet:'):
return url
else:
return add_domain(domain, url, secure=False)
views.add_domain = add_domain_if_we_should
Next, add it to your INSTALLED_APPS so that you can patch the function.
settings.py
INSTALLED_APPS = [
'django_overrides',
...
]
This is a bit gnarly, but here's a potential solution if you don't want to give up on the Django framework:
The problem is that the method add_domain is buried deep in a huge method within syndication framework, and I don't see a clean way to override it. Since this method is used for both the feed URL and the feed items, a monkey patch of add_domain would need to consider this.
Django source:
https://github.com/django/django/blob/master/django/contrib/syndication/views.py#L178
Steps:
1: Subclass the Feed class you're using and do a copy-paste override of the huge method get_feed
2: Modify the line:
link = add_domain(
current_site.domain,
self._get_dynamic_attr('item_link', item),
request.is_secure(),
)
To something like:
link = self._get_dynamic_attr('item_link', item)
I did end up digging through the syndication source code and finding no easy way to override it and did some hacky monkey patching. (Unfortunately I did it before I saw the answers posted here, all of which I assume will work about as well as this one)
Here's how I did it:
def item_link(self, item):
# adding http:// means the internal get_feed won't modify it
return "http://"+item.magnet_link
def get_feed(self, obj, request):
# hacky way to bypass the domain handling
feed = super().get_feed(obj, request)
for item in feed.items:
# strip that http:// we added above
item['link'] = item['link'][7:]
return feed
For future readers, this was as of Django 2.0.1. Hopefully in a future patch they allow support for protocols like magnet.

Django context processors and URL arguments

I have some code that is repeated at the start of my Django views. It basically just adds some variables to the context, but based on the URL argument, e.g.
def someView(request, id):
target = Target.objects.get(id=id)
# name will be added to ctx
name = target.name
(there are more attributes added and other attributes from related models, but this gives the general idea --- There are quite a few lines of repeat code at the start of each view)
I thought I could make my code more DRY by taking advantage of Django's context processors, but it would seem these don't access to the URL arguments?
Is there another way to avoid these repeat lines? Maybe middleware or something else?
You can access the URL parameters via request through the resolver_match attribute. So for instance you can do request.resolver_match.kwargs['id'] to get the ID kwarg.

Tastypie resource name clash

I am currently using tastypie with 2 apps. Each of those apps has a model called Group. They operate very differently, and the only similarity is the name 'Group'.
When only one or the other app is added to the urls file, then it works like a charm. However, as soon as I add both apps, then there's a name clash, and the get_resource_uri() method returns the wrong string. Here is some code:
urls.py
from myapp1.resources import GroupResource as gr_a
from myapp2.resources import GroupResource as gr_b
myapp1_api = Api(api_name='1.0')
myapp1_api.register(gr_a())
myapp2_api = Api(api_name='1.0')
myapp2_api.register(gr_b())
on line 37 of the current api.py file in the tastypie repo I see this code:
if resource_name is None:
raise ImproperlyConfigured("Resource %r must define a 'resource_name'." % resource)
self._registry[resource_name] = resource
Since both of my Group resources have the resource_name of 'group', they get registered on top of each other, even though they are registered at separate urls. Apart from changing the actual resource name, is there a way around this name clash?
Update
The uris would look something like this:
/myapp1/1.0/group/
/myapp2/1.0/group/
Ideally I don't want myapp1 and myapp2 to know about each other (ie the Group class is distinct). The workaround for this is to change myapp2.Group to myapp2.MyGroup (to avoid the name clash), but its really not that elegant.
In all my resources I had the resource_name blank, since I was happy with the default name. Also I wanted a url such as /myapp/1.0/group/ and not /myapp/1.0/myapp/group/
What I've now done is change all the resource_name attributes to the form "myapp/group" and bound them all to an empty url. This gave me a nice such as: /1.0/myapp/group/ while making sure there are no name clashes in the resources.

Capturing URL parameters in request.GET

I am currently defining regular expressions in order to capture parameters in a URL, as described in the tutorial. How do I access parameters from the URL as part the HttpRequest object?
My HttpRequest.GET currently returns an empty QueryDict object.
I'd like to learn how to do this without a library, so I can get to know Django better.
When a URL is like domain/search/?q=haha, you would use request.GET.get('q', '').
q is the parameter you want, and '' is the default value if q isn't found.
However, if you are instead just configuring your URLconf**, then your captures from the regex are passed to the function as arguments (or named arguments).
Such as:
(r'^user/(?P<username>\w{0,50})/$', views.profile_page,),
Then in your views.py you would have
def profile_page(request, username):
# Rest of the method
To clarify camflan's explanation, let's suppose you have
the rule url(regex=r'^user/(?P<username>\w{1,50})/$', view='views.profile_page')
an incoming request for http://domain/user/thaiyoshi/?message=Hi
The URL dispatcher rule will catch parts of the URL path (here "user/thaiyoshi/") and pass them to the view function along with the request object.
The query string (here message=Hi) is parsed and parameters are stored as a QueryDict in request.GET. No further matching or processing for HTTP GET parameters is done.
This view function would use both parts extracted from the URL path and a query parameter:
def profile_page(request, username=None):
user = User.objects.get(username=username)
message = request.GET.get('message')
As a side note, you'll find the request method (in this case "GET", and for submitted forms usually "POST") in request.method. In some cases, it's useful to check that it matches what you're expecting.
Update: When deciding whether to use the URL path or the query parameters for passing information, the following may help:
use the URL path for uniquely identifying resources, e.g. /blog/post/15/ (not /blog/posts/?id=15)
use query parameters for changing the way the resource is displayed, e.g. /blog/post/15/?show_comments=1 or /blog/posts/2008/?sort_by=date&direction=desc
to make human-friendly URLs, avoid using ID numbers and use e.g. dates, categories, and/or slugs: /blog/post/2008/09/30/django-urls/
Using GET
request.GET["id"]
Using POST
request.POST["id"]
Someone would wonder how to set path in file urls.py, such as
domain/search/?q=CA
so that we could invoke query.
The fact is that it is not necessary to set such a route in file urls.py. You need to set just the route in urls.py:
urlpatterns = [
path('domain/search/', views.CityListView.as_view()),
]
And when you input http://servername:port/domain/search/?q=CA. The query part '?q=CA' will be automatically reserved in the hash table which you can reference though
request.GET.get('q', None).
Here is an example (file views.py)
class CityListView(generics.ListAPIView):
serializer_class = CityNameSerializer
def get_queryset(self):
if self.request.method == 'GET':
queryset = City.objects.all()
state_name = self.request.GET.get('q', None)
if state_name is not None:
queryset = queryset.filter(state__name=state_name)
return queryset
In addition, when you write query string in the URL:
http://servername:port/domain/search/?q=CA
Do not wrap query string in quotes. For example,
http://servername:port/domain/search/?q="CA"
def some_view(request, *args, **kwargs):
if kwargs.get('q', None):
# Do something here ..
For situations where you only have the request object you can use request.parser_context['kwargs']['your_param']
You have two common ways to do that in case your URL looks like that:
https://domain/method/?a=x&b=y
Version 1:
If a specific key is mandatory you can use:
key_a = request.GET['a']
This will return a value of a if the key exists and an exception if not.
Version 2:
If your keys are optional:
request.GET.get('a')
You can try that without any argument and this will not crash.
So you can wrap it with try: except: and return HttpResponseBadRequest() in example.
This is a simple way to make your code less complex, without using special exceptions handling.
I would like to share a tip that may save you some time.
If you plan to use something like this in your urls.py file:
url(r'^(?P<username>\w+)/$', views.profile_page,),
Which basically means www.example.com/<username>. Be sure to place it at the end of your URL entries, because otherwise, it is prone to cause conflicts with the URL entries that follow below, i.e. accessing one of them will give you the nice error: User matching query does not exist.
I've just experienced it myself; hope it helps!
These queries are currently done in two ways. If you want to access the query parameters (GET) you can query the following:
http://myserver:port/resource/?status=1
request.query_params.get('status', None) => 1
If you want to access the parameters passed by POST, you need to access this way:
request.data.get('role', None)
Accessing the dictionary (QueryDict) with 'get()', you can set a default value. In the cases above, if 'status' or 'role' are not informed, the values ​​are None.
If you don't know the name of params and want to work with them all, you can use request.GET.keys() or dict(request.GET) functions
This is not exactly what you asked for, but this snippet is helpful for managing query_strings in templates.
If you only have access to the view object, then you can get the parameters defined in the URL path this way:
view.kwargs.get('url_param')
If you only have access to the request object, use the following:
request.resolver_match.kwargs.get('url_param')
Tested on Django 3.
views.py
from rest_framework.response import Response
def update_product(request, pk):
return Response({"pk":pk})
pk means primary_key.
urls.py
from products.views import update_product
from django.urls import path
urlpatterns = [
...,
path('update/products/<int:pk>', update_product)
]
You might as well check request.META dictionary to access many useful things like
PATH_INFO, QUERY_STRING
# for example
request.META['QUERY_STRING']
# or to avoid any exceptions provide a fallback
request.META.get('QUERY_STRING', False)
you said that it returns empty query dict
I think you need to tune your url to accept required or optional args or kwargs
Django got you all the power you need with regrex like:
url(r'^project_config/(?P<product>\w+)/$', views.foo),
more about this at django-optional-url-parameters
This is another alternate solution that can be implemented:
In the URL configuration:
urlpatterns = [path('runreport/<str:queryparams>', views.get)]
In the views:
list2 = queryparams.split("&")
url parameters may be captured by request.query_params
It seems more recommended to use request.query_params. For example,
When a URL is like domain/search/?q=haha, you would use request.query_params.get('q', None)
https://www.django-rest-framework.org/api-guide/requests/
"request.query_params is a more correctly named synonym for request.GET.
For clarity inside your code, we recommend using request.query_params instead of the Django's standard request.GET. Doing so will help keep your codebase more correct and obvious - any HTTP method type may include query parameters, not just GET requests."