I have some links that include Persian texts, such as:
http://sample.com/fields/طب%20نظامی
And in the view function I want to access to Persian part, so:
url = request.path_info
key = re.findall('/fields/(.+)', url)[0]
But I get the following error:
IndexError at /fields/
list index out of range
Actually, the problem is with the index zero because it can not see anything there! It should be noted that it is a Django project on IIS Server and I have successfully tested it with other servers and the local server. I think it has some thing related to IIS. Moreover I have tried to slugify the url without success. I can encode urls successfully, but I think it is not the actual answer to this question.
Based on the comments:
I checked the request.path too and the same problem. It contains:
/fields/
I implemented a sample django project in local server and here is my views:
def test(request):
t = request.path
return HttpResponse(t)
The results:
http://127.0.0.1:8000/تست/
/تست/
Without any problem.
Based on the #sytech comment, I have created a middlware.py in my app directory:
from django.core.handlers.wsgi import WSGIHandler
class SimpleMiddleware(WSGIHandler):
def __call__(self, environ, start_response):
print(environ['UNENCODED_URL'])
return super().__call__(environ, start_response)
and in settings.py:
MIDDLEWARE = [
...
'apps.middleware.SimpleMiddleware',
]
But I am getting the following error:
__call__() missing 1 required positional argument: 'start_response'
Assuming you don't have another problem in your rewrite configuration, on IIS, depending on your rewrite configuration, you may need to access this through the UNENCODED_URL variable which will contain the unencoded value.
This can be demonstrated in a simple WSGI middleware:
from django.core.handlers.wsgi import WSGIHandler
class MyHandler(WSGIHandler):
def __call__(self, environ, start_response):
print(environ['UNENCODED_URL'])
return super().__call__(environ, start_response)
You would see the unencoded URL and the path part that's in Persian would be passed %D8%B7%D8%A8%2520%D9%86%D8%B8%D8%A7%D9%85%DB%8C. Which you can then decode with urllib.parse.unquote
urllib.parse.unquote('%D8%B7%D8%A8%2520%D9%86%D8%B8%D8%A7%D9%85%DB%8C')
# طب%20نظامی
If you wanted, you could use a middleware to set this as an attribute on the request object or even override the request.path_info.
You must be using URL rewrite v7.1.1980 or higher for this to work.
You could also use the UNENCODED_URL directly in the rewrite rule, but that may result in headaches with routing.
I can encode urls successfully, but I think it is not the actual answer to this question.
Yeah, that is another option, but may result in other issues like this: IIS10 URL Rewrite 2.1 double encoding issue
You can do this by using python split() method
url = "http://sample.com/fields/طب%20نظامی"
url_key = url.split(sep="/", maxsplit=4)
url_key[-1]
output : 'طب%20نظامی'
in this url is splited by / which occurs 4 time in string so it will return a list like this
['http:', '', 'sample.com', 'fields', 'طب%20نظامی']
then extract result like this url_key[-1] from url_key
you can Split the URL by :
string = http://sample.com/fields/طب%20نظامی
last_part = string. Split("/")[-1]
print(last_part)
output :< طب%20نظامی >
slugify(last_part)
or
slugify(last_part, allow_unicode=True)
I guess This Will Help You :)
My website generates json files in saves them with the user specified name: test.json
I couldn't immediately access it, so I assumed that I would have to write a URL pattern and view to see this file.
I want something akin to this :
url(r'^(?P<Controller>).json$', views.loadjson, name='loadjson')
If you need to serve static JSON files then doing so through Django is not the best way - if that's the case use something like Nginx.
If you wan't Django to generate those files and/or for e.g. an authentication mechanism in front of it then ok for Django.
To your question:
# urls.py
url(r'^(?P<json_file>[\w]+).json$', views.loadjson, name='loadjson')
It seems like you forgotten about [\w]+ which I guess is the right pattern for your needs.
the view function will be called with a WSGIRequest instance and the json_file argument (from your url pattern), from there you should be able to do whatever you like with the file. It's a good idea to return in as an application/json content type as those are supposed to be JSON.
# views.py
from django.http import HttpResponse
def loadjson(request, json_file):
# open, generate, fetch the json file
# for e.g.:
json_content = read_file(json_file)
return HttpResponse(
json_content,
content_type='application/json',
status=200
)
I have a django project that needs to search 2 different models and one of the models has 3 types that I need to filter based on. I have haystack installed and working in a basic sense (using the default url conf and SearchView for my model and the template from the getting started documentation is returning results fine).
The problem is that I'm only able to get results by using the search form in the basic search.html template and I'm trying to make a global search bar work with haystack but I can't seem to get it right and I'm not having a lot of luck with the haystack documentation. I found another question on here that led me to the following method in my search app.
my urls.py directs "/search" to this view in my search.views:
def search_posts(request):
post_type = str(request.GET.get('type')).lower()
sqs = SearchQuerySet().all().filter(type=post_type)
view = search_view_factory(
view_class=SearchView,
template='search/search.html',
searchqueryset=sqs,
form_class=HighlightedSearchForm
)
return view(request)
The url string that comes in looks something like:
http://example.com/search/?q=test&type=blog
This will get the query string from my global search bar but returns no results, however if I remove the .filter(type=post_type) part from the sqs line I will get search results again (albeit not filtered by post type). Any ideas? I think I'm missing something fairly obvious but I can't seem to figure this out.
Thanks,
-Sean
EDIT:
It turns out that I am just an idiot. The reason why my filtering on the SQS by type was returning no results was because I didn't have the type field included in my PostIndex class. I changed my PostIndex to:
class PostIndex(indexes.SearchIndex, indexes.Indexable):
...
type = indexes.CharField(model_attr='type')
and rebuilt and it all works now.
Thanks for the response though!
def search_posts(request):
post_type = str(request.GET.get('type')).lower()
sqs = SearchQuerySet().filter(type=post_type)
clean_query = sqs.query.clean(post_type)
result = sqs.filter(content=clean_query)
view = search_view_factory(
view_class=SearchView,
template='search/search.html',
searchqueryset=result,
form_class=HighlightedSearchForm
)
return view(request)
I need to get django to send an email which contains a URL like this
http://www.mysite.org/history/
Where 'history' is obtained like so:
history_url = urlresolvers.reverse('satchmo_order_history')
history_url is a parameter that I pass on to the function that sends the email, and it correctly produces '/history/'. But how do I get the first part? (http://www.mysite.org)
Edit 1
Is there anything wrong or unportable about doing it like this? :
history = urlresolvers.reverse('satchmo_order_history')
domain = Site.objects.get_current().domain
history_url = 'http://' + domain + history
If you have access to an HttpRequest instance, you can use HttpRequest.build_absolute_uri(location):
absolute_uri = request.build_absolute_uri(relative_uri)
In alternative, you can get it using the sites framework:
import urlparse
from django.contrib.sites.models import Site
domain = Site.objects.get_current().domain
absolute_uri = urlparse.urljoin('http://{}'.format(domain), relative_uri)
Re: Edit1
I tend to use urlparse.join, because it's in the standard library and it's technically the most Pythonic way to combine URIs, but I think that your approach is fine too.
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."