I want to pass some string for some urls to my views in django
Suppose i have
path('someurl/', someview , name='someurl'),
I want to pass some string to someview, when this url is called so is this possible
path('someurl/', someview(somevar="test") , name='someurl'),
and then i have the view
def someview(request, somevar):
access somevar here
Is this possible in Django urls.
If you wish to accept parameter from the client, update your path as below:
path('someurl/<str:somevar>/', someview , name='someurl')
And view now can accept extra parameter:
def someview(request, somevar):
# now you can use somevar
With this definition, if client requests somevar/urlparam/, "urlparam" will be passed to you view function.
Otherwise if you want to provide your own argument, Django doesn't provide the way to do it directly in url definition. But, since that variable is your own one, why don't assign (or compute) that in view? I mean:
def someview(request):
somevar = "test" # or you may call some function for dynamic assignment
# now somevar exists in this scope, so you can use it as you want
yes it is possible. You need to define those parameters in the url as pseudo path:
path('articles/<int:year>/<int:month>/', views.month_archive),
There is also an option to use request.GET and request.POST to access the optional parameters list in the url:
request.POST.get('<par name here>','<default value here>')
request.GET.get('<par name here>','<default value here>')
Another thing you may find useful is this question.
I have a function based view that I only use to update a session:
def admin_privileges(request):
// toggle request.session['is_admin']
return
Typically I use this to toggle a session variable between True and False.
E.g. in a template:
Toggle admin privileges
How can I pass a variable and amend return in admin_privileges to return the user to the original view it was requested from?
I don't want to use anything on the front end, and I can't use HTTP_REFERRER as it's not always set.
I thought of passing something via the URL from the referring view?
There are 2 ways:
You can use request.META.get('HTTP_REFERER') in your view which will return previous page.
or (more safely, check this link to learn why)
You can pass the link (request.path for current page) as a parameter to admin_privileges link so:
Toggle admin privileges
In the view, you can get that parameter using request.GET and redirect using Django's redirect utilities.
NoReverseMatch at /natrium/script/4c55be7f74312bfd435e4f672e83f44374a046a6aa08729aad6b0b1ab84a8274/
Reverse for 'run_details' with arguments '()' and keyword arguments '{'script_text': u'print "happy"', 'run_id': '6b2f9127071968c099673254fb3efbaf'}' not found.
This is an excerpt of my views.py
run_id = new_run.run_id
if not run_id:
raise AssertionError("bad run id")
# I tried with args=[run_id, clean['script_text']] too
return HttpResponseRedirect(reverse('run_details', kwargs={'run_id':run_id, 'script_text':clean['script_text']}))
which in turns calling this view function
def run_details(request, run_id, script_text):
"""
Displays the details of a given run.
"""
run = Run(run_id)
run.update(request.user)
codebundle = CodeBundle(run.cbid)
codebundle.update(request.user)
return render_response(request, "graphyte/runs/run_script.html",
{'run':run, 'codebundle':codebundle, 'files':run.artifacts, 'bundle':codebundle,
'source_code': script_text
})
Now this is my urls.py. The actual redirect views is in another app (kinda insane, but whatever...).
urlpatterns = patterns("webclient.apps.codebundles.views",
# many.....
url(r"^cb/newfolder/$", 'codebundle_newfolder', name="codebundle_newfolder"),
)
urlpatterns += patterns('webclient.apps.runs.views',
url(r"^run_details/(?P<run_id>\w+)/$", 'run_details', name="run_details"),)
This is getting really nasty for the last three hours. I am not sure what's going on. Can someone help me debug this?
Thanks.
The original plan did not have script_text, and I used args=['run_id'] only. It works. In other words, remove script_text from the two views everything will work.
EDIT
Sorry for the confusion. Script text is just a context variable that I need to pass to the reverse destination, and from there I render my template. The URLs should only display the run_id.
No, you can't really pass an 'extra keyword' to the view function when redirecting. I'll try to explain why.
When you return HttpResponseRedirect, Django returns a response with a 302 status code, and the new location.
HTTP/1.1 302 Found
Location: http://www.example.com/new-url/
Your browser will then usually fetch the new url, but that's a separate request. If your view needs a keyword, it needs to be included in that response somehow, unless you store state in the session. Your two options are
Include the extra keyword in the url:
http://www.example.com/new-url/keyword-value/
Include the extra keyword as a GET parameter
http://www.example.com/new-url/?keyword=keyword-value.
Then in your view, grab the keyword with keyword=request.GET['keyword']. Note that the keyword is no longer a kwarg in the view signature.
A third approach is to stick the keyword into the session before you redirect, then grab it out the session in the redirected view. I would advise against doing this because it's stateful and can cause odd results when users refresh pages etc.
Your run_details url doesn't accept a kwarg named script_text at all -- remove that from your reverse kwargs.
How can I redirect from view to another url, passing my queryset to another view?
I tried this:
return simple.redirect_to(request, 'some_url', **{'queryset': results})
and this
return redirect('some_url', queryset=results )
but it does not work....
How can i do it?
Gabi.
How are you expecting this to work? Redirection happens by getting the browser to request another URL. Anything you want to pass as a parameter to the redirection must therefore go into the URL you're redirecting to. It simply doesn't make sense to put a queryset into a URL parameter.
Presumably you could pass whatever arguments you used to get the queryset in the first place, but that's a lot of extra work.
Do you really need to redirect at all? What about simply calling the new view from your original one, and returning its response?
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."