How can I pass parameters via URL as query parameters to avoid multiple and complicated url patterns?
For example, instead of making a complicated url like
example.com/page/12/red/dog/japan/spot
or something like that, and then a corresponding entry in urls.py that will parse that url and direct it to a view, I want to simply get a url where I can freely add or remove parameters as needed similar to the ugly way
example.com/page?id=12&color=red&animal=dog&country=Japan&name=spot
Then in urls.py simply have something like
path('page/<parameter_dictionary>', views.page, name='page' parameters='parameter_dictionary)
If I have to use url patterns, how can I account for urls that have parameters that may or may not fit the pattern, such as sometimes
"/page/12/red/dog/Japan/spot" -> path('page/<int:id>/<str:color>/<str:animal>/<str:country>/<str:name>', views.page, name='page'),
"/page/12/dog/red/Japan/"-> path('page/<int:id>/<str:animal>/<str:color>/<str:country>', views.page, name='page')
"/page/dog/red/Japan/"-> path('page/<str:animal>/<str:color>/<str:country>', views.page, name='page')
I would like to just have anything sent to http://example.com/page/
go to views.page(), and then be accessible by something like
animal = request.GET['animal']
color = request.GET['color']
id = request.GET['id']
etc. so examples below would all work via one entry in urls.py
example.com/page?id=12&animal=dog&country=Japan&name=spot
example.com/page?id=12&color=red&animal=dog&name=spot
example.com/page?id=12&country=Japan&color=red&animal=dog&name=spot
You are looking for queryparameters and you are almost done with it. The following code is untested but should kinda work:
def page(request):
animal = request.GET.get("animal",None) # default None if not present
color = request.GET.get("color",None)
return render(request,'some_html.html')
# urls.py:
path('page/', views.page, name='page')
You access the queryparameters via the passed request object request.GET. This is a dict like object. The main difference is that this object handles multi keys.
For example if you pass the these params ?a=1&a=2 to your url, it converts request.GET.getlist("a") # Returns ["1","2"] to a list.
request.GET.get("a") returns the last passed value "2" as #Kbeen mentioned in comments,. Read more about QueryDict here.
Also be sure to know the difference and best practice for url parameters and queryparameters. Example Stackoverflow post
Edit: Added request.GET.getlist()
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."
I am working on a project where I'm asking the user to enter a Company name in a form and before submitting the form I have button that redirects the user to the following URL and passes in the Company name that the user entered in the form:
http://.../sync/CompanyName
I'm trying to now call the sync() function and pass in the above Company name as a parameter.
This is the urls.py:
path('sync/<str:company>', views.sync, name='sync')
I am able to redirect it to the sync() view since I have some test lines in there that are being printed.
This is the views.py:
def sync(request,company):
print(company)
.
.
.
The issue is that the company name being printed above comes back as undefined. Am I missing something here?
How can I to do this without doing a POST request? I have also tried to get the company name using something like this: print(request.GET.get(company)) when my URL was http://.../sync/company=CompanyName but that didn't work either - still came back as undefined.
I think it should work just fine, but add a slash at the end of the url(as per Django's design philosophies):
path('sync/<str:company>/', views.sync, name='sync')
^
Also, if you are getting undefined, then it is working fine. Because undefined is probably coming from your JavaScript code, not Python. If it were not working, then it would have thrown a error or you would have gotten None if you had passed company=None as keyword argument in the sync view function.
If you want to use http://.../sync/?company=CompanyName(ie using URL querystring), then you can use request.GET to get the company name. For example,
# url
path('sync/', views.sync, name='sync')
#view
def sync(request):
print(request.GET.get('company', None))
FYI, GET methods should be idempotent, meaning it should not make any DB changes. So, consider using POST method if you intend to do changes in DB.
how do i redirect a registered user to his/her db.table.id 'view' without going through smartgrid in web2py?
i have tried using:
redirect(URL(f='first', args=['mydata/view', 'mydata/%s', %request.vars.name]))
where mydata is the view for my table db.mydata and 'first' is my function.
It always returns to the smartgrid interface.
There are two problems. First, the final URL argument must be the record ID, but it looks like you are instead using a name (i.e., request.vars.name). Second, by default, the grid uses signed URLs, so you must either disable the signatures (not recommended) or add a user signature to the URL you generate. So, the link should be something like this:
redirect(URL(f='first', args=['mydata', 'view', 'mydata', request.vars.id],
user_signature=True))
Also, note that in the args list, each element can (and generally should) be a separate URL arg. So, instead of ['mydata/view', ...], it should be ['mydata', 'view', ...].
Code:
class Telegram(tornado.web.RequestHandler):
def my_f(self,number):
return number
def get(self,number):
self.write( self.my_f(number))
application = tornado.web.Application([
(r"/number/(.*?)", Telegram),
])
Using this piece of code, i can trigger Telegram, providing it with something from the (.*?) part.
Question is: i need to make POST queries like:
/number/messenger=telegram&phone=3332223332211
so that I can grab messenger parameter and phone parameter, and trigger the right class with provided phone number (like Telegram with 3332223332211)
POST requests (usually) have a body, so if you want everything in the URL you probably want a GET instead of a POST.
The normal way to pass arguments is by form-encoding them. That starts with a ? and looks like this: /number?messenger=telegram&phone=12345. To use arguments like this in Tornado, you use self.get_argument("messenger") instead of an argument to the get() method.
A second way of passing parameters is to put them in the "path" part of the URL, without a question mark. This is when you use (.*?) in your routing pattern and an argument to get(). Use this when you want to avoid the question mark for some reason (usually aesthetics).
You can also combine the two: pass the messenger parameter in the URL as you've done here, and add ?number=12345 and use get_argument. But unless you really care about what your URLs look like, I recommend the first form.