I am passing a variable in my URL:
mydomain.com/app/?next_page=my_page
I can get this variable in a view with:
def mypage(request):
var = request.GET['next_page']
Is it best practice to also change the URL to require next_page? Something along the lines of:
path('app/?nextpage=<str>', mypage, name='my_page')
What's best practice? If so, what's the correct syntax for this (I know the example is incorrect)?
It depends on your needs.
Do not define a fixed url route; if you use the query parameters for filtering and there is more than one possible parameter
Example: "app/photos?size=100x100" and "app/photos/?color=blue"
Define a fixed url route; if it will be the same for each and every page, like details of a particular page:
Example: "app/orders/123123123" and "app/orders/123123123"
Btw, the correct syntax is:
path(app/<str:next_page>/, mypage, name="my_page")
You should take a look at path patterns. Enforcing a GET parameter in a path is not really a good practice. So if you want to require a username for example you can use:
path('bio/<username>/', views.bio, name='bio'),
You can find more patterns in Django documentation to catch strings, slugs, integers etc.
And in views you should define your function as such:
def mypage(request, username):
...code...
About GET:
Keep in mind that request.GET["value"] will raise a ValueError if that parameter does not exist. So you can catch that error to inform user that they are missing a parameter. (This will make this parameter obligatory.)
You can also use request.GET.get("value") which will return None if the key does not exist. If you want to use a default parameter you can use of course, request.GET.get("value", "default")
You can use as many parameters as you want in your link with or without path patterns. Their values will be stored in request.GET
Related
I am trying to pass the first part of a django url to a view, so I can filter my results by the term in the url.
Looking at the documentation, it seems quite straightforward.
However, I have the following urls.py
url('<colcat>/collection/(?P<name>[\w\-]+)$', views.collection_detail, name='collection_detail'),
url('<colcat>/', views.collection_view, name='collection_view'),
In this case, I want to be able to go to /living and have living be passed to my view so that I can use it to filter by.
When trying this however, no matter what url I put it isn't being matched, and I get an error saying the address I put in could not be matched to any urls.
What am I missing?
<colcat> is not a valid regex. You need to use the same format as you have for name.
url('(?P<colcat>[\w\-]+)/collection/(?P<name>[\w\-]+)$', views.collection_detail, name='collection_detail'),
url('(?P<colcat>[\w\-]+)/$', views.collection_view, name='collection_view'),
Alternatively, use the new path form which will be much simpler:
path('<str:colcat>/collection/<str:name>', views.collection_detail, name='collection_detail'),
path('<str:colcat>/', views.collection_view, name='collection_view'),
Say you have a url like this:
/cats/?filter=kittens
Is it possible to build a django url pattern that forces the use of the querystring?
Currently I have:
url(r'^/cats/$', views.CatsListView.as_view(), name='cats')
Now I want to add the querystring and get a different view, something like this:
url(r'^/cats/?filter=(?P<filter>.+?)$', views.CatsFilteredListView.as_view(), name='cats-filtered')
Is it possible to do something like this and still keep the querystring in the GET parameter of the request?
Remember that this is just a testcase, I, and you should too, know that filtering like probably this isn't the way to go..
Short answer: no, it's not possible. Django url patterns match only on the "path" componant of the url, period.
No, it's not possible to do this. If you really need two separate views, you can write a view that dispatches the correct view.
def cat_list_view(request, *args, **kwargs):
if 'filter' in request.GET:
return cat_list_filter_view(request, *args, **kwargs)
else:
return cat_list_unfiltered_view(request, *args, **kwargs)
However, for your example of CatsListView and CatsFilteredListView there is probably a better way to combine the views. For example you might be able to do the filtering in the get_queryset method.
While it is true that you cannot manipulate the path component like that, you can pass a dictionary in. It's a 3rd unnamed argument.
This approach can be useful if you want to use the same view for multiple resources, and pass data to configure its behaviour in each case (below we supply a different template in each case).
path('url/', views.my_reused_view, {'my_template_name': 'some_path'}, name='aurl'),
path('anotherurl/', views.my_reused_view, {'my_template_name': 'another_path'}, name='anotherurl'),
Note: Both extra options and named captured patterns are passed to the view as named arguments. If you use the same name for both a captured pattern and an extra option then only the captured pattern value will be sent to the view (the value specified in the additional option will be dropped).
Courtesy of https://developer.mozilla.org/en-US/docs/Learn/Server-side/Django/Generic_views
I'm just learning Django, and am getting stuck with some url logic. I'm trying to allow either a category name or id in the url:
...
url(r'^(?P<booze_q>\w+|\d+)/$','glasses.views.booze'),
...
And then in thew view, only deal with that result once. However, if the url is a string - in this case, Whiskey, I get an error for trying to pass a string where an int is expected. This is the closest I've gotten so far:
def booze(request, booze_q):
booze = get_object_or_404(Booze,Q(pk=booze_q)|Q(name=booze_q))
return render_to_response('booze/detail.html', {'booze': booze})
But this returns an error: invalid literal for int() with base 10: 'Whiskey'
I'm sure it's a pretty easy thing, but this is my first Django app, so any help would be appreciated.
tl;dr: End result, I'd like mysite.com/1/ or mysite.com/Whiskey/ to both call the glasses.views.booze view, and get the object with id=1 or name=Whiskey
This is a common scenario you'll encounter quite often, which is typically handled by resorting to multiple arguments and having views behave differently based on which of the view arguments are then present or not.
What you do is first define a URL pattern that uniquely matches each specific case and then let Django's URL resolver set the arguments accordingly based on which of the patterns was matched.
Here's an example with a class based view, that performs two different queries based on which of the two keyword arguments, booze_id or booze_name, is set:
url(r'^(?P<booze_id>\d+)/$', BoozeDetailView.as_view()),
url(r'^(?P<booze_name>\w+)/$', BoozeDetailView.as_view()),
class BoozeDetailView(DetailView):
model = Booze
def get_object(self):
booze_id = self.kwargs.get('booze_id', None)
booze_name = self.kwargs.get('booze_name', None)
if booze_id:
return self.model.objects.get(id=booze_id)
else:
return self.model.objects.get(name=booze_name)
You will always get a string, even if the string contains a number.
1) You should not have a parameter that could be either an id or something else. One day you will enter an item whose name is a number and your app will fail.
2) When querying for pk with a string django automatically tries to convert it into an integer. You'll have to handle the non-pk case before constructing that query.
In django, when a URL is matched, the match group is passed as a second parameter to the view function, the first being a HttpRequest object. For example, with a URL patter like this
'/foo/(\d{2})/', 'app.views.handler'
the handler routine will have
def handler(request, value):
where value will contain a two digit number (as a string).
My question is: is value also contained in the request object, and if yes, how can I get it (of course, parsing the URL from the request object is not an option, too impractical).
Thanks
I'm not going to debate the merit of your idea. Just try to answer your question:
No there is no way, other than applying the regex to the URL again, to get at the url parameter.
Your view will be the first point where the parameter list will be available. Why don't you just create a wrapper object to encapsulate your request and your parameter list at that point?
Just pass that around...
Can you give any reason why you would need this?
I don't see why parsing the url path is 'impractical', given that you've already got a regexp that works, in your urlconf.
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."