Capturing url parameters in request.GET

Question

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.

Answer 1

When url is like: domain/search/?q=haha, Then 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

Answer 2

This is not exactly what you asked for, but this snippet is helpful for managing query_strings in templates.

Answer 3

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')
• a in 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/

Answer 4

You have two common ways to do that in case your url looks like that:

https://domain/method/?a=x&b=y

v1:

If specific key is mandatory you can use:

key_a = request.GET['a']

This will return a value of a if key exists and an Exception if not.

v2:

If your keys are optional:

request.GET.get('a')

You can try that without any argument 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.

Answer 5

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!

Answer 6

def some_view(request, *args, **kwargs):
    if kwargs.get('q', None):
        # Do something here ..