Determine the requested content type?

Question

I'd like to write a Django view which serves out variant content based on what's requested. For example, for "text/xml", serve XML, for "text/json", serve JSON, etc. Is there a way to determine this from a request object? Something like this would be awesome:

def process(request):
    if request.type == "text/xml":
        pass
    elif request.type == "text/json":
        pass
    else:
        pass

Is there a property on HttpRequest for this?

Answer 1

'Content-Type' header indicates media type send in the HTTP request. This is used for requests that have a content (POST, PUT).

'Content-Type' should not be used to indicate preferred response format, 'Accept' header serves this purpose. To access it in Django use: HttpRequest.META.get('HTTP_ACCEPT')

See more detailed description of these headers

Answer 2

As said in other answers, this information is located in the Accept request header. Available in the request as HttpRequest.META['HTTP_ACCEPT'].

However there is no only one requested content type, and this header often is a list of accepted/preferred content types. This list might be a bit annoying to exploit properly. Here is a function that does the job:

import re

def get_accepted_content_types(request):
    def qualify(x):
        parts = x.split(';', 1)
        if len(parts) == 2:
            match = re.match(r'(^|;)q=(0(\.\d{,3})?|1(\.0{,3})?)(;|$)',
                             parts[1])
            if match:
                return parts[0], float(match.group(2))
        return parts[0], 1

    raw_content_types = request.META.get('HTTP_ACCEPT', '*/*').split(',')
    qualified_content_types = map(qualify, raw_content_types)
    return (x[0] for x in sorted(qualified_content_types,
                                 key=lambda x: x[1], reverse=True))

For instance, if request.META['HTTP_ACCEPT'] is equal to "text/html;q=0.9,application/xhtml+xml,application/xml;q=0.8,*/*;q=0.7". This will return: ['application/xhtml+xml', 'text/html', 'application/xml', '*/*'] (not actually, since it returns a generator).

Then you can iterate over the resulting list to select the first content type you know how to respond properly.

Note that this function should work for most cases but do not handle cases such as q=0 which means "Not acceptable".

Sources: HTTP Accept header specification and Quality Values specification

Answer 3

in django 1.10, you can now use, request.content_type, as mentioned here in their doc