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. I

Answer 1

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