Django HTTP Request to curl command (for replay)

req_to_curl.py

"""
Convert a Django HTTPRequest object (or dictionary of such a request) into a
cURL command.
"""
def get_curl(request):
    def convert(word, delim='-'):
        return delim.join(x.capitalize() or '_' for x in word.split('_'))

    def get_headers(request):
        headers = {
            convert(name[5:]): value
            for name, value in request.META.items()
            if name.startswith('HTTP_')
        }
        return headers
    headers = get_headers(request)
    url = '{server}{path}?{query}'.format(
        server=request.META['SERVER_NAME'],
        path=request.META['PATH_INFO'],
        query=request.META['QUERY_STRING'],
    )
    return 'curl {headers} "{url}"'.format(
        headers=' '.join(('-H "%s: %s"' % (h, v)) for h, v in headers.items()),
        url=url)
        
"""
Laurent Lyaudet : I simplified the original code of asfaltboy to make it work in my use case.
(I had request.META directly available in DRF.)
Copy the function above where needed and add something like that in your view.
logger.info(f"AAAAAAAAAAAAAA {get_curl(request)}")
If it does not fit your need,
you may open an issue here and I'll try to help.
But you may find a faster answer back in asfaltboy original code.
"""