parsing an xml URL? [question]

[Madsvj]

Is there anyway supported way for parsing through an url instead of a local file?

So something like this:

url = "http://regnskaber.virk.dk/17438023/eGJybHN0b3JlOi8vWC1DOUZFREM2OC0yMDE0MDUyMV8xMzU2MDdfMTA2L3hicmw.xml"

xbrl_parser = XBRLParser(precision=0)

xbrl = xbrl_parser.parse(file(url))

(probably with out the file())

[GBULLR]

was this ever made possible?

[greedo]

Functionality for this has not yet been added