HTTPError... read the response body?
Diez B. Roggisch
deets at nospam.web.de
Tue Mar 3 11:22:33 CET 2009
Stuart Davenport schrieb:
> On Mar 2, 11:50 pm, Wojtek Walczak <gmin... at bzt.bzt> wrote:
>> On Mon, 2 Mar 2009 14:29:12 -0800 (PST), Stuart Davenport wrote:
>>> I am trying to connect to a web service but I am getting HTTP 400, I
>>> am not too concerned about the HTTP error - but what I'd like to know
>>> if there is anyway I can read the response body in the HTTP 400 or 500
>>> case? Does the HTTPError allow this? or the urllib2 in anyway?
>> HTTP error 400 means 'bad request', so it's almost certainly
>> your mistake.
>>> #url, body, headers
>>> rq = urllib2.Request(args, args, headers)
>> Is args a valid URL? And are you sure that you want to send
>> something to the web serwer? By specifying the second argument
>> (args) you're asking python to send HTTP "POST" request,
>> not "GET".
>> Wojtek Walczak,http://tosh.pl/gminick/
> Hi Wojtek,
> Yes, the args is a valid url (I've printed it out to check) and
> thats correct, I do want to initiate a POST request by passing over
> the SOAP Envelope in the request. All I'd like to know if there is
> anyway to read the body of an HTTP Error, so for instance, if I were
> to recieve a 404 - I'd like to see the content that the server returns
> with a 404 error.
HTTPError.read() works for me.
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