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Editorial Team
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Asked: 2026-05-13T05:28:45+00:00

I’m working with an API which I post files to. However, when I receive

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I’m working with an API which I post files to. However, when I receive the response, the HTTP status code is a 202. This is to be expected, but in addition the API will also respond with XML content.

So in my try/except block urllib2.urlopen will result in a raised urllib2.HTTPError and destroying the XML content.

try:
    response = urllib2.urlopen(req)
except urllib2.HTTPError, http_e:
    if http_e.code == 202:
        print 'accepted!'
        pass

print response.read() # UnboundLocalError: local variable 'response' referenced before assignment

How can I expect the 202 and keep the response content, but not raise an error?

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1 Answer

  1. Editorial Team

    Edit

    Being silly, I forgot to inspect the exception that is returned by urllib2. It features all of the properties I’ve been waxing on about for httplib. This should do the trick for you:

    try:
    urllib2.urlopen(req)
except urllib2.HTTPError, e:
    print "Response code",e.code # prints 404
    print "Response body",e.read() # prints the body of the response...
                                   # ie: your XML
    print "Headers",e.headers.headers

    Original

    In this case, given that you’re using HTTP as your transport protocol, you’ll probably have more luck with the httplib library:

    >>> import httplib
>>> conn = httplib.HTTPConnection("www.stackoverflow.com")
>>> conn.request("GET", "/dlkfjadslkfjdslkfjd.html")
>>> r = conn.getresponse()
>>> r.status
301
>>> r.reason
'Moved Permanently'
>>> r.read()
'<head><title>Document Moved</title></head>\n<body><h1>Object Moved</h1>
 This document may be found   
 <a HREF="http://stackoverflow.com/dlkfjadslkfjdslkfjd.html">here</a></body>'

    You can further use r.getheaders() and so forth to inspect other aspects of the response.

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