Hi Everyone,
Thanks for all your suggestions. I read up on gzip and urllib and also
learned in the process that I could use urllib2 as its the latest form of
that library.
Herewith my solution: I don't know how elegant it is, but it works just
fine.
def get_contests():
url = '
http://xml.matchbook.com/xmlfeed/feed?sport-id=&vendor=TEST&sport-name=&short-name=Po
'
req = urllib2.Request(url)
req.add_header('accept-encoding','gzip/deflate')
opener = urllib2.build_opener()
response = opener.open(req)
compressed_data = response.read()
compressed_stream = StringIO.StringIO(compressed_data)
gzipper = gzip.GzipFile(fileobj=compressed_stream)
data = gzipper.read()
current_path = os.path.realpath(MEDIA_ROOT + '/xml-files/d.xml')
data_file = open(current_path, 'w')
data_file.write(data)
data_file.close()
xml_data = ET.parse(open(current_path, 'r'))
contest_list = []
for contest_parent_node in xml_data.getiterator('contest'):
contest = Contest()
for contest_child_node in contest_parent_node:
if (contest_child_node.tag == "name" and
contest_child_node.text is not None and contest_child_node.text != ""):
contest.name = contest_child_node.text
if (contest_child_node.tag == "league" and
contest_child_node.text is not None and contest_child_node.text != ""):
contest.league = contest_child_node.text
if (contest_child_node.tag == "acro" and
contest_child_node.text is not None and contest_child_node.text != ""):
contest.acro = contest_child_node.text
if (contest_child_node.tag == "time" and
contest_child_node.text is not None and contest_child_node.text != ""):
contest.time = contest_child_node.text
if (contest_child_node.tag == "home" and
contest_child_node.text is not None and contest_child_node.text != ""):
contest.home = contest_child_node.text
if (contest_child_node.tag == "away" and
contest_child_node.text is not None and contest_child_node.text != ""):
contest.away = contest_child_node.text
contest_list.append(contest)
try:
os.remove(current_path)
except:
pass
return contest_list
Many thanks!
On Tue, May 24, 2011 at 12:35 PM, Stefan Behnel <stefan...@behnel.de> wrote:
> Sithembewena Lloyd Dube, 24.05.2011 11:59:
>
> I am trying to parse an XML feed and display the text of each child node
>> without any success. My code in the python shell is as follows:
>>
>> >>> import urllib
>> >>> from xml.etree import ElementTree as ET
>>
>> >>> content = urllib.urlopen('
>>
>> http://xml.matchbook.com/xmlfeed/feed?sport-id=&vendor=TEST&sport-name=&short-name=Po
>> ')
>> >>> xml_content = ET.parse(content)
>>
>> I then check the xml_content object as follows:
>>
>> >>> xml_content
>> <xml.etree.ElementTree.ElementTree instance at 0x01DC14B8>
>>
>
> Well, yes, it does return an XML document, but not what you expect:
>
> >>> urllib.urlopen('URL see above').read()
> "<response>\r\n <error-message>you must add 'accept-encoding' as
> 'gzip,deflate' to the header of your request</error-message>\r
> \n</response>"
>
> Meaning, the server forces you to pass an HTTP header to the request in
> order to receive gzip compressed data. Once you have that, you must
> decompress it before passing it into ElementTree's parser. See the
> documentation on the gzip and urllib modules in the standard library.
>
> Stefan
>
>
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--
Regards,
Sithembewena Lloyd Dube
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