我有以下XML,我想用Python解析ElementTree:
<rdf:RDF xml:base="http://dbpedia.org/ontology/"
xmlns:rdf="http://www.w3.org/1999/02/22-rdf-syntax-ns#"
xmlns:owl="http://www.w3.org/2002/07/owl#"
xmlns:xsd="http://www.w3.org/2001/XMLSchema#"
xmlns:rdfs="http://www.w3.org/2000/01/rdf-schema#"
xmlns="http://dbpedia.org/ontology/">
<owl:Class rdf:about="http://dbpedia.org/ontology/BasketballLeague">
<rdfs:label xml:lang="en">basketball league</rdfs:label>
<rdfs:comment xml:lang="en">
a group of sports teams that compete against each other
in Basketball
</rdfs:comment>
</owl:Class>
</rdf:RDF>
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我想找到所有owl:Class标签,然后提取其中所有rdfs:label实例的值.我使用以下代码:
tree = ET.parse("filename")
root = tree.getroot()
root.findall('owl:Class')
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由于命名空间,我收到以下错误.
SyntaxError: prefix 'owl' not found in prefix map
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我尝试在http://effbot.org/zone/element-namespaces.htm上阅读该文档,但由于上述XML具有多个嵌套命名空间,因此我仍然无法正常工作.
请告诉我如何更改代码以查找所有owl:Class标签.
我正在尝试编写一个与last.fm API进行交互的小脚本.
我有一些使用经验ElementTree,但我之前使用它的方式似乎不起作用,它返回一个空列表.
我删除了API密钥,因为我不确切知道它应该是多么私密,并举例说明了我收到的XML.
与API交互的类:
from xml.etree import ElementTree
import urllib
import urllib2
class Last_fmWrapper(object):
def __init__(self):
self.last_fm_api_key = '*****************************'
self.api_url = 'http://ws.audioscrobbler.com/2.0/'
def get_now_playing(self, last_fm_user, method):
self.last_fm_user = last_fm_user
self.method = method
parameters = {'user':self.last_fm_user, 'api_key':self.last_fm_api_key, 'method':self.method}
encoded_parameters = urllib.urlencode(parameters)
request = urllib2.Request(self.api_url, encoded_parameters)
response = urllib2.urlopen(request)
api_results = ElementTree.parse(response).findall('track')
# why does api_results == []?
def compare_tasteometer(self):
pass
def register_user(self):
pass
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调用get_now_playing方法Last_fmWrapper():
from last_fm_wrapper import Last_fmWrapper
last_fm = Last_fmWrapper()
now_playing = last_fm.get_now_playing('BiriBiriRG', 'user.getRecentTracks')
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