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python - sparql.query().convert() 中 SPARQLWrapper 的 URLError

转载 作者:太空宇宙 更新时间:2023-11-04 01:05:18 29 4
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我尝试使用一个小的 Python 脚本来测试我的 SPARQL 请求。然而,只是下一个简单的代码是行不通的。

from SPARQLWrapper import SPARQLWrapper, JSON
import rdflib
#connect to the sparql point
sparql = SPARQLWrapper("http://localhost:3030/sparql")
#SPARQL request
sparql.setQuery("""
PREFIX rdf: <http://www.w3.org/1999/02/22-rdf-syntax-ns#>
PREFIX rdfs: <http://www.w3.org/2000/01/rdf-schema#>
PREFIX owl: <http://www.w3.org/2002/07/owl#>
PREFIX xsd: <http://www.w3.org/2001/XMLSchema#>
PREFIX rme: <http://www.semanticweb.org/reminer/>

SELECT ?o
WHERE { ?s ?p ?o }
LIMIT 1
""")
sparql.setReturnFormat(JSON)
results = sparql.query().convert()

for result in results["results"]["bindings"]:
print(result["o"]["value"])

我的代码在转换步骤卡住了很长时间,然后给我一个 URLError。当我停止脚本时,看到下一条消息:

HTTPError                                 Traceback (most recent call last)
<ipython-input-6-2ab63307a418> in <module>()
18 """)
19 sparql.setReturnFormat(JSON)
---> 20 results = sparql.query().convert()
21
22 for result in results["results"]["bindings"]:

/Users/francocy/anaconda/lib/python3.4/site-packages/SPARQLWrapper/Wrapper.py in query(self)
533 @rtype: L{QueryResult} instance
534 """
--> 535 return QueryResult(self._query())
536
537 def queryAndConvert(self):

/Users/francocy/anaconda/lib/python3.4/site-packages/SPARQLWrapper/Wrapper.py in _query(self)
513 raise EndPointInternalError(e.read())
514 else:
--> 515 raise e
516
517 def query(self):

/Users/francocy/anaconda/lib/python3.4/site-packages/SPARQLWrapper/Wrapper.py in _query(self)
503
504 try:
--> 505 response = urlopener(request)
506 return response, self.returnFormat
507 except urllib.error.HTTPError as e:

/Users/francocy/anaconda/lib/python3.4/urllib/request.py in urlopen(url, data, timeout, cafile, capath, cadefault, context)
159 else:
160 opener = _opener
--> 161 return opener.open(url, data, timeout)
162
163 def install_opener(opener):

/Users/francocy/anaconda/lib/python3.4/urllib/request.py in open(self, fullurl, data, timeout)
467 for processor in self.process_response.get(protocol, []):
468 meth = getattr(processor, meth_name)
--> 469 response = meth(req, response)
470
471 return response

/Users/francocy/anaconda/lib/python3.4/urllib/request.py in http_response(self, request, response)
577 if not (200 <= code < 300):
578 response = self.parent.error(
--> 579 'http', request, response, code, msg, hdrs)
580
581 return response

/Users/francocy/anaconda/lib/python3.4/urllib/request.py in error(self, proto, *args)
505 if http_err:
506 args = (dict, 'default', 'http_error_default') + orig_args
--> 507 return self._call_chain(*args)
508
509 # XXX probably also want an abstract factory that knows when it makes

/Users/francocy/anaconda/lib/python3.4/urllib/request.py in _call_chain(self, chain, kind, meth_name, *args)
439 for handler in handlers:
440 func = getattr(handler, meth_name)
--> 441 result = func(*args)
442 if result is not None:
443 return result

/Users/francocy/anaconda/lib/python3.4/urllib/request.py in http_error_default(self, req, fp, code, msg, hdrs)
585 class HTTPDefaultErrorHandler(BaseHandler):
586 def http_error_default(self, req, fp, code, msg, hdrs):
--> 587 raise HTTPError(req.full_url, code, msg, hdrs, fp)
588
589 class HTTPRedirectHandler(BaseHandler):

HTTPError: HTTP Error 403: Forbidden

我在 2.7 和 3.4 中遇到了同样的情况。编辑:我将连接从 Wifi 更改为 Intranet。我的脚本适用于 DBpedia Sparql 端点,但是当我在本地服务器上请求时出现 Http 错误。这似乎是代理或访问我本地服务器的问题。

预先感谢您的帮助。

最佳答案

对于 2019 年来到这里的用户,对于访问维基数据 sparql 端点的错误,维基数据执行严格的用户代理策略,请参阅此存档(感谢 Pere)Wikidata Project chat ,它表示应用程序发送信息性 header 表示行为良好的非机器人脚本,另请参阅 user-agent policy维基媒体。

根据documentation ,我们可以使用 agent 实例变量设置用户代理。用户代理 HTTP header 在 MDN web docs 中描述。 .最后,您可以将类对象初始化为,

sparql = SPARQLWrapper("https://query.wikidata.org/sparql", agent="Mozilla/5.0 (X11; Linux x86_64) AppleWebKit/537.11 (KHTML, like Gecko) Chrome/23.0.1271.64 Safari/537.11")

希望这对您有所帮助!

关于python - sparql.query().convert() 中 SPARQLWrapper 的 URLError,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/30755625/

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