URLError SPARQLWrapper at sparql.query().convert()

Question

我尝试使用一个小python脚本来测试我的SPARQL请求。然而，只是下一个简单的代码不起作用。

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"])

在转换步骤中，我的代码会冻结很长时间，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

我的情况和2.7和3.4都一样。编辑:我将连接从Wifi更改为intranet。我的脚本在DBpedia Sparql端点上运行良好，但是当我在本地服务器上请求时有一个Http错误。这似乎是代理或访问本地服务器的问题。

提前谢谢你的帮助。

Answer 1

如果您试图通过python脚本在本地fuseki服务器上执行一些SPARQL请求，则可能会受到某些代理问题的干扰。要解决这个问题，可以使用urllib的自动检测属性。

from SPARQLWrapper import SPARQLWrapper, JSON, XML
#import urllib.request module. Don't forget for Python 3.4 the urllib has been split into several different modules.
import urllib.request

#if the arg is empty in ProxyHandler, urllib will find itself your proxy config.
proxy_support = urllib.request.ProxyHandler({})
opener = urllib.request.build_opener(proxy_support)
urllib.request.install_opener(opener)

#connect to the sparql point
sparql = SPARQLWrapper("http://localhost:3030/yourOwnDb/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#>

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

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

Answer 2

对于2019年来到这里的用户，对于访问Wikidata sparql端点的错误，Wikidata强制执行严格的用户代理策略，请参阅这个存档的维基数据项目聊天 (谢Pere)，其中表示发送信息头的应用程序表明非bot脚本行为良好，也请参见维基媒体的用户代理策略。

根据文档，我们可以使用agent实例变量来设置用户代理.用户代理HTTP报头在MDN网络文档中描述.最后，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")

希望这能有所帮助！

Answer 3

正如错误告诉您的那样：

作业超时

似乎当您运行代码时，无法从连接中找到dbpedia.org。

现在运行您的代码，它将立即为我返回以下内容：

http://www.openlinksw.com/schemas/virtrdf#QuadMapFormat

因此，在生产过程中，您可能希望抓住那个URLError并以某种方式处理它。

更新一个又一个问题编辑：

目前，SPARQLWrapper依赖于urllib2来执行它的请求，所以如果您在代理后面，您应该能够使用urllib2的ProxyHandler (如这里 )。

proxy = urllib2.ProxyHandler({'http': '192.168.x.x'})
opener = urllib2.build_opener(proxy)
urllib2.install_opener(opener)
# and then:
results = sparql.query().convert()