Python PycURL的替代品？

下面是一段上传文件的代码：

  file_size = os.path.getsize('Tea.rdf')
  f = file('Tea.rdf')
  c = pycurl.Curl()
  c.setopt(pycurl.URL, 'http://localhost:8080/openrdf-sesame/repositories/rep/statements')
  c.setopt(pycurl.HTTPHEADER, ["Content-Type: application/rdf+xml;charset=UTF-8"])
  c.setopt(pycurl.PUT, 1)
  c.setopt(pycurl.INFILE, f)
  c.setopt(pycurl.INFILESIZE, file_size)
  c.perform()
  c.close()

现在，我一点也不喜欢这种PycURL体验。你能建议其他的选择吗？也许urllib2或httplib也可以这样做？你能写一些代码来显示它吗

非常感谢

使用：

---

您的示例已转换为httplib：

import httplib

host = 'localhost:8080'
path = '/openrdf-sesame/repositories/rep/statements'
path = '/index.html'
headers = {'Content-type': 'application/rdf+xml;charset=utf-8'}

f = open('Tea.rdf')
conn = httplib.HTTPConnection(host)
conn.request('PUT', path, f, headers)
res = conn.getresponse()
print res.status, res.reason
print res.read()

---

是的，pycurl的API设计很糟糕，cURL很强大。它比urllib/urllib2有更多的未来

也许你想试着用人类的卷发。这是python卷曲包装器。您可以从源代码或通过pip:pip-install-human\u-curl安装它

例如：

>>> import human_curl as hurl
>>> r = hurl.put('http://localhost:8080/openrdf-sesame/repositories/rep/statements',
... headers = {'Content-Type', 'application/rdf+xml;charset=UTF-8'},
... files = (('my_file', open('Tea.rdf')),))
>>> r
    <Response: 201>

>>将人形卷曲作为投掷导入
>>>r=投掷。投掷（'http://localhost:8080/openrdf-sesame/repositories/rep/statements'，
…headers={'Content-Type'，'application/rdf+xml；charset=UTF-8'}，
…文件=（“我的文件”，打开（'Tea.rdf'），）
>>>r

---

您还可以读取响应标题、cookies等

问题是，现在我正试图发布一个针对此存储的查询。。。卷发给我带来了很多麻烦。所以，我只是想知道是否有其他方法可以做到这一点。再次感谢

>>> import human_curl as hurl
>>> r = hurl.put('http://localhost:8080/openrdf-sesame/repositories/rep/statements',
... headers = {'Content-Type', 'application/rdf+xml;charset=UTF-8'},
... files = (('my_file', open('Tea.rdf')),))
>>> r
    <Response: 201>