PycURL的替代品？

Question

这里有一些上传文件的代码：

  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。你能给我其他的建议吗？也许urllib2或httplib可以做同样的事情？你能写一些显示它的代码吗？

非常感谢！

Answer 1

使用httplib2

import httplib2
http = httplib2.Http()

f = open('Tea.rdf')
body = f.read()
url = 'http://localhost:8080/openrdf-sesame/repositories/rep/statements'
headers = {'Content-type': 'application/rdf+xml;charset=utf-8'}
resp, content = http.request(url, 'PUT', body=body, headers=headers)
# resp will contain headers and status, content the response body

Answer 2

是的，pycurl有一个糟糕的API设计，cURL是强大的。它有更多的未来，然后是urllib/urllib2。

也许你想尝试使用human_curl。这是python卷曲包装器。您可以从源代码https://github.com/lispython/human_curl或通过pip: pip install human_curl安装它。

示例：

>>> 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>

您还可以读取响应头、cookie等

Answer 3

您的示例转换为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()