我熟悉PHP中的CURL,但我第一次使用pycurl在Python中使用它.

我一直收到错误：

Exception Type:     error
Exception Value:    (2, '')

我不知道这可能是什么意思.这是我的代码：

data = {'cmd': '_notify-synch',
        'tx': str(request.GET.get('tx')),
        'at': paypal_pdt_test
        }

post = urllib.urlencode(data)

b = StringIO.StringIO()

ch = pycurl.Curl()
ch.setopt(pycurl.URL, 'https://www.sandbox.paypal.com/cgi-bin/webscr')
ch.setopt(pycurl.POST, 1)
ch.setopt(pycurl.POSTFIELDS, post)
ch.setopt(pycurl.WRITEFUNCTION, b.write)
ch.perform()
ch.close()

错误是指ch.setopt行(pycurl.POSTFIELDS,post)

解决方法:

看来你的pycurl安装(或curl库)以某种方式被损坏了.从curl错误代码文档：

CURLE_FAILED_INIT (2)
Very early initialization code failed. This is likely to be an internal error or problem.

您可能需要重新安装或重新编译curl或pycurl.

但是,要像你一样做一个简单的POST请求,你实际上可以使用python的“urllib”而不是CURL：

import urllib

postdata = urllib.urlencode(data)

resp = urllib.urlopen('https://www.sandbox.paypal.com/cgi-bin/webscr', data=postdata)

# resp is a file-like object, which means you can iterate it,
# or read the whole thing into a string
output = resp.read()

# resp.code returns the HTTP response code
print resp.code # 200

# resp has other useful data, .info() returns a httplib.HTTPMessage
http_message = resp.info()
print http_message['content-length']  # '1536' or the like
print http_message.type  # 'text/html' or the like
print http_message.typeheader # 'text/html; charset=UTF-8' or the like


# Make sure to close
resp.close()

要打开https：// URL,您可能需要安装PyOpenSSL：
http://pypi.python.org/pypi/pyOpenSSL

一些distibutions包括这个,其他的通过你最喜欢的包管理器提供它作为额外的包.

编辑：你有没有打电话给pycurl.global_init()？我仍然建议尽可能使用urllib / urllib2,因为您的脚本将更容易移动到其他系统.