Python 获取在urllib2 http请求中发送的标头

我知道如何收到邮件头

resp = urllib2.urlopen('http://www.google.com')
print resp.info()

但是，如何访问发送到的标题http://www.google.com”“是吗

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我通常使用wireshark分析并查看实际发送的内容，但我希望能够在脚本中访问这些信息。

有关如何检查实际请求的信息，请参见此处：如何访问类外的相同标头？除了打印出来之外，我还需要对标题做些什么。不幸的是，没有使用globals是很难做到的，因为opener.open会返回一个由HTTPResponse的HTTPMessage创建的AddInfo URL（这可能对您毫无意义）无论如何，您可以将标题附加到响应中，并可能以req作为前缀。我正在添加代码作为另一个答案，或者有一种方法可以在注释中格式化？

import httplib
import urllib2

class CustomHTTPConnection(httplib.HTTPConnection):
    def request(self, method, url, body=None, headers={}):
        print headers
        self._send_request(method, url, body, headers)

class CustomHTTPHandler(urllib2.AbstractHTTPHandler):
    def http_open(self, req):
        return self.do_open(CustomHTTPConnection, req)
    http_request = urllib2.AbstractHTTPHandler.do_request_

if __name__ == '__main__':
    opener = urllib2.OpenerDirector()
    opener.add_handler(CustomHTTPHandler())
    res = opener.open('http://www.google.it/')

import httplib
import urllib2

class CustomHTTPConnection(httplib.HTTPConnection):
    def request(self, method, url, body=None, headers={}):
        self.req_headers = headers
        self._send_request(method, url, body, headers)
    def getresponse(self, buffering=False):
        resp = httplib.HTTPConnection.getresponse(self, buffering)
        for key, value in self.req_headers.items():
            resp.msg.headers.append('req_%s: %s\r\n' % (key, value))
        return resp

class CustomHTTPHandler(urllib2.AbstractHTTPHandler):
    def http_open(self, req):
        resp = self.do_open(CustomHTTPConnection, req)
        return resp
    http_request = urllib2.AbstractHTTPHandler.do_request_

if __name__ == '__main__':
    opener = urllib2.OpenerDirector()
    opener.add_handler(CustomHTTPHandler())
    res = opener.open('http://www.google.it/')
    info = res.info()
    print info