Question

我知道如何获取标题RECEIVED

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

但是如何将标题SENT访问为“http://www.google.com”？

我通常使用wireshark来分析和查看实际发送的内容，但我希望能够在我的脚本中访问这些信息。

Answer 1

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/')

Answer 2

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

获取在urllib2 http请求中发送的标头

2 个答案: