我想發送請求帶上 headers 頭部，請問為什麼報錯了？
<code>Traceback (most recent call last):
File "D:\python\get-email-by-tieba.py", line 49, in <module>
main()
File "D:\python\get-email-by-tieba.py", line 6, in main
getThreadByTid()
File "D:\python\get-email-by-tieba.py", line 36, in getThreadByTid
req = urllib2.urlopen(url, post_data, headers)
File "C:\Python27\lib\urllib2.py", line 126, in urlopen
return _opener.open(url, data, timeout)
File "C:\Python27\lib\urllib2.py", line 391, in open
response = self._open(req, data)
File "C:\Python27\lib\urllib2.py", line 409, in _open
'_open', req)
File "C:\Python27\lib\urllib2.py", line 369, in _call_chain
result = func(*args)
File "C:\Python27\lib\urllib2.py", line 1173, in http_open
return self.do_open(httplib.HTTPConnection, req)
File "C:\Python27\lib\urllib2.py", line 1142, in do_open
h.request(req.get_method(), req.get_selector(), req.data, headers)
File "C:\Python27\lib\httplib.py", line 946, in request
self._send_request(method, url, body, headers)
File "C:\Python27\lib\httplib.py", line 987, in _send_request
self.endheaders(body)
File "C:\Python27\lib\httplib.py", line 940, in endheaders
self._send_output(message_body)
File "C:\Python27\lib\httplib.py", line 803, in _send_output
self.send(msg)
File "C:\Python27\lib\httplib.py", line 755, in send
self.connect()
File "C:\Python27\lib\httplib.py", line 736, in connect
self.timeout, self.source_address)
File "C:\Python27\lib\socket.py", line 557, in create_connection
sock.settimeout(timeout)
File "C:\Python27\lib\socket.py", line 222, in meth
return getattr(self._sock,name)(*args)</code>我想發送請求帶上 headers 頭部，請問為什麼報錯了？
<code>Traceback (most recent call last):
File "D:\python\get-email-by-tieba.py", line 49, in <module>
main()
File "D:\python\get-email-by-tieba.py", line 6, in main
getThreadByTid()
File "D:\python\get-email-by-tieba.py", line 36, in getThreadByTid
req = urllib2.urlopen(url, post_data, headers)
File "C:\Python27\lib\urllib2.py", line 126, in urlopen
return _opener.open(url, data, timeout)
File "C:\Python27\lib\urllib2.py", line 391, in open
response = self._open(req, data)
File "C:\Python27\lib\urllib2.py", line 409, in _open
'_open', req)
File "C:\Python27\lib\urllib2.py", line 369, in _call_chain
result = func(*args)
File "C:\Python27\lib\urllib2.py", line 1173, in http_open
return self.do_open(httplib.HTTPConnection, req)
File "C:\Python27\lib\urllib2.py", line 1142, in do_open
h.request(req.get_method(), req.get_selector(), req.data, headers)
File "C:\Python27\lib\httplib.py", line 946, in request
self._send_request(method, url, body, headers)
File "C:\Python27\lib\httplib.py", line 987, in _send_request
self.endheaders(body)
File "C:\Python27\lib\httplib.py", line 940, in endheaders
self._send_output(message_body)
File "C:\Python27\lib\httplib.py", line 803, in _send_output
self.send(msg)
File "C:\Python27\lib\httplib.py", line 755, in send
self.connect()
File "C:\Python27\lib\httplib.py", line 736, in connect
self.timeout, self.source_address)
File "C:\Python27\lib\socket.py", line 557, in create_connection
sock.settimeout(timeout)
File "C:\Python27\lib\socket.py", line 222, in meth
return getattr(self._sock,name)(*args)</code>
urlopen的參數到底怎麼傳遞可以看看手冊
樓下幾位的答案是不準確的，錯在參數要加上鍵，urllib2.open(url,data=data,headers=header)類似這樣的
貼了那麼多track，卻沒有貼報的錯，心塞
更新，補充下@雲語的答案，你可以看到他給出的官方文檔裡面也沒有提到有headers這個參數，但之所以可以傳headers並且需要寫成headers=的形式是因為這樣寫類似於寫一個dict然後被處理成requests物件傳給urlopen。而如果確實不能處理headers這個參數，那也會報錯“typeerror:urlopen got an unexpected keyword argument headers”，所以我才說你貼了很多traceback卻沒有把最後一行報的錯貼出來。
參數傳錯了
去翻翻手冊, 就知道urllib2的urlopen第三個參數並不是headers, 並且同時根本也沒有headers參數.
然後構造urllib2的Request對象的第三個參數才是headers, 所以你需要先建構一個Request物件, 然後urllib2.urlopen的參數傳遞這個Request物件
對於urllib2, 要加請求頭,要這樣寫
<code>request = urllib2.Request(uri)
request.add_header('User-Agent', 'fake-client')
response = urllib2.urlopen(request)</code>題主那種寫法不對,建議你看下requests庫,寫法就像你那種寫法,比較好用.
建議用requests吧
<code>import requests
url = ''
data = {}
headers = {}
g = requests.get(url, data=data, headers=headers)
p = requests.post(url, data=data, headers=headers)
print g.text, p.text</code>
