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>