在Python中，如何使用urllib来查看网站是404还是200？

如何通过urllib获取头文件的代码？

Answer 1

getcode（）方法（在python2.6中添加）返回与响应一起发送的HTTP状态码，如果URL没有HTTP URL，则返回None。

>>> a=urllib.urlopen('http://www.google.com/asdfsf') >>> a.getcode() 404 >>> a=urllib.urlopen('http://www.google.com/') >>> a.getcode() 200 

Answer 2

你也可以使用urllib2 ：

 import urllib2 req = urllib2.Request('http://www.python.org/fish.html') try: resp = urllib2.urlopen(req) except urllib2.HTTPError as e: if e.code == 404: # do something... else: # ... except urllib2.URLError as e: # Not an HTTP-specific error (eg connection refused) # ... else: # 200 body = resp.read() 

请注意， HTTPError是存储HTTP状态码的URLError的子类。

Answer 3

对于python 3：

 import urllib.request, urllib.error url = 'http://www.google.com/asdfsf' try: conn = urllib.request.urlopen(url) except urllib.error.HTTPError as e: # Return code error (eg 404, 501, ...) # ... print(e.code) except urllib.error.URLError as e: # Not an HTTP-specific error (eg connection refused) # ... print('URLError') else: # 200 # ... print('good') 

Answer 4

 import urllib2 try: fileHandle = urllib2.urlopen('http://www.python.org/fish.html') data = fileHandle.read() fileHandle.close() except urllib2.URLError, e: print 'you got an error with the code', e