如何通过urllib获取headers的代码?
问问题
134870 次
4 回答
185
getcode() 方法(在 python2.6 中添加)返回随响应发送的 HTTP 状态代码,如果 URL 不是 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
于 2009-11-13T00:45:14.387 回答
89
您也可以使用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 (e.g. connection refused)
# ...
else:
# 200
body = resp.read()
请注意,它是存储 HTTP 状态代码HTTPError
的子类。URLError
于 2009-11-13T00:46:01.280 回答
48
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 (e.g. 404, 501, ...)
# ...
print('HTTPError: {}'.format(e.code))
except urllib.error.URLError as e:
# Not an HTTP-specific error (e.g. connection refused)
# ...
print('URLError: {}'.format(e.reason))
else:
# 200
# ...
print('good')
于 2016-04-27T17:20:18.110 回答
6
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
于 2011-07-08T19:25:41.567 回答