在python中获取HTTPS标头
我正在处理 HTTPS，我想获取 live.com 的 HTTP 标头
import urllib2
try:
email="[email protected]"
response = urllib2.urlopen("https://signup.live.com/checkavail.aspx?chkavail="+email+"&tk=1258056184535&ru=http%3a%2f%2fmail.live.com%2f%3frru%3dinbox&wa=wsignin1.0&rpsnv=11&ct=1258055283&rver=6.0.5285.0&wp=MBI&wreply=http:%2F%2Fmail.live.com%2Fdefault.aspx&lc=1036&id=64855&bk=1258055288&rollrs=12&lic=1")
print 'response headers: "%s"' % response.info()
except IOError, e:
if hasattr(e, 'code'): # HTTPError
print 'http error code: ', e.code
elif hasattr(e, 'reason'): # URLError
print "can't connect, reason: ", e.reason
else:
raise
，所以我不需要标头中的所有信息，
如果您询问脚本的作用，我只需要 Set-Cookie 信息： 从该病毒 CheckAvail= 获取金额，
它用于检查电子邮件是否可在 hotmail 中使用，方法是编辑后
感谢帮助..修复仅获取 Set-Cookie 后，我遇到了问题当我得到cookie 无法获取 CheckAvil= 在浏览器中打开它并打开我得到的源代码后，我得到了很多没有 `CheckAvil= 的信息！看图片 
i'm dealing with HTTPS and i want to get HTTP header for live.com
import urllib2
try:
email="[email protected]"
response = urllib2.urlopen("https://signup.live.com/checkavail.aspx?chkavail="+email+"&tk=1258056184535&ru=http%3a%2f%2fmail.live.com%2f%3frru%3dinbox&wa=wsignin1.0&rpsnv=11&ct=1258055283&rver=6.0.5285.0&wp=MBI&wreply=http:%2F%2Fmail.live.com%2Fdefault.aspx&lc=1036&id=64855&bk=1258055288&rollrs=12&lic=1")
print 'response headers: "%s"' % response.info()
except IOError, e:
if hasattr(e, 'code'): # HTTPError
print 'http error code: ', e.code
elif hasattr(e, 'reason'): # URLError
print "can't connect, reason: ", e.reason
else:
raise
so i don't want all the information from headers i just want Set-Cookie information
if you asking what is script do : it's for checking if email avilable to use in hotmail by get the amount from this viralbe CheckAvail=
after edit
thanks for help .. after fixing get only Set-Cookie i got problem it's when i get cookie not get CheckAvil= i got a lot information without `CheckAvil= after open it in browser and open the source i got it !! see the picture
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response.info()返回的对象是mimetools.Message（如urllib2docs)，它是rfc822.Message，它有一个getheader()方法。因此，您可以执行以下操作：
但是，如果您要检查邮件，我建议您使用 POP3 或 IMAP（如果可用）（Python 附带了这两种模块）。
The object returned by
response.info()is an instance ofmimetools.Message(as described by theurllib2docs), which is a subclass ofrfc822.Message, which has agetheader()method.So you can do the following:
However, if you are checking for mail, I would recommend you to use POP3 or IMAP if available (Python comes with modules for both).
这是因为您的 http 响应中包含“Httponly”，这意味着，由于规范的原因，只有 http 连接可以查看。
It's because 'Httponly' in your http response, which meant, because of the specifications, only http connection can view.