代理检查python [英] Proxy Check in python
问题描述
我在python中编写了一个使用cookie和POST / GET的脚本。我还在脚本中包含了代理支持。但是,当一个人进入死代理代理时,脚本崩溃。有没有办法在运行我的其余脚本之前检查代理是否死/活?
I have written a script in python that uses cookies and POST/GET. I also included proxy support in my script. However, when one enters a dead proxy proxy, the script crashes. Is there any way to check if a proxy is dead/alive before running the rest of my script?
此外,我注意到有些代理不处理cookie / POST标题正确。有没有办法解决这个问题?
Furthermore, I noticed that some proxies don't handle cookies/POST headers properly. Is there any way to fix this?
推荐答案
最简单的方法就是从urllib中捕获IOError异常:
The simplest was is to simply catch the IOError exception from urllib:
try:
urllib.urlopen(
"http://example.com",
proxies={'http':'http://example.com:8080'}
)
except IOError:
print "Connection error! (Check proxy)"
else:
print "All was fine"
此外,来自此博客文章 - 检查状态代理地址(略有改进):
Also, from this blog post - "check status proxy address" (with some slight improvements):
import urllib2
import socket
def is_bad_proxy(pip):
try:
proxy_handler = urllib2.ProxyHandler({'http': pip})
opener = urllib2.build_opener(proxy_handler)
opener.addheaders = [('User-agent', 'Mozilla/5.0')]
urllib2.install_opener(opener)
req=urllib2.Request('http://www.example.com') # change the URL to test here
sock=urllib2.urlopen(req)
except urllib2.HTTPError, e:
print 'Error code: ', e.code
return e.code
except Exception, detail:
print "ERROR:", detail
return True
return False
def main():
socket.setdefaulttimeout(120)
# two sample proxy IPs
proxyList = ['125.76.226.9:80', '213.55.87.162:6588']
for currentProxy in proxyList:
if is_bad_proxy(currentProxy):
print "Bad Proxy %s" % (currentProxy)
else:
print "%s is working" % (currentProxy)
if __name__ == '__main__':
main()
请记住,如果代理停机,这可能会使脚本占用时间翻倍(因为您将拥有等待两个连接超时)..除非y你必须要知道代理有问题,处理IOError更清晰,更简单,更快..
Remember this could double the time the script takes, if the proxy is down (as you will have to wait for two connection-timeouts).. Unless you specifically have to know the proxy is at fault, handling the IOError is far cleaner, simpler and quicker..
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