无效的 url 抛出异常 - python [英] Invalid url's throw an exception - python

问题描述

导入httplib导入 urlparsedef getUrl(url):尝试:部分 = urlparse.urlsplit(url)服务器 = 零件 [1]路径 = 部分 [2]obj = httplib.HTTPConnection(server,80)obj.connect()obj.putrequest('HEAD',path)obj.putheader('接受','*/*')obj.endheaders()响应 = obj.getresponse()contentType = response.getheader("content-type", "unknown")obj.close()如果 response.status !=200:打印错误"别的:打印真棒"除了例外，e:打印e

我编写了上面的代码来检查给定的 URL 是否有效.但不知何故，当我测试它时，对于每个无效的 url，它都会引发异常.

<预><代码>>>>getUrl('http://www.idlebrfffain.com')[Errno -2] 名称或服务未知

Python 版本:

chaitu@ubuntu:~$ python -V蟒蛇 2.6.4

谁能帮我找出错误的确切位置?

解决方案

你必须抓住 socket.error:

导入httplib、socket导入 urlparsedef getUrl(url):部分 = urlparse.urlsplit(url)服务器 = 零件 [1]路径 = 部分 [2]obj = httplib.HTTPConnection(server,80)尝试:obj.connect()除了 socket.gaierror:打印主机 %s 不存在" % 服务器返回除了socket.error:打印无法连接到 %s:%s."%(服务器，80)返回obj.putrequest('HEAD',path)obj.putheader('接受','*/*')obj.endheaders()响应 = obj.getresponse()contentType = response.getheader("content-type", "unknown")obj.close()如果 response.status !=200:打印错误"别的:打印真棒"getUrl('http://www.idlebrfffain.com') # 不是注册域名getUrl('http://8.8.8.8') # 不是 http 服务器

仅 try: except: 围绕特定行并且仅当您知道会发生什么时.Python 会向您显示未捕获异常的回溯，以便您轻松找出问题所在.

import httplib
import urlparse

def getUrl(url):
   try:
     parts = urlparse.urlsplit(url)
     server = parts[1]
     path = parts[2]
     obj = httplib.HTTPConnection(server,80)
     obj.connect()
     obj.putrequest('HEAD',path)
     obj.putheader('Accept','*/*')
     obj.endheaders()
     response = obj.getresponse()
     contentType = response.getheader("content-type", "unknown")
     obj.close()
     if response.status !=200:
       print 'Error'
     else:
       print 'Awesome'
   except Exception, e:
     print e

I wrote the code above to check if a given URL is valid or not. But somehow when I test it, for every invalid url it throws an exception.

>>> getUrl('http://www.idlebrfffain.com')
[Errno -2] Name or service not known

Python version:

chaitu@ubuntu:~$ python -V
Python 2.6.4

Can anyone help me find out where exactly is the mistake?

解决方案

You have to catch socket.error:

import httplib, socket
import urlparse

def getUrl(url):
    parts = urlparse.urlsplit(url)
    server = parts[1]
    path = parts[2]
    obj = httplib.HTTPConnection(server,80)

    try:
        obj.connect()
    except socket.gaierror:
        print "Host %s does not exist" % server
        return
    except socket.error:
        print "Cannot connect to %s:%s." % (server, 80)
        return

    obj.putrequest('HEAD',path)
    obj.putheader('Accept','*/*')
    obj.endheaders()
    response = obj.getresponse()
    contentType = response.getheader("content-type", "unknown")
    obj.close()
    if response.status !=200:
        print 'Error'
    else:
        print 'Awesome'


getUrl('http://www.idlebrfffain.com') # not a registered domain
getUrl('http://8.8.8.8') # not a http server

Only try: except: around specific lines and only if you know what happens. Python will show you tracebacks for uncaught exceptions, so you can find out where the problem is with ease.

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