In Python how do I use urllib to see if a website is 404 or 200

Question

How to get the code of the headers through urllib

User · Answer

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

User · Answer

You can use urllib2 as well   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     Note that HTTPError is a subclass of URLError which stores the HTTP status code

User · Answer

For 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

User · Answer

The getcode   method  Added in python2 6  returns the HTTP status code that was sent with the response  or None if the URL is no HTTP URL    gt  gt  gt  a urllib urlopen  http   www google com asdfsf    gt  gt  gt  a getcode   404  gt  gt  gt  a urllib urlopen  http   www google com     gt  gt  gt  a getcode   200

[python] In Python, how do I use urllib to see if a website is 404 or 200?

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