What s the best way to get an HTTP response code from a URL

Question

I   m looking for a quick way to get an HTTP response code from a URL  i e  200  404  etc    I   m not sure which library to use

User · Answer

Update using the wonderful requests library  Note we are using the HEAD request  which should happen more quickly then a full GET or POST request    import requests try      r   requests head  https   stackoverflow com       print r status code        prints the int of the status code  Find more at httpstatusrappers com    except requests ConnectionError      print  failed to connect

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Here s an httplib solution that behaves like urllib2   You can just give it a URL and it just works   No need to mess about splitting up your URLs into hostname and path   This function already does that   import httplib import socket def get link status url             Gets the HTTP status of the url or returns an error associated with it   Always returns a string          https False   url re sub r           r  1  url    url url split     3    if len url   gt  3      path     url 3    else      path       if url 0      http        port 80   elif url 0      https        port 443     https True   if     in url 2       host url 2  split      0      port url 2  split      1    else      host url 2    try      headers   User-Agent   Mozilla 5 0  X11  Ubuntu  Linux x86 64  rv 26 0  Gecko 20100101 Firefox 26 0                 Host  host                    if https        conn httplib HTTPSConnection host host port port timeout 10      else        conn httplib HTTPConnection host host port port timeout 10      conn request method  HEAD  url path headers headers      response str conn getresponse   status      conn close     except socket gaierror e      response  Socket Error   d    s     e 0  e 1     except StandardError e      if hasattr e  getcode   and len e getcode     gt  0        response str e getcode        if hasattr e   message   and len e message   gt  0        response str e message      elif hasattr e   msg   and len e msg   gt  0        response str e msg      elif type        type e         response e     else        response  Exception occurred without a good error message   Manually check the URL to see the status   If it is believed this URL is 100  good then file a issue for a potential bug     return response

User · Answer

In future  for those that use python3 and later  here s another code to find response code   import urllib request  def getResponseCode url       conn   urllib request urlopen url      return conn getcode

User · Answer

Here s a solution that uses httplib instead   import httplib  def get status code host  path               This function retreives the status code of a website by requesting         HEAD data from the host  This means that it only requests the headers          If the host cannot be reached or something else goes wrong  it returns         None instead              try          conn   httplib HTTPConnection host          conn request  HEAD   path          return conn getresponse   status     except StandardError          return None   print get status code  stackoverflow com     prints 200 print get status code  stackoverflow com     nonexistant     prints 404

User · Answer

The urllib2 HTTPError exception does not contain a getcode   method  Use the code attribute instead

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Addressing  Niklas R s comment to  nickanor s answer   from urllib error import HTTPError import urllib request  def getResponseCode url       try          conn   urllib request urlopen url          return conn getcode       except HTTPError as e          return e code

User · Answer

You should use urllib2  like this   import urllib2 for url in   http   entrian com     http   entrian com does-not-exist         try          connection   urllib2 urlopen url          print connection getcode           connection close       except urllib2 HTTPError  e          print e getcode      Prints    200  from the try block    404  from the except block

[python] What’s the best way to get an HTTP response code from a URL?

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