CODEWITHSUNDEEP
pythonurllib2
Ez0r
Ez0r

Reputation: 190

urllib2.Request check if URL is reachable

So I have the following code to verify certain url is correct, I only need 200 response so I made a script works fine but it's too slow (:

import urllib2
import string
def my_range(start, end, step):
    while start <= end:
        yield start
        start += step
url = 'http://exemple.com/test/'
y = 1
for x in my_range(1, 5, 1):
 y =y+1 
 url+=str(y)
 print url 
 req = urllib2.Request(url)
 try:
    resp = urllib2.urlopen(req)
 except urllib2.URLError, e:
    if e.code == 404:
        print "404"
    else:
        print "not 404"
 else:
    print "200"
 url = 'http://exemple.com/test/'
body = resp.read()

in this example i assume that i have the following directories in my local host with this results

http://exemple.com/test/2
200
http://exemple.com/test/3
200
http://exemple.com/test/4
404
http://exemple.com/test/5
404
http://exemple.com/test/6
404

so I searched how to do it quicker i found this code :

import urllib2
request = urllib2.Request('http://www.google.com/')
response = urllib2.urlopen(request)
if response.getcode() == 200:
   print "200"

it's seems quicker but when i test it with a 404 like (http://www.google.com/111) it's gives me this result :

Traceback (most recent call last):
  File "C:\Python27\res.py", line 3, in <module>
    response = urllib2.urlopen(request)
  File "C:\Python27\lib\urllib2.py", line 126, in urlopen
    return _opener.open(url, data, timeout)
  File "C:\Python27\lib\urllib2.py", line 400, in open
    response = meth(req, response)
  File "C:\Python27\lib\urllib2.py", line 513, in http_response
    'http', request, response, code, msg, hdrs)
  File "C:\Python27\lib\urllib2.py", line 438, in error
    return self._call_chain(*args)
  File "C:\Python27\lib\urllib2.py", line 372, in _call_chain
    result = func(*args)
  File "C:\Python27\lib\urllib2.py", line 521, in http_error_default
    raise HTTPError(req.get_full_url(), code, msg, hdrs, fp)
urllib2.HTTPError: HTTP Error 404: Not Found

any ideas guys ? and thanks very much for any help :)

Upvotes: 1

Views: 5737

Answers (1)

Stephan
Stephan

Reputation: 3093

HTTPError is defined as a family of exceptions so that you can use Try/Except in cases like this:

import urllib2
request = urllib2.Request('http://www.google.com/')
try:
    response = urllib.urlopen(request)
    # do stuff..
except urllib2.HTTPError: # 404, 500, etc..
    pass

You can also add a further except clause for urllib2.URLError which covers other (non-HTTP) errors such as timeouts.

Upvotes: 4

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