Python: Urllib.urlopen nonnumeric port

for the following cod开发者_如何学Pythone

theurl = "https://%s:%s@members.dyndns.org/nic/update?hostname=%s&myip=%s&wildcard=NOCHG&mx=NOCHG&backmx=NOCHG" % (username, password, hostname, theip)

conn = urlopen(theurl) # send the request to the url
print(conn.read())  # read the response
conn.close()   # close the connection

i get the following error

File "c:\Python31\lib\http\client.py", line 667, in _set_hostport
    raise InvalidURL("nonnumeric port: '%s'" % host[i+1:])

Any Ideas???

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You probably need to url-encode the password. You'll see an error like that if the password happens to contain a '/'.

Here's a local example (actual values redacted):

>>> opener
<urllib.FancyURLopener instance at 0xb6f0e2ac>
>>> opener.open('http://admin:somepass@example.com')
<addinfourl at 3068618924L whose fp = <socket._fileobject object at 0xb6e7596c>>
>>> opener.open('http://admin:somepass/a@example.com')
*** InvalidURL: nonnumeric port: 'somepass'

Encode the password:

>>> opener.open('http://admin:somepass%2Fa@example.com')

You can use urllib.quote('somepass/a', safe='') to do the encoding.

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I agree with muckabout, this is the problem. You're probably used to using this in a browser, which would cause the browser to authenticate with the host. You should probably drop everything before the first @ sign.

have a look at urllib docs, specifically FancyURLOpener which might resolve your issue with authentication.

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The error message shows that there is some issue with the url that you are preparing. Print and check if this is a valid url.

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The ':' in the HTTP URL is assumed to precede a port number. You are placing an account name which is not numeric. It must be an integer port value.