CODEWITHSUNDEEP
pythonregex
tubos
tubos

Reputation: 99

list index error from regex

I have following function that gives me errors. It should extract the IP from the website url Sometimes it works sometimes i get following error:

File "test.py", line 30, in getIPextA
address = grab[0]                    # get address as a string
IndexError: list index out of range

Function:

def getIPextA():
"""
Version A Get external ip from "http://checkip.dyndns.org/"
"""
site=geturldata("http://checkip.dyndns.org/")
if site =="" : return [0,0,0,0]
grab = re.findall('\d{2,3}.\d{2,3}.\d{2,3}.\d{2,3}',site)
address = grab[0]                    # get address as a string
return map(int,address.split('.'))   # as an integer list

Upvotes: 0

Views: 123

Answers (2)

Borodin
Borodin

Reputation: 126722

I suggest something like this instead

    match = re.search(r' (\d{1,3}) \. (\d{1,3}) \. (\d{1,3}) \. (\d{1,3}) ', site, flags=re.X)
    return match.groups() if match else [0, 0, 0, 0]

which allows between one and three digits in each octet, and has the dots . escaped properly as well. The flags parameter allows me to add insignificant whitespace so that the regex is more readable. Each octet is cointained in parentheses, so the MatchObject will return a list of four decimal strings if you call its groups method.

But I don't know what your geturldata function is so I can't help any more.

Upvotes: 0

ProgramFOX
ProgramFOX

Reputation: 6390

You get the error when the IP address could not match the regex. The reason is probably that you sometimes have an IP containing a one-digit part. These are also valid IPs. You should also escape the dot, because a dot in a regex means that it can match any char.

grab = re.findall('\d{1,3}\.\d{1,3}\.\d{1,3}\.\d{1,3}',site)

Upvotes: 1

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