How to get ipv4 address from a url which contains http via socket in python [duplicate]

Question

def statusCheck(URL,PORT):
    sock = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
    result = sock.connect_ex((URL,PORT))
    return result

Here URL = "http://127.0.0.1"

When I am calling statusCheck function with below parameter,

I got error.

BASE_URL = "http://127.0.0.1"
PORT = 7000


res = statusCheck(BASE_URL,PORT)
print (res)

Error:

Traceback (most recent call last):

File "test.py", line 15, in <module>

res = statusCheck(BASE_URL,PORT)

File "test.py", line 12, in statusCheck

result = sock.connect_ex((URL,PORT))

File "/usr/lib64/python2.7/socket.py", line 224, in meth

return getattr(self._sock,name)(*args)

socket.gaierror: [Errno -2] Name or service not known

When I use URL = "127.0.0.1". It work fine.

So my question is how I can get ipv4 address from a url which contains http://

Sunitha · Accepted Answer · 2018-09-08 08:49:17Z

1

Use str.split('/') to split the url based on the delimiter / and grab the 3rd element

>>> BASE_URL = "http://127.0.0.1"
>>> BASE_URL.split('/', 3)[2]
'127.0.0.1'

answered Sep 8, 2018 at 8:49

Sunitha

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