Get IP from argv fail

Question

I'm trying to get the ip address of a website given from args. When I try with the website directly in the source code like 'url='https://google.com' it works but when I try with 'url = sys.argv[1]' it fails.

When I print the 'url = sys.argv[1]' I get the desired website. I tried to str(url) it but it doesn't work neither.

Here's the code :

import socket
import sys

# Params
url = sys.argv[1]
# url = str(sys.argv[1])

print (type(url))   # I get the desired url

s = socket.socket()

# Get IP
ip = socket.gethostbyname(url)

# Print Infos
print ('IP Adress : ' + ip + '\n' + 15*'-')

s.close()

Do you have any idea?

Thank you, it's driving me crazy.

Answer 1

It is because you're passing in the https//:; you need to remove it:

In [3]: ip = socket.gethostbyname("http://google.com")
---------------------------------------------------------------------------
gaierror                                  Traceback (most recent call last)
<ipython-input-3-7466d856e904> in <module>()
----> 1 ip = socket.gethostbyname("http://google.com")

Instead, try:

In [4]: ip = socket.gethostbyname("google.com")

In [5]: ip
Out[5]: '172.217.25.238'

Note that you'll also want to remove any trailing slashes, for example, remove / in google.com/.

If you look at man gethostbyname you'll see that you're making a DNS request:

The gethostbyname() function returns a structure of type hostent for the given host name. Here name is either a hostname or an IPv4 address in standard dot notation (as for inet_addr(3)).

So, you need to make sure you cleanup anything you pass to that function call.