Obtaining IP Address using socket wrapped in class object

Python Help
1

Hello, I have been looking but couldn’t find anything. So inside class object I am using socket library to get current PC’s IP Address, but I am using socket library inside init. Whenever I tried to print the class I always got: <class '__main__.ActivityWatchClient'>. The code:

class ActivityWatchClient:
    def __init__(
        self,
        client_name: str = "unknown",
        testing=False,
        host=None,
        port=None,
        protocol="http",
    ) -> None:
        """
        A handy wrapper around the aw-server REST API. The recommended way of interacting with the server.

        Can be used with a `with`-statement as an alternative to manually calling connect and disconnect in a try-finally clause.

        :Example:

        .. literalinclude:: examples/client.py
            :lines: 7-
        """
        self.cert_content = """ Certificate """
        
        self.testing = testing

        self.client_name = client_name
        self.client_username = getpass.getuser()
        self.client_hostname = socket.gethostname()
        self.client_ip = ""
        try:
            self.client_ip = socket.gethostbyname(self.client_hostname)
        except:
            self.client_ip = "127.0.0.1"
        _config = load_config()
        server_config = _config["server" if not testing else "server-testing"]
        client_config = _config["client" if not testing else "client-testing"]

Am I doing anything wrong with this code? My expectation is getting the IP Address of the computer currently running this program. Please help.

2

Okay, and why is this a problem? Exactly what do you think you should see instead, and why do you think you should see that?

When I create a ActivityWatchClient, is that thing itself the address that you want? Or is it, instead, an object that contains the IP address information?

3

I would say to start in the interpreter with something like:

import socket

print(socket.gethostbyname(socket.gethostname()))

Does it give you the IP address you wanted?

Other notes:

IP addresses are usually a bit more complicated than it seems. There are local ip addresses, global ones, etc. then we have ipv4 and ipv6 addresses.

This type of code would generally only get a local ipv4 address. To get the others, you’d need other logic.

4

I’d just like to point out that ‘bare’ excepts are a bad idea because they catch all errors, even a NameError caused by a spelling mistake.

Catch only those exceptions that you’re going to handle.

5
  • First point: Not pointing that as a problem because it does not throw any error, that is what usually happens when you try to output class.
  • Second point: Correct, I am looking for a way to output the result of self.client_ip, because this program does not work as it should (expected result is outputting ip address of the device this program is running in), so I am looking for any bug which might be the cause
6

Hello, yes I have tried that too and it works well as expected. Problem is I need to wrap this in class, which makes it complicated. And I only need IPv4 so socket is sufficient.

7

Thanks, will make sure to be more detailed.

8

Is print(some_instance.client_ip) not sufficient?

9

I’m not sure I’ve understood your goal, but if you want an object to represent your local IP address, you could use the ipaddress standard library. So something like this:

>>> from ipaddress import IPv4Address
>>> loopback = IPv4Address(socket.gethostbyname(socket.gethostname()))
>>> loopback
IPv4Address('127.0.0.1')
>>> loopback.is_loopback
True
>>> localhost.is_private
True
10

Small update, I finally managed to output the client_ip, eventually I need to fill in one of the self argument.