jsonip

You need to have two

Thanks to @T.Aoukar for the hint I managed to get the IP by creating a callable cloud function to return the user's IP address.

Here's the code it might help:

There are many questions to answer.

Addresses 127.X.X.X are reserved for the loopback interface, most common one is 127.0.0.1. The loopback is a virtual, but important interface and as you have probably guessed, it is usable on the local machine only. You cannot use 127.X.X.X address to make two hosts to communicate with each other.

Addresses 192.168.X.X (and also 10.X.X.X and 172.16-31.X.X.) are reserved for local LANs. They are not valid on the Internet.

You cannot use these addresses to make two hosts to communicate with each other over the public Internet (unless you create a tunnel, an advanced networking topic)

Almost everybody uses them, because we ran out of IPv4 addresses long time ago, they were difficult to get, expensive, etc. Also such hosts are isolated from the Internet, they can be reached only via a router that translates addresses. Such router feature is called NAT. A typical router has one valid Internet address and all connections to the Internet appear as coming from the router. If you contant a service like jsonip.com from a PC, you get your router's address, not your PC's address.

See also: Finding local IP addresses using Python's stdlib

To make your program working, make it to accept connections on all interfaces. See the first example in the socket docs. On Linux, use port numbers >= 1024. Ports < 1024 are reserved, not available to regular users.

Final point is that a firewall may prevent connections to your server. It depends on your system and setup.