How to connect Ipython with own frontend?

from jupyter_client import KernelManager

def main():
    # Create a new kernel manager
    km = KernelManager(kernel_id='9912')
    km.start_kernel()

    # Create a client to interact with the kernel
    kc = km.client()
    kc.start_channels()

    # Ensure the client is connected before executing code
    kc.wait_for_ready()

    # Execute the code
    code = 'print("Hello, World!")'
    msg_id = kc.execute(code)

    # Wait for the result and display it
    while True:
        try:
            msg = kc.get_iopub_msg(timeout=1)
            print(msg)
            content = msg["content"]

            # When a message with the text stream comes and it's the result of our execution
            if msg["msg_type"] == "stream" and content["name"] == "stdout":
                print(content["text"])
                break
        except KeyboardInterrupt:
            print("Interrupted by user.")
            break
        except:
            # If no messages are available, we'll end up here, but we can just continue and try again.
            pass

    # Cleanup
    kc.stop_channels()
    km.shutdown_kernel()

if __name__ == '__main__':
    main()

{
    "shell_port": 63101,
    "iopub_port": 63102,
    "stdin_port": 63103,
    "control_port": 63106,
    "hb_port": 63105,
    "ip": "127.0.0.1"
  }

NOTE: When using the `ipython kernel` entry point, Ctrl-C will not work.

To exit, you will have to explicitly quit this process, by either sending
"quit" from a client, or using Ctrl-\ in UNIX-like environments.

To read more about this, see https://github.com/ipython/ipython/issues/2049

To connect another client to this kernel, use:
    --existing kernel-9912.json