Voice commands without wakeword

Help
1

Hi there,

I’m new in this subject.
I’m from germany (maybe is my english pronunciation to bad for wakeword recognition) and I tested many default wakewords and engines with an english profile, but the wakeword recognition ist very bad. My saved sentences working as expected, after click the wake up button.

I would like to try a python script, that listen for the voice command directly without wakeword.
But, I couldn’t find any detailed information for this.

Maybe someone has an idea about this.

For voice test I have a small python script:

import websocket
import requests
import json

Intents are passed through here

def on_message(ws, message):
data = json.loads(message)
print("Erkannter Spracheingabe Ausdruck: ", data[“intent”][“name”])
print(data)

if ("forward" == data["intent"]["name"]):
   print(" ")
   print("Okay, ich fahre vorwärts.")
elif ("backward" == data["intent"]["name"]):
   print(" ")
   print("Okay, ich fahre rückwärts.")
elif ("turnright" == data["intent"]["name"]):
   print(" ")
   print("Okay, ich fahre nach rechts.")
elif ("turnleft" == data["intent"]["name"]):
   print(" ")
   print("Okay, ich fahre nach links.")
elif ("stop" == data["intent"]["name"]):
   print(" ")
   print("Okay, ich halte an.")

def on_error(ws, error):
print(error)

def on_close(ws):
print("\nDisconnected\n")

def on_open(ws):
print("\nConnected\n")

Start web socket client

if name == “main”:
ws = websocket.WebSocketApp(“ws://localhost:12101/api/events/intent”,
on_message = on_message,
on_error = on_error,
on_close = on_close)
ws.on_open = on_open
ws.run_forever()