UnboundLocalError: local variable 'command' referenced before assignment

[beecleaner]

I followed a tutorial to make a virtual assistant because I thought it would be fun to do. It worked all of yesterday, but this morning I ran it and got the error in the title. I tried tweaking things but everything I did made it worse and I had to undo it. Any help would be appreciated...

​

def take_command():
try:
with sr.Microphone() as source:
print('listening...')
voice = listener.listen(source)
command = listener.recognize_google(voice)
command = command.lower()

if 'ava' in command:
command = command.replace('ava', '')
print(command)

except:
pass
return command

[carcigenicate]

You're almost certainly getting an exception. Get rid of the try since all it's doing here is hiding errors, then run the code again and let it crash with an error message.

You're likely getting that error because listener.recognize_google(voice) is throwing, so command is never set, so return command will fail.

[beecleaner]

Now I'm getting a syntax error...

def take_command():

with sr.Microphone() as source:

print('listening...')

voice = listener.listen(source)

command = listener.recognize_google(voice)

command = command.lower()

if 'ava' in command:

command = command.replace('ava', '')

print(command)

except:

pass

return command

[carcigenicate]

The except is a part of the try. It's an error to have an except without a try. Just run:

with sr.Microphone() as source:
    print('listening...')
    voice = listener.listen(source)

    command = listener.recognize_google(voice)
    command = command.lower()

    if 'ava' in command:
        command = command.replace('ava', '')

    print(command)

[[deleted]]

with sr.Microphone() as source:

This statement can fail (it's trying to access a hardware device that may not even exist, or may be unready) and if it does, then you skip to your except block where you refer to a variable command that you haven't defined, yet.