Error

[Marsi591]

I get this error at every Model

[2023-06-16 13:41:17] [ERROR ] discord.app_commands.tree: Ignoring exception in command 'create'
Traceback (most recent call last):
File "C:\Users\Desktop\gpt4free-discord-main\main.py", line 66, in say
completion = client.send_message(models[model],
KeyError: 'claude'

During handling of the above exception, another exception occurred:

Traceback (most recent call last):
File "C:\Users\AppData\Local\Programs\Python\Python39\lib\site-packages\discord\app_commands\commands.py", line 842, in _do_call
return await self._callback(interaction, **params) # type: ignore
File "C:\Users\Desktop\gpt4free-discord-main\main.py", line 76, in say
await interaction.response.send_message(f'an error occured: {e}')
File "C:\Users\AppData\Local\Programs\Python\Python39\lib\site-packages\discord\interactions.py", line 751, in send_message
raise InteractionResponded(self._parent)
discord.errors.InteractionResponded: This interaction has already been responded to before

The above exception was the direct cause of the following exception:

Traceback (most recent call last):
File "C:\Users\AppData\Local\Programs\Python\Python39\lib\site-packages\discord\app_commands\tree.py", line 1248, in _call
await command._invoke_with_namespace(interaction, namespace)
File "C:\Users\AppData\Local\Programs\Python\Python39\lib\site-packages\discord\app_commands\commands.py", line 867, in _invoke_with_namespace
return await self._do_call(interaction, transformed_values)
File "C:\Users\AppData\Local\Programs\Python\Python39\lib\site-packages\discord\app_commands\commands.py", line 860, in _do_call
raise CommandInvokeError(self, e) from e
discord.app_commands.errors.CommandInvokeError: Command 'create' raised an exception: InteractionResponded: This interaction has already been responded to before
ERROR:discord.app_commands.tree:Ignoring exception in command 'create'
Traceback (most recent call last):
File "C:\Users\Desktop\gpt4free-discord-main\main.py", line 66, in say
completion = client.send_message(models[model],
KeyError: 'claude'

During handling of the above exception, another exception occurred:

Traceback (most recent call last):
File "C:\Users\AppData\Local\Programs\Python\Python39\lib\site-packages\discord\app_commands\commands.py", line 842, in _do_call
return await self._callback(interaction, **params) # type: ignore
File "C:\Users\Desktop\gpt4free-discord-main\main.py", line 76, in say
await interaction.response.send_message(f'an error occured: {e}')
File "C:\Users\AppData\Local\Programs\Python\Python39\lib\site-packages\discord\interactions.py", line 751, in send_message
raise InteractionResponded(self._parent)
discord.errors.InteractionResponded: This interaction has already been responded to before

The above exception was the direct cause of the following exception:

Traceback (most recent call last):
File "C:\Users\AppData\Local\Programs\Python\Python39\lib\site-packages\discord\app_commands\tree.py", line 1248, in _call
await command._invoke_with_namespace(interaction, namespace)
File "C:\Users\AppData\Local\Programs\Python\Python39\lib\site-packages\discord\app_commands\commands.py", line 867, in _invoke_with_namespace
return await self._do_call(interaction, transformed_values)
File "C:\Users\AppData\Local\Programs\Python\Python39\lib\site-packages\discord\app_commands\commands.py", line 860, in _do_call
raise CommandInvokeError(self, e) from e
discord.app_commands.errors.CommandInvokeError: Command 'create' raised an exception: InteractionResponded: This interaction has already been responded to before

[QuartzWarrior]

Hey! by chance are you entering "claude" for the model when doing the /say command? KeyError means that whatever model was inputted was not found in the models dictionary. If you had entered "claude", well it is obviously not there. try "claude-v1" or "claude-instant" next time.

Models Dictionary

models = {
    'claude-v1': 'a2_2',
    'claude-instant': 'a2',
    'claude-instant-100k': 'a2_100k',
    'sage': 'capybara',
    'gpt-4': 'beaver',
    'gpt-3.5-turbo': 'chinchilla',
}