The play function does not accept the url argument, what should I do?

Question

Python

Hello, I'm writing a discord music bot and I've run into a problem. I wrote the video queue and play code separately, and now I'm trying to combine them. To do this, I entered the url argument in brackets to def play, but it complains and displays an error

TypeError: play() missing 1 required keyword-only argument: 'url'

What to do?
Now my def start looks like this: Thanks in advance, and sorry if this is a dumb question.
async def play(ctx, *, command = None, url):

Answer 1

Hey! I have a counter question: do you write play in the main file or in cog?
There are 2 solutions:
- First, if you have kogi

@commands.command()
    async def play(self, ctx: commands.Context, url: str, *args: str):
        '''Adds a song to the queue either by YouTube URL or YouTube Search.'''

- Second, if you have everything in the main file

@client.command()
async def play(ctx, url : str):

Good luck writing the bot!