只打印查询

Question

我正在使用 Discords API 制作一个 Discord 机器人。我需要在文本中打印某些部分。例如，如果输入命令是!play songname，它应该打印songname；如果输入是 !play "query"，它应该打印 "query"。

到目前为止，我有以下代码：

import discord
from discord.ext import commands
from discord.ext.commands import Bot


TOKEN = # redacted

client = commands.Bot(command_prefix = '!')
 
@client.event
async def on_ready():
  print("Qaki.us-MB-J&L")


@client.command(pass_content=True)
async def play (ctx):

如何让它打印命令的适当部分？

编辑：

 
import discord
import time
import spotipy
from discord.ext import commands
from discord.ext.commands import Bot
from spotipy.oauth2 import SpotifyClientCredentials

Word = 'word'

sp = spotipy.Spotify(auth_manager=SpotifyClientCredentials(client_id="ID", client_secret="ID"))

TOKEN = ("ID")

bot = commands.Bot(command_prefix = '!')
 
Songname = 'word'

@bot.event
async def on_ready():
  print("Qaki.us-MB-J&L")
  print(Songname)

@bot.command()
async def test(ctx, arg):
  Songname = (arg)
  
  if Word == arg:
    await ctx.send('Fail')
    return
  
    await ctx.send('Now playing {}'.format (arg))
  
  
  results = sp.search(q= arg, limit=20)
  for idx, track in enumerate(results['tracks']['items']):
    print(idx, track['name'])

      

bot.run(TOKEN)

找出args是什么后开始工作

Answer 1

您想要做的是使用*, arg。这显示在 discord.py documentation 中。使用此命令将为您提供发起命令的人提供的完整“查询”。

@client.command()
async def test(ctx, *, arg):
    await ctx.send(arg) # sending in channel that the command was initiated in
    print(arg) # printing to console

要查看此内容的其他部分，例如特定参数、单字参数等，请从 the top of the linked page 开始。