这里是否有重复的“或”语句(python)的替代方法?

因此,在使用if message.content.startswith((prefix + smalltalk0[0]) or (prefix + smalltalk0[1]) or (prefix + smalltalk0[2])):时,这里似乎有问题,而if语句似乎只在第一个条件下工作:(prefix + smalltalk0[0]),(我在其中写“ sab hi”,它会做出响应。但是由于某种原因,它不会响应“ sab hello”或“ sab hey”。)

有人可以在这里帮助我吗,我猜我的错误与使用倍数或语句有关,但请更正其他可能出错的地方,非常感谢!

这是我所有的代码(是的,我确实知道help命令还没有完全完成并且可以运行,大声笑,我只是被位于代码末尾的一小段讨论区所困扰!):

import discord
import random
import asyncio

TOKEN = '*insert token here'

client = discord.Client()

prefix = ("sab ")



@client.event
async def on_message(message):
    if message.author == client.user:
        return

    if message.content.startswith(prefix + "help"):
        _help = "{0.author.mention}...Check your PMs".format(message)
        await message.channel.send("Prefix: 'sab '(yes,including the space after)\nGENERAL COMMANDS\n------------\n'help': Sends some stuff on how to use the bot.\n'tell a joke': Tells a rubbish joke...".format(author))
        await message.channel.send(_help)
    if message.content.startswith(prefix + "tell a joke"):



        joke = ["What did the grape do when he got stepped on?","Why couldnt the bicycle stand up by itself?","How did the frog die?","What do you call a can opener that doesn't work?","Why does he want his job to be cleaning mirrors?","What does a clock do when it's hungry?","Why do sea-gulls fly over the sea?"]

        answer = ["He let out a little wine!","Because it's two-tired!","He Kermit suicide!","A can't opener!","Because it's something he can really see himself doing!","It goes back four seconds!","Because if they flew over the bay they would be bagels!"]

        y = [0,1,2,3,4,5,6]
        x = random.choice(y)
        jokenum = joke[x]
        answernum = answer[x]
        msg = jokenum.format(message)
        msg2 = answernum.format(message)
        await asyncio.sleep(1)
        await message.channel.send(msg)
        await asyncio.sleep(4)
        await message.channel.send(msg2)

    colours = ["blue","orange","yellow","red","green" ]
    p = [0,4] 
    z = random.choice(p)
    colournum = colours[z]
    colourcorrect = str(colournum)
    if message.content.startswith(prefix + "play eye spy"):
        await message.channel.send("Eye spy with my little eyes,a certain colour!(Guess it!)")
    if colourcorrect in message.content:
        await message.channel.send("Correct," + colourcorrect + "!")







    smalltalk0 = ["hi","hello","hey"]
    q = [0,2]
    s = random.choice(q)
    smalltalk = smalltalk0[s]
    if message.content.startswith((prefix + smalltalk0[0]) or (prefix + smalltalk0[1]) or (prefix + smalltalk0[2])):
        await message.channel.send(smalltalk)

@client.event
async def on_ready():
    print('Sabios ready')
client.run(TOKEN)
shao1014 回答:这里是否有重复的“或”语句(python)的替代方法?

首先,这不是or的工作方式。实际上,您实际上需要重复startswith()调用,并用or分隔不同的调用:

if (
    message.content.startswith(prefix + smalltalk0[0])
    or message.content.startswith(prefix + smalltalk0[1])
    or message.content.startswith(prefix + smalltalk0[2])
):
    await message.channel.send(smalltalk)

但是您确实可以使用any来简化它:

if any(message.content.startswith(prefix + x) for x in smalltalk0):
    await message.channel.send(smalltalk)

或利用startswith()接受元组进行检查的事实:

if message.content.startswith(tuple(prefix + x for x in smalltalk0)):
    await message.channel.send(smalltalk)

但是,此功能中的 all 检查都使用前缀。那么,为什么一开始就不做这样的事情?

if message.author == client.user:
    return
if not message.content.startswith(prefix):
    return
content = message.content[len(prefix):]

像这样,您会得到content,其中仅包含前缀后的消息,然后可以简单地检查例如为此:

if message.content.startswith(tuple(smalltalk0)):
    ...
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