如何检查不和谐机器人是否是用户输入命令

Question

这是我的代码

import discord
import asyncio
import sys
import commands
import pickle
import os
import random
import array
#if len(sys.argv) != 2:
#    print('Usage: python3 main.py [token]')

commandz=("yes","no","hi","hi")
try:
    with open(os.path.join(os.path.dirname(__file__), "token.pickle"), 'rb') as file:
        token = pickle.load(file)
    if len(token) == 59:
        key = int(random.random() * 10000000000000000)
        print('Found saved token in stored.py, use phrase tokenreset'+str(key), 'to undo this.')#youll need to add the command
    else:
        raise
except:
    #code tokenreset with admin permissions
    token = input('What is your Discord bot token? (found on Discord developer page): ')
    with open(os.path.join(os.path.dirname(__file__), "token.pickle"), 'wb') as file:
        pickle.dump(token, file)
client = discord.Client()

@client.event
async def on_ready():
    print('Logged in as' + '[' + client.user.id + ']' + client.user.name)
    print('--------')


@client.event
async def on_message(message):
        await handle_command(message)


async def handle_command(message):
    print('Noticed: ' + message.content)
    if message.content == 'tokenreset'+str(key):
        await client.send_message(message.channel, 'code accepted')
    i = 0
    if str(message.author) == "NOTAKOALAINVENEZUELA#6895":
        i = 1
    x = 0
    await client.send_message(message.channel, message.author)
    while i == 0:
        if commandz[x] in message.content:
            x = x + 1
            await client.send_message(message.channel, commandz[x])
            i = i + 1
        else:
            if x == len(commandz) - 2:
                i = i + 1
            else:
                x = x + 2
client.run(token)

结果是我的机器人一遍又一遍地触发自己的命令，我已经尽一切努力让它无法识别自己的消息，而我束手无策。

我认为由于某种原因 message.author 是一种无法用于比较的奇怪变量，我对编程和 python 非常陌生，所以我不确定。

Answer 1

你可以尝试使用这个来检查作者是否是机器人：

if message.author.id === message.server.me.id

这样您就可以确保机器人不会自行触发。