如何发送不一致的消息。从事件循环外部（即从python电报bot线程）复制？

Question

我想使用python telegram bot和discord库制作一个机器人，在discord和telegram之间进行通信。py（版本1.0.0）。然而，问题在于这种不和谐。py使用异步函数和python电报bot线程。使用下面的代码，在discord中发布消息时一切正常（bot将消息正确地发送到telegram），但是另一种方法不起作用（bot从telegram获取消息并将其发送到discord）。我以前在尝试运行discords频道时遇到语法/运行时错误问题。在同步函数中发送函数（因此，要么仅返回生成器对象，要么抱怨无法在同步函数中使用wait）。然而，与此同时，python telegram bot的MessageHandler需要一个同步函数，因此当给他一个异步函数时，python抱怨异步函数从未调用“wait”。我现在尝试使用asgiref库中的async_to_sync方法从MessageHandler运行我的asyncbroadcastMsg，但是代码仍然不会将消息发送到discord！它似乎正确地调用了函数，但只有在第print行（“我到达这里”）之前才能调用。没有显示错误，也没有不一致的消息弹出。我想这与我必须在discord中将函数注册为任务这一事实有关。然而，py事件循环只有在botDiscord之前发生时才起作用。run（TOKENDISCORD）已经执行，当然之前必须执行。所以，把我的问题归结为一个问题：我如何才能与不和谐互动。来自另一个线程（来自电报MessageHandler）的py事件循环。或者，如果这是不可能的：我如何才能发送不和谐的消息。不要陷入不和谐。py事件循环？

谢谢你的帮忙

import asyncio
from asgiref.sync import async_to_sync
from telegram import Message as TMessage
from telegram.ext import (Updater,Filters,MessageHandler)
from discord.ext import commands
import discord

TChannelID = 'someTelegramChannelID'
DChannel = 'someDiscordChannelObject'

#%% define functions / commands
prefix = "?"
botDiscord = commands.Bot(command_prefix=prefix)
discordChannels = {}


async def broadcastMsg(medium,channel,message):
    ''' 
    Function to broadcast a message to all linked channels.
    '''
    if isinstance(message,TMessage):
        fromMedium = 'Telegram'
        author = message.from_user.username
        channel = message.chat.title
        content = message.text
    elif isinstance(message,discord.Message):
        fromMedium = 'Discord'
        author = message.author
        channel = message.channel.name
        content = message.content
    # check where message comes from        
    textToSend = '%s wrote on %s %s:\n%s'%(author,fromMedium,channel,content)
    # go through channels and send the message
    if 'telegram' in medium:
        # transform channel to telegram chatID and send
        updaterTelegram.bot.send_message(channel,textToSend)

    elif 'discord' in medium:
        print('I get to here')
        await channel.send(textToSend)

    print("I do not get there")


@botDiscord.event
async def on_message(message):
    await broadcastMsg('telegram',TChannelID,message)

def on_TMessage(bot,update):
    # check if this chat is already known, else save it
    # get channels to send to and send message
    async_to_sync(broadcastMsg)('discord',DChannel,update.message)


#%% initialize telegram and discord bot and run them
messageHandler = MessageHandler(Filters.text, on_TMessage)
updaterTelegram = Updater(token = TOKENTELEGRAM, request_kwargs={'read_timeout': 10, 'connect_timeout': 10})
updaterTelegram.dispatcher.add_handler(messageHandler)
updaterTelegram.start_polling()
botDiscord.run(TOKENDISCORD)  

Answer 1

您可以尝试将它们拆分为两个单独的*。py文件。

t2d.py#电报到迪斯科尔

import subprocess
from telethon import TelegramClient, events, sync

api_id = '...'
api_hash = '...'

with TelegramClient('name', api_id, api_hash) as client:     
    @client.on(events.NewMessage()) #inside .NewMessage() you can put specific channel like: chats="test_channel"
    async def handler(event):        
        print('test_channel raw text: ', event.raw_text) #this row is not necessary

        msg = event.raw_text

        subprocess.call(["python", "s2d.py", msg])

    client.run_until_disconnected()

s2d。py#发送到discord

import discord, sys

my_secret = '...'

clientdiscord = discord.Client()

@clientdiscord.event
async def on_ready():
    #print('We have logged in as {0.user}'.format(clientdiscord)) #this row is not necessary

    channel = clientdiscord.get_channel(123456789) # num of channel where you want to write message

    msg = sys.argv[1] #grab message
    
    msg = 's2d: ' + msg #only for test, you can delete this row 
    
    await channel.send(msg)

    quit() # very important quit this bot 

    
clientdiscord.run(my_secret)

它会稍微慢一点（子进程会造成延迟），但很容易解决

Answer 2

我怎样才能发送不和谐的信息。不要陷入不和谐。py事件循环？

要从事件循环线程外部安全地调度协同路由，请使用asyncio。运行线程安全的协同程序：

_loop = asyncio.get_event_loop()

def on_TMessage(bot, update):
    asyncio.run_coroutine_threadsafe(
        broadcastMsg('discord', DChannel, update.message), _loop)