messageid - 在热解图中什么是 message_id？

Question

我需要导入（从 pyrogram.types 导入消息）但是我应该给“消息”什么值？据我所知，我应该给它“message_id”，但我不知道如何获取它，

import asyncio
import os
from datetime import datetime

import ffmpeg
import pyrogram
import pytgcalls
from pyrogram import Client, filters
from pyrogram.types import Message
from pytgcalls import GroupCallFactory, GroupCallFileAction, GroupCall

GROUP_CALL = None
SECONDS_TO_RECORD = 120
CLIENT_TYPE = pytgcalls.GroupCallFactory.MTPROTO_CLIENT_TYPE.PYROGRAM
API_HASH = ""
API_ID = 
Proxy = dict(
        hostname="",
        port=
    )
pyro_client = pyrogram.Client(
        os.environ.get('SESSION_NAME', 'pytgcalls'),
        api_hash=os.environ.get('API_HASH', API_HASH),
        api_id=os.environ.get('API_ID', API_ID),
    )

client_message = Message(client=pyro_client,message_id=???)


@Client.on_message(
    filters.group & filters.text & filters.outgoing & ~filters.edited & filters.command('record', prefixes='!')
)
async def record_from_voice_chat(client: Client, m: Message):
    global GROUP_CALL
    if not GROUP_CALL:
        GROUP_CALL = GroupCallFactory(client,CLIENT_TYPE ,path_to_log_file='c:/users').get_file_group_call()

    GROUP_CALL.add_handler(network_status_changed_handler, GroupCallFileAction.NETWORK_STATUS_CHANGED)
    await GROUP_CALL.start(m.chat.id)
    await m.delete()