Pyrogram CallBack函数

Question

根据Pyrogram库文档，我试图用一个可调用的函数来设置"phone_code“参数，但是我总是得到这个错误，有人能帮我吗？

from pyrogram import Client
import time

def codePhone(phone_number):
    print('numero')
    sleep (10)
    return '2154'

app = Client("my_account",phone_number='39**********',phone_code=codePhone)
with app:
    app.send_message("me", "Hi!")

错误：

Pyrogram v1.0.7, Copyright (C) 2017-2020 Dan <https://github.com/delivrance>
Licensed under the terms of the GNU Lesser General Public License v3 or later (LGPLv3+)

The confirmation code has been sent via SMS
Traceback (most recent call last):
  File "/home/Topnews/Test/tdlib/prova3.py", line 10, in <module>
    with app:
  File "/usr/local/lib/python3.8/dist-packages/pyrogram/client.py", line 248, in __enter__
    return self.start()
  File "/usr/local/lib/python3.8/dist-packages/pyrogram/sync.py", line 56, in async_to_sync_wrap
    return loop.run_until_complete(coroutine)
  File "/usr/lib/python3.8/asyncio/base_events.py", line 616, in run_until_complete
    return future.result()
  File "/usr/local/lib/python3.8/dist-packages/pyrogram/methods/utilities/start.py", line 57, in start
    await self.authorize()
  File "/usr/local/lib/python3.8/dist-packages/pyrogram/client.py", line 327, in authorize
    signed_in = await self.sign_in(self.phone_number, sent_code.phone_code_hash, self.phone_code)
  File "/usr/local/lib/python3.8/dist-packages/pyrogram/methods/auth/sign_in.py", line 61, in sign_in
    r = await self.send(
  File "/usr/local/lib/python3.8/dist-packages/pyrogram/methods/advanced/send.py", line 77, in send
    r = await self.session.send(
  File "/usr/local/lib/python3.8/dist-packages/pyrogram/session/session.py", line 441, in send
    return await self._send(data, timeout=timeout)
  File "/usr/local/lib/python3.8/dist-packages/pyrogram/session/session.py", line 364, in _send
    message = self.msg_factory(data)
  File "/usr/local/lib/python3.8/dist-packages/pyrogram/session/internals/msg_factory.py", line 37, in __call__
    len(body)
  File "/usr/local/lib/python3.8/dist-packages/pyrogram/raw/core/tl_object.py", line 76, in __len__
    return len(self.write())
  File "/usr/local/lib/python3.8/dist-packages/pyrogram/raw/functions/auth/sign_in.py", line 81, in write
    data.write(String(self.phone_code))
  File "/usr/local/lib/python3.8/dist-packages/pyrogram/raw/core/primitives/string.py", line 31, in __new__
    return super().__new__(cls, value.encode())
AttributeError: 'function' object has no attribute 'encode'

Answer 1

对不起，我想我不完全理解这里的要求，但是从外观上看，你需要让它工作。因此，在这种情况下，您需要了解传递给函数的参数是什么。

由于codePhone函数接受一个参数(phone_number)，但在客户机类中，您只是将其作为回调引用发送，这似乎是不正确的。

此外，根据模块(Pyrogram)的文档，here指出phone_code参数需要是字符串(最有可能是数字字符串-> '2154')。所以试着用它来代替。

如果这对你有效，请让我知道。

TL;DR:使用

app = Client("my_account",phone_number='39**********',phone_code='2154')

Answer 2

您使用的是版本1.0.7，但support for callback functions for parameters for Client was removed in ver. 0.16.0

删除了对客户端参数的回调函数的支持，如phone_number、phone_code、password、…支持更简单和顺序的授权流程。

目前，它希望您为phone_code提供str。

Answer 3

Client类中的phone_number参数接受字符串，而不是函数。如果要自动执行电话号码登录过程：

from pyrogram import Client
from pyrogram.errors import SessionPasswordNeeded
import os

api_id    = input('API ID: ')
api_hash  = input('API HASH: ')
phone     = input('Phone Number (with +): ')

def connect():
    try:
        client = Client(phone, api_id, api_hash)
        client.connect()
        code   = client.send_code(phone)
        try:
            signed_in = client.sign_in(phone, code.phone_code_hash, input('Verification Code: '))
            client.accept_terms_of_service(signed_in.id)
        except SessionPasswordNeeded:
            client.check_password(input('Two-Step Verification password: '))
        client.disconnect()
    except Exception as ex:
        os.remove(f'{phone}.session')
        print(f'Error !\n\t`{type(ex).__name__}`\n\t{ex}')
        return

    with client as app:
        app.send_message('me', "Hey it's YOU!")

if __name__ == '__main__':
    connect()