通过用户名邀请用户进入电报频道

Question

我试图通过用户名将用户添加到我的频道。我正在使用python 3.6 telethon库和pythonanywhere服务器：

api_hash = 'b7**'
phone = '+7***'
client = TelegramClient('new_ses', api_id, api_hash)
client.connect()
client = TelegramClient('session_neworig', api_id, api_hash,)
client.connect()

from telethon.tl.functions.channels import InviteToChannelRequest
from telethon.tl.functions.contacts import ResolveUsernameRequest
from telethon.tl.types import InputPeerChannel ,InputPeerUser,InputUser
from telethon.tl.functions.channels import JoinChannelRequest

chann=client.get_entity('channelname') #its public channel
print(chann.id)
1161823752

print(chann.access_hash)
8062085565372622341

time.sleep(30)
chan=InputPeerChannel(chann.id, chann.access_hash)
user = client(ResolveUsernameRequest('Chai***'))


print(user.users[0].id)
193568760
print(user.users[0].access_hash)
-4514649540347033311
time.sleep(1*30)

user=InputUser(user.users[0].id,user.users[0].access_hash,)

client.invoke(InviteToChannelRequest(chan,[user]))

这项艰巨的工作，我得到了 -telethon.errors.rpc_error_list.PeerFloodError: (PeerFloodError(...), 'Too many requests')

我究竟做错了什么？如何避免呢？这段代码为我工作，但是添加了20个用户后，我就大吃一惊了：

 from telethon.helpers import get_input_peer

client.invoke(InviteToChannelRequest(
    get_input_peer(client.get_entity(chan),
    [get_input_peer(client.get_entity(user))]
))

请帮忙，如何通过用户名添加200个用户而没有任何禁止，也许还有另一种通过python做到这一点的方法？另一个库还是通过api？

Answer 1

from telethon.sync import TelegramClient
from telethon.tl.functions.channels import InviteToChannelRequest
from telethon.tl.types import InputPeerChannel ,InputUser
from telethon.tl.functions.contacts import ResolveUsernameRequest

api_id = ******
api_hash = '******'
phone = '+2519******'
client = TelegramClient(phone, api_id, api_hash);
client.connect()


chann=client.get_entity('channelname') 
channel_id = chann.id;
channel_access_hash = chann.access_hash

chanal=InputPeerChannel(channel_id, channel_access_hash)
user = client(ResolveUsernameRequest('mrmi6'))

user=InputUser(user.users[0].id,user.users[0].access_hash,)
client(InviteToChannelRequest(chanal,[user]))
print('action completted')