我当前的代码是
@client.event
async def on_member_join(member):
serverchannel = member.server.default_channel
msg = "Wuss poppin', {0}. Welcome to {1}".format(member.mention, member.server.name)
await client.send_message(member.server.default_channel, msg)`
@client.event
async def on_member_remove(member):
serverchannel = member.server.default_channel
msg = "Well. Cya, {0}!".format(member.mention)
await client.send_message(serverchannel, msg)
以及其他必要的行(如 import discord
等)和其他命令.
As well as the other necessary lines (like import discord
etc.) and other commands.
当成员加入或离开时,我收到此错误:
When a member joins or leaves, I get this error:
Ignoring exception in on_member_join
Traceback (most recent call last):
File "C:UsersWillLAppDataLocalProgramsPythonPython36libsite-packagesdiscordclient.py", line 307, in _run_event
yield from getattr(self, event)(*args, **kwargs)
File "C:/Users/WillL/Desktop/MemzBot-master/index.py", line 102, in on_member_join
await client.send_message(member.server.default_channel, msg)
File "C:UsersWillLAppDataLocalProgramsPythonPython36libsite-packagesdiscordclient.py", line 1145, in send_message
channel_id, guild_id = yield from self._resolve_destination(destination)
File "C:UsersWillLAppDataLocalProgramsPythonPython36libsite-packagesdiscordclient.py", line 289, in _resolve_destination
raise InvalidArgument(fmt.format(destination))
discord.errors.InvalidArgument: Destination must be Channel, PrivateChannel, User, or Object. Received NoneType
我不明白为什么,我错过了什么?
I don't understand why, what am I missing?
默认频道不再存在.这意味着两件事:
server.default_channel
可能返回 NoneType.server
发送消息可能会引发 discord.NotFound
.server.default_channel
may return a NoneType. server
may raise discord.NotFound
.这些情况将发生在 8 月 3 日之后创建的任何服务器上,或者任何已删除其默认"频道的服务器上.
These cases will happen on any servers created after August 3rd, or any servers who have deleted their "default" channel.
您需要自己选择一个频道,或者要求服务器为欢迎消息设置一个频道.您仍然可以尝试发送到服务器,但请确保您有一个失败的后备选项.
You need to either choose a channel yourself, or ask the server to set a channel for the welcome message. You can still attempt to send to the server, but make sure you have a fall back option if it fails.
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