从 guild.fetch_members() 获取列表

Question

我想要一个线程成员列表

使用时：

channel.send(str(await message.channel.fetch_members()))

我明白了：

[<ThreadMember id=432287480599937024 thread_id=1185095420892958781 joined_at=datetime.datetime(2023, 12, 15, 5, 46, 6, 357000, tzinfo=datetime.timezone.utc)>, <ThreadMember id=1069317582559121419 thread_id=1185095420892958781 joined_at=datetime.datetime(2023, 12, 15, 5, 46, 7, 385000, tzinfo=datetime.timezone.utc)>]

如何将其转换为列表？

PS：我尝试执行 thread.members 但它不会返回任何人，并且根据文档：

A list of thread members in this thread.

This requires Intents.members to be properly filled. Most of the time however, this data is not provided by the gateway and a call to fetch_members() is needed.

Answer 1

我不知道我是否理解你的问题。你似乎迷失在术语中。

我相信您想发送对线程成员的提及，每行一个成员：

# this is the list of thread members
thread_members = await message.channel.fetch_members()

text = "These are the members of the Thread:\n"
for member in thread_members:
    text += f"- {member.mention}\n"

await message.reply(text)

Answer 2

fetch_members()

将返回ThreadMemeber

对象

的列表，每个对象代表线程的一个成员并包括它们的ID，它们所属的ID，以及他们加入线程的日期时间，提示您的输出。

您需要迭代

ThreadMember

对象并从中提取信息。

正如您提到的

thread.members

应该返回线程成员列表，但是，如果没有，您需要启用

Intents.Members

，或者网关不提供数据

thread.members

它将返回一个空列表，需要使用

fetch_members()

# Assuming `thread` is your thread object

members = await thread.fetch_members()
member_ids = [member.id for member in members]


# send the list of member IDs as a message 
await channel.send(str(member_ids))