如何获得多个在线会员？

Question

我有一个命令，对成员和机器人进行计数，并分别输出它们。然后我想输出在线用户的数量。这个是可能的吗？

此命令获取成员和bot计数。

if message.content.startswith('<count'):    
        membersInServer = message.guild.members
        channel = message.channel
        # Filter to the list, returns a list of bot-members
        botsInServer = list(filter(filterOnlyBots, membersInServer))
        botsInServerCount = len(botsInServer)
        # (Total Member count - bot count) = Total user count
        usersInServerCount = message.guild.member_count - botsInServerCount
        msg = discord.Embed(title="Amount of Human Members in this Discord:", description=usersInServerCount, color=0x00FD00)
        msg.add_field(name="Amount of Bot Users in this Discord:",value=botsInServerCount, inline=False)
        await channel.send(embed=msg)
def filterOnlyBots(member):
    return member.bot

我已经试过client.member.status了，那只会回到网上

Answer 1

每个成员具有状态属性。，您可以使用它来检查是否有状态脱机。

然后，可以通过脱机状态筛选membersInServer。

        onlineMembersInServer = list(filter(filterOnlyOnlineMembers, membersInServer))

        onlineMembersCount = len(onlineMembersInServer)

# Somewhere...
def filterOnlyOnlineMembers(member):
    return member.status != 'offline'

请注意，它计算在线用户和机器人。

如果希望只对联机用户进行筛选，则可以将筛选器更改为：

# Set the filter to be a non-offline member, and the member not being a bot.
def filterOnlyOnlineMembers(member):
    return member.status != 'offline' and not member.bot

请注意，如果服务器很大，这可能会导致性能问题。

编辑

正如Patrick所提到的，你可以把它做成1条线。

sum(member.status!=discord.Status.offline and not member.bot for member in message.guild.members)