不一致当用户开始流媒体时，py使bot发送消息

Question

当我在twitch上开始直播时，我正试图让我的机器人向指定频道发送消息。到目前为止，我还在忙于从我的状态中获取正确的“活动”，这表示我正在流媒体。到目前为止，我得到的是：

class Stream(commands.Cog):
    def __init__(self, bot):
        self.bot = bot
        
    
    @commands.Cog.listener()
    async def on_ready(self):
        print("Streamingstatus loaded")
    
    
    @commands.Cog.listener()
    async def on_member_update(self, before, after):
        channel = self.bot.get_channel (751954378772185149)
        if after.activity is not None and after.activity.name == '#Twitch or so?' and after.id == 254375451084980224 and guild.id == 670015807551438874:
                print(before.activity)
                print(after.activity)
                if after.activity.title.startswith(#streaming) and before.activity.title != after.activity.title:
                        await channel.send("Spleens went live on twitch right now! Watch him making a fool out of himself!")

提前感谢任何帮助！^^

Answer 1

您的问题是您正在检查活动。name而正确的方法是检查活动。键入。此外，要检查用户是否正在流媒体，您需要检查活动是否正常。类型是不协调。ActivityType。流媒体。我不完全确定您试图在代码的某些行中实现什么，但要检查用户是否正在流式处理，请执行以下操作：

@commands.Cog.listener()
async def on_member_update(self, before, after):
    try:
        activity_type = after.activity.type
    except:
        pass
    # Here we check if the user is streaming
    if activitity_type is discord.ActivityType.streaming:
        # Do X if he is streaming
    else:
        # Do Y if he is not streaming