Disnake/discord.py 无限期等待响应

Question

当我使用我的 python 机器人创建按钮并处理回调或发送消息并等待 discord 中的反应时，这似乎是时间有限的。 有时大约 1 小时后，机器人不再记录反应。 当然，一旦我重新启动机器人，连接就会丢失，它不会再注册交互了。

但是，我在 discord 中看到机器人总是对按钮做出反应，无论该按钮是多久之前创建的。 有没有办法做到这一点？ 我是否必须定期将机器人“重新连接”到它创建的按钮？

简单的例子：

class ButtonView(disnake.ui.View):
    def __init__(self):
        super().__init__(timeout=None)

    @disnake.ui.button(label="Hi", style=ButtonStyle.red)
    async def first_button(
        self, button: disnake.ui.Button, interaction: disnake.MessageInteraction
    ):
        await interaction.response.send_message("Button clicked.")

class Test(commands.Cog):
    def __init__(self, bot: commands.Bot):
        self.bot = bot
       
    @commands.slash_command() 
    async def test(self, inter):
        await inter.send("Button!", view=ButtonView())

-> 在此示例中，机器人在经过一段时间后不再对按钮单击做出反应，或者我重新启动了机器人。

Answer 1

你可以这样做：

import disnake
from disnake.ext import commands


# Define a simple View that persists between bot restarts
# In order a view to persist between restarts it needs to meet the following conditions:
# 1) The timeout of the View has to be set to None
# 2) Every item in the View has to have a custom_id set
# It is recommended that the custom_id be sufficiently unique to
# prevent conflicts with other buttons the bot sends.
# For this example the custom_id is prefixed with the name of the bot.
# Note that custom_ids can only be up to 100 characters long.
class PersistentView(disnake.ui.View):
    def __init__(self):
        super().__init__(timeout=None)

    @disnake.ui.button(
        label="Green", style=disnake.ButtonStyle.green, custom_id="persistent_view:green"
    )
    async def green(self, button: disnake.ui.Button, interaction: disnake.MessageInteraction):
        await interaction.response.send_message("This is green.", ephemeral=True)

    @disnake.ui.button(label="Red", style=disnake.ButtonStyle.red, custom_id="persistent_view:red")
    async def red(self, button: disnake.ui.Button, interaction: disnake.MessageInteraction):
        await interaction.response.send_message("This is red.", ephemeral=True)

    @disnake.ui.button(
        label="Grey", style=disnake.ButtonStyle.grey, custom_id="persistent_view:grey"
    )
    async def grey(self, button: disnake.ui.Button, interaction: disnake.MessageInteraction):
        await interaction.response.send_message("This is grey.", ephemeral=True)


class PersistentViewBot(commands.Bot):
    def __init__(self):
        super().__init__(command_prefix=commands.when_mentioned)
        self.persistent_views_added = False

    async def on_ready(self):
        if not self.persistent_views_added:
            # Register the persistent view for listening here.
            # Note that this does not send the view to any message.
            # In order to do this you need to first send a message with the View, which is shown below.
            # If you have the message_id you can also pass it as a keyword argument, but for this example
            # we don't have one.
            self.add_view(PersistentView())
            self.persistent_views_added = True

        print(f"Logged in as {self.user} (ID: {self.user.id})")
        print("------")


bot = PersistentViewBot()


@bot.command()
@commands.is_owner()
async def prepare(ctx: commands.Context):
    """Starts a persistent view."""
    # In order for a persistent view to be listened to, it needs to be sent to an actual message.
    # Call this method once just to store it somewhere.
    # In a more complicated program you might fetch the message_id from a database for use later.
    # However this is outside of the scope of this simple example.
    await ctx.send("What's your favourite colour?", view=PersistentView())


bot.run("token")

此代码来自disnake 存储库

Answer 2

我认为您的互联网很可能存在连接问题，并且您在没有注意到的情况下断开连接。

这发生在我身上，所以我添加了 on_disconnect 和 on_resumed 机器人事件，它们只是简单的打印语句，以便我能够检查这是否是问题的根源。

bot: commands.Bot = commands.Bot(command_prefix='.', intents=intents)
bot.time = time.time()

@bot.event
async def on_ready():
    print(f"Logged in as {bot.user} (ID: {bot.user.id})")
    print('TesterBot is ready starting at ' + time.ctime(bot.time))

@bot.event
async def on_disconnect():
    uptimedelta = time.time() - bot.time
    print('TesterBot is disconnected at ' + time.ctime(bot.time) + '. Testerbot has been up for ' + str(datetime.timedelta(seconds=uptimedelta)))

@bot.event
async def on_resumed():
    uptimedelta = time.time() - bot.time
    print("TesterBot reconnected " + time.ctime(bot.time) + '. Testerbot has been up for ' + str(datetime.timedelta(seconds=uptimedelta)))

@bot.event
async def on_connect():
    print('TesterBot is connected starting at ' + time.ctime(time.time()))

像这样的基本东西有助于揭示问题在于我的机器正在断开连接。 这是我的互联网的物理问题，而不是编码错误或对 api 或库的误解。

Answer 3

等待 inter.response.defer（在此处输入图像描述）