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无限期等待回应
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Stack Overflow用户
提问于 2022-05-15 14:52:29
回答 3查看 1.3K关注 0票数 2

当我创建一个按钮并处理回调或发送消息,等待与我的python不一致的反应,这似乎是有限的时间。有时在1小时后,机器人就不再登记反应了。可以肯定的是,一旦我重新启动机器人,连接就会丢失,并且它将不再注册交互。

然而,我已经看到机器人在不和谐的情况下,总是对一个按钮做出反应,不管这个按钮是在多久以前创建的。有办法做到这一点吗?我是否必须定期“重新连接”机器人到它创建的按钮?

简单的例子:

代码语言:javascript
复制
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())

在本例中,->将不再对按钮单击作出反应,过了一段时间后,或者我重新启动了机器人。

EN

回答 3

Stack Overflow用户

回答已采纳

发布于 2022-05-15 15:33:51

你可以这样做:

代码语言:javascript
复制
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")

这段代码来自disnake储存库

票数 1
EN

Stack Overflow用户

发布于 2022-05-16 17:47:54

我认为最有可能的是你的互联网有连接问题,而且你在没有注意到的情况下断开了连接。

这种情况发生在我身上,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,这样我就能够检查这是否是问题的根源。

代码语言:javascript
复制
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()))

就像这样的基本东西帮助我们发现问题在于我的机器断开了连接。这是我的互联网的一个物理问题,而不是编码错误或对api或库的误解。

票数 0
EN

Stack Overflow用户

发布于 2022-08-10 07:44:50

等待inter.response.defer(在这里输入图像描述)

票数 0
EN
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接:

https://stackoverflow.com/questions/72249395

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