discord.ui.查看超时

Question

按下时间已到的按钮时，提示“时间已到，请重复写入的命令”。如何在discord.py上实现它？

我尝试阅读文档，但没有任何帮助，我知道有些机器人有这样的功能

我尝试过这样的

import discord
from discord.ext import commands

class MyView(discord.ui.View):
    def __init__(self, timeout=10):
        super().__init__(timeout=timeout)
        self.timeout_reached = False

    async def on_timeout(self):
        self.timeout_reached = True

    async def on_error(self, error, item, interaction: discord.Interaction):
        if self.is_finished():
            await interaction.followup.send('Times up!', ephemeral=True)
        else:
            await interaction.followup.send('An error occurred!', ephemeral=True)

    @discord.ui.button(label='Random Button', style=discord.ButtonStyle.primary)
    async def random_button(self, interaction: discord.Interaction, button: discord.ui.Button):
        if self.is_finished():
            await interaction.response.send_message('Times up!', ephemeral=True)
        else:
            await interaction.response.send_message('The button has been pressed!', ephemeral=True)

class ExampleCog(commands.Cog):
    def __init__(self, bot):
        self.bot = bot

    @commands.command()
    async def random_button(self, ctx):
        view = MyView()
        await ctx.send('Press the button!', view=view)

async def setup(bot):
    await bot.add_cog(ExampleCog(bot))

Answer 1

遗憾的是，

on_timeout

方法不需要任何参数。您可以将最新的交互存储在实例变量中，然后使用它来发送消息。

class MyView(discord.ui.View):
    def __init__(self, timeout: int = 10) -> None:
        super().__init__(timeout=timeout)
        self.latest_interaction = None

    async def on_timeout(self) -> None:
        if self.latest_interaction is not None:
            await self.latest_interaction.followup.send("Retry")

    async def interaction_check(self, interaction: Interaction) -> bool:
        result = await super().interaction_check(interaction)
        # store the interaction in our instance variables
        self.latest_interaction = interaction
        if not result:
            await interaction.response.defer()
        return result   

    async def on_error(self, error, item, interaction: discord.Interaction) -> None:
        await interaction.followup.send('An error occurred!', ephemeral=True)

    @discord.ui.button(label='Random Button', style=discord.ButtonStyle.primary)
    async def random_button(self, interaction: discord.Interaction, button: discord.ui.Button) -> None:
        ...

参考资料：

• View.on_timeout

• View.interaction_check