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pythondiscord.pypycord

Python py-cord / discord.py(2.0.0): Button interaction failed

If I click on my Button everything works as expected, but in Discord is's showing:

enter image description here

This is the Button Code:

# Start Button
start_b = Button(label="Starten", style=discord.ButtonStyle.gray)

async def start_callback(interaction: discord.Interaction):
    if not is_server_running():
        msg = await interaction.channel.send(embed=discord.Embed(title='START: Starte Server'))
        start_server()
        await msg.delete()
    else:
        msg = await interaction.channel.send(embed=discord.Embed(title='Server läuft noch'))
        time.sleep(5)
        await msg.delete()

start_b.callback = start_callback

...

view.add_item(start_b)
await bot.get_channel(Some ID).send(embed=discord.Embed(title="Dashboard"), view=view)

I guess I have to acknowledge the button interaction somehow. Is there a way to do this without sending reply messages?

Solution

  • Finally I found it 🥳

    You have to add this at the top of your callback:

    await interaction.response.defer()

    In my example this would be:

    # Start Button
start_b = Button(label="Starten", style=discord.ButtonStyle.gray)

async def start_callback(interaction: discord.Interaction):

await interaction.response.defer() # <---- Here

    if not is_server_running():
        msg = await interaction.channel.send(embed=discord.Embed(title='START: Starte Server'))
        start_server()
        await msg.delete()
    else:
        msg = await interaction.channel.send(embed=discord.Embed(title='Server läuft noch'))
        time.sleep(5)
        await msg.delete()

start_b.callback = start_callback

...

view.add_item(start_b)
await bot.get_channel(Some ID).send(embed=discord.Embed(title="Dashboard"), view=view)