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由于交互,无法访问消息id。response.send_message()返回none,且不一致.py

discord.py discord python

GaGex · 技术社区 · 2 年前

@bot.tree.command(name='create_dungeon')
async def create_dungeon(interaction: discord.Interaction):
    new_doc_id = create_document(str(interaction.user))
    if new_doc_id == False:
        await interaction.response.send_message("Can't create more than one dungeon per user!")
    else:
        created_doc_query = maplestory_collection.find_one({"_id": new_doc_id})
        dps_players_count = len(created_doc_query["Dps"])
        tank_players_count = len(created_doc_query["Tank"])
        healer_players_count = len(created_doc_query["Healers"])
        class_types_max_players_and_counts = [
            ("Dps", dps_players_count, 10),
            ("Healer", healer_players_count, 3),
            ("Tank", tank_players_count, 4)
        ]

        embed = discord.Embed(title="Dungeon", description=f'Leader: {interaction.user} - ID {interaction.user.id}')
        for class_type, count, max_players in class_types_max_players_and_counts:
            embed.add_field(name=f"{class_type}", value=f"{max_players}/{count}")
        
        view = MyView(doc_id=new_doc_id, original_message_id=None, creator_id=interaction.user.id)
        view.create_buttons(buttons_info=buttons_info)
        
        message = await interaction.response.send_message(view=view, embed=embed)
        # message returns None even though it should return a Message object

Message var应该返回消息对象,但不返回任何对象?我试图获取消息的id,然后将其设置为view属性 original_message_id 但它输出错误TypeError