如何修复Discord.py中‘TypeError: Menu2.button_callback()缺少必填参数button’错误？

错误出在Menu2类手动绑定按钮回调的方式上：当使用@discord.ui.button装饰器时，discord.py会自动将按钮实例作为第二个参数传入回调函数，但手动给button.callback赋值时，回调函数只会收到interaction这一个参数，导致你定义的button_callback要求的第二个button参数缺失，触发TypeError。

以下是两种可行的解决方法：

方案一：修改回调函数参数，移除button参数 #

如果不需要在回调里直接调用按钮实例，可直接删除button参数；若需要按钮相关信息，也可通过interaction.component获取：

class Menu2(View):
    def __init__(self, ctx):
        super().__init__()
        self.ctx = ctx
      
        options = ["A", "B", "C", "D", "E", "F", "G", "H", "I", "J"]
        for option in options:
            button = discord.ui.Button(label=option, style=discord.ButtonStyle.green, emoji="📜")
            button.callback = self.button_callback
            self.add_item(button)
        
    async def interaction_check(self, interaction: discord.Interaction):
        if interaction.user == self.ctx.author:
            return True
        await interaction.response.send_message("You can't interact with this menu.", ephemeral=True)
        return False

    # 移除button参数，或通过interaction.component获取按钮信息
    async def button_callback(self, interaction: discord.Interaction):
        # 如需按钮信息，可取消注释下面这行
        # button = interaction.component
        await interaction.response.send_message('You Clicked it')

方案二：用functools.partial绑定button参数 #

如果必须在回调中直接使用按钮实例，可借助functools.partial将按钮对象绑定到回调函数中：

首先导入functools模块：

import functools

然后修改Menu2类的初始化逻辑：

class Menu2(View):
    def __init__(self, ctx):
        super().__init__()
        self.ctx = ctx
      
        options = ["A", "B", "C", "D", "E", "F", "G", "H", "I", "J"]
        for option in options:
            button = discord.ui.Button(label=option, style=discord.ButtonStyle.green, emoji="📜")
            # 用partial将button参数绑定到回调函数
            button.callback = functools.partial(self.button_callback, button=button)
            self.add_item(button)
        
    async def interaction_check(self, interaction: discord.Interaction):
        if interaction.user == self.ctx.author:
            return True
        await interaction.response.send_message("You can't interact with this menu.", ephemeral=True)
        return False

    # 保留原参数签名
    async def button_callback(self, interaction: discord.Interaction, button: discord.ui.Button):
        await interaction.response.send_message(f'You Clicked {button.label}!')

内容的提问来源于stack exchange，提问作者Jake Noguera