Discord机器人模态框报告命令报错：'Context'无'response'属性

问题代码 #

我正在开发Discord机器人，编写了一个基于模态框的报告命令，代码如下：

class ReportModal(discord.ui.Modal, title="Репорт юзер"):
    user_name = discord.ui.TextInput(label="Юзер дискорд наме", placeholder="eg. Jgfhfhjfkfk#0000", required=True, max_length=10000, style=discord.TextStyle.short)
    user_id = discord.ui.TextInput(label="Юзер дискорд наме2", placeholder="eg. Jgfhfhjfkfk#0000", required=True, max_length=10000, style=discord.TextStyle.long)

@bot.command(name="report", description="Репорт юзер")
async def report(interaction: discord.Interaction):
    modal = ReportModal()
    await interaction.response.send_modal(modal)

报错信息 #

运行时出现如下错误：

File "/home/container/.local/lib/python3.11/site-packages/discord/ext/commands/core.py", line 235, in wrapped
    ret = await coro(*args, **kwargs)
          ^^^^^^^^^^^^^^^^^^^^^^^^^^^
  File "/home/container/main.py", line 26, in report
    await interaction.response.send_modal(modal)
          ^^^^^^^^^^^^^^^^^^^^
AttributeError: 'Context' object has no attribute 'response'

The above exception was the direct cause of the following exception:

Traceback (most recent call last):
  File "/home/container/.local/lib/python3.11/site-packages/discord/ext/commands/bot.py", line 1350, in invoke
    await ctx.command.invoke(ctx)
  File "/home/container/.local/lib/python3.11/site-packages/discord/ext/commands/core.py", line 1029, in invoke
    await injected(*ctx.args, **ctx.kwargs)  # type: ignore
    ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
  File "/home/container/.local/lib/python3.11/site-packages/discord/ext/commands/core.py", line 244, in wrapped
    raise CommandInvokeError(exc) from exc
discord.ext.commands.errors.CommandInvokeError: Command raised an exception: AttributeError: 'Context' object has no attribute 'response'

解决方案 #

问题根源 #

你错误地给**前缀命令（@bot.command）**的回调函数声明了Interaction类型参数。前缀命令实际传入的是Context对象，它没有response属性，而send_modal是Interaction对象的专属方法，因此触发报错。

方法一：改用斜杠命令（推荐） #

Discord官方更推荐使用斜杠交互命令，修改步骤如下：

1. 将@bot.command替换为@bot.tree.command
2. 确保机器人启动时同步命令到Discord服务器
3. 保留Interaction类型参数

修改后的完整代码：

import discord
from discord.ext import commands

intents = discord.Intents.default()
intents.message_content = True
bot = commands.Bot(command_prefix="!", intents=intents)

class ReportModal(discord.ui.Modal, title="Репорт юзер"):
    user_name = discord.ui.TextInput(label="Юзер дискорд наме", placeholder="eg. Jgfhfhjfkfk#0000", required=True, max_length=100, style=discord.TextStyle.short)
    user_id = discord.ui.TextInput(label="Юзер ID/Тег", placeholder="eg. 1234567890 или Jgfhfhjfkfk#0000", required=True, max_length=100, style=discord.TextStyle.short)

    # 添加模态框提交后的处理逻辑
    async def on_submit(self, interaction: discord.Interaction):
        await interaction.response.send_message(f"Репорт отправлен!\nЮзер: {self.user_name.value}\nID/Тег: {self.user_id.value}", ephemeral=True)

@bot.tree.command(name="report", description="Репорт юзер")
async def report(interaction: discord.Interaction):
    modal = ReportModal()
    await interaction.response.send_modal(modal)

# 同步命令到服务器（全局同步可能需要1小时， guild_ids指定服务器可即时生效）
@bot.event
async def on_ready():
    await bot.tree.sync()  # 全局同步
    # await bot.tree.sync(guild=discord.Object(id=你的服务器ID))  # 指定服务器同步
    print(f"Бот {bot.user} запущен!")

bot.run("你的机器人令牌")

方法二：前缀命令中通过按钮触发模态框（不推荐） #

如果坚持使用前缀命令，需要先发送带按钮的消息，用户点击按钮后触发模态框：

import discord
from discord.ext import commands

intents = discord.Intents.default()
intents.message_content = True
bot = commands.Bot(command_prefix="!", intents=intents)

class ReportButton(discord.ui.View):
    @discord.ui.button(label="Отправить репорт", style=discord.ButtonStyle.danger)
    async def report_button(self, interaction: discord.Interaction, button: discord.ui.Button):
        modal = ReportModal()
        await interaction.response.send_modal(modal)

class ReportModal(discord.ui.Modal, title="Репорт юзер"):
    user_name = discord.ui.TextInput(label="Юзер дискорд наме", placeholder="eg. Jgfhfhjfkfk#0000", required=True, max_length=100, style=discord.TextStyle.short)
    user_id = discord.ui.TextInput(label="Юзер ID/Тег", placeholder="eg. 1234567890 или Jgfhfhjfkfk#0000", required=True, max_length=100, style=discord.TextStyle.short)

    async def on_submit(self, interaction: discord.Interaction):
        await interaction.response.send_message(f"Репорт отправлен!\nЮзер: {self.user_name.value}\nID/Тег: {self.user_id.value}", ephemeral=True)

@bot.command(name="report", description="Репорт юзер")
async def report(ctx: commands.Context):
    await ctx.send("Нажмите кнопку ниже для отправки репорта:", view=ReportButton())

@bot.event
async def on_ready():
    print(f"Бот {bot.user} запущен!")

bot.run("你的机器人令牌")

内容的提问来源于stack exchange，提问作者zootmn15