Discord.py随机表情包命令开发遇404错误求助

你在开发Discord.py的随机表情包命令时，两种实现方式都触发了相同错误：

discord.app_commands.errors.CommandInvokeError: Command 'test' raised an exception: NotFound: 404 Not Found (error code: 10015): Unknown Webhook

你的两段命令代码：

async def meme(interaction: discord.Interaction):
    content = requests.get("https://meme-api.com/gimme").text
    data = json.loads(content,)
    meme = discord.Embed(title=f"{data['title']}")
    await interaction.followup.send(embed=meme)

和

@bot.tree.command()
async def test(interaction: discord.Interaction):
    embed = discord.Embed(title='Meme', description=None)

    async with aiohttp.ClientSession() as cs:
        async with cs.get('https://www.reddit.com/r/wholesomememes/new.json?sort=hot') as r:
            res = await r.json()
            
            embed.set_image(url=res['data']['children'] [random.randint(0, 25)]['data']['url'])
            
            await interaction.followup.send(embed=embed, content=None)

错误原因 #

问题出在你直接调用了interaction.followup.send()，但没有先对初始交互发送响应。Discord要求Slash命令必须在3秒内完成初始响应（比如发送确认消息或延迟通知），否则后续的followup请求会找不到对应的Webhook会话，从而触发404错误。

修正方案 #

给两段代码都加上初始响应逻辑即可解决：

第一段代码修正版

async def meme(interaction: discord.Interaction):
    # 先发送延迟响应，告知Discord命令正在处理
    await interaction.response.defer()
    # 建议替换同步requests为异步aiohttp，避免阻塞事件循环
    async with aiohttp.ClientSession() as cs:
        async with cs.get("https://meme-api.com/gimme") as r:
            data = await r.json()
    meme = discord.Embed(title=f"{data['title']}")
    await interaction.followup.send(embed=meme)

第二段代码修正版

@bot.tree.command()
async def test(interaction: discord.Interaction):
    # 先发送延迟响应
    await interaction.response.defer()
    embed = discord.Embed(title='Meme')

    async with aiohttp.ClientSession() as cs:
        async with cs.get('https://www.reddit.com/r/wholesomememes/new.json?sort=hot') as r:
            res = await r.json()
            
            embed.set_image(url=res['data']['children'][random.randint(0, 25)]['data']['url'])
            
            await interaction.followup.send(embed=embed)

额外提示 #

• 如果你的命令能在3秒内完成处理，也可以直接用await interaction.response.send_message()发送最终结果，无需使用followup
• 务必确保初始响应（defer()或send_message()）在任何耗时操作（比如网络请求）之前执行，避免超过Discord的3秒响应超时限制

内容的提问来源于stack exchange，提问作者Fraser