如何将 ID 更改为 Ping (PyCharm, Python 3.9)
How to change ID to Ping (PyCharm, Python 3.9)
我做了这个狙击命令(来自 Whosebug 的一个问题)但是当机器人回答时,标题说的是被删除消息的作者的 ID,而不是 ping 或名字。我该如何更改?
代码:
snipe_message_content = None
snipe_message_author = None
snipe_message_id = None
@bot.event
async def on_message_delete(message):
global snipe_message_content
global snipe_message_author
global snipe_message_id
snipe_message_content = message.content
snipe_message_author = message.author.id
snipe_message_id = message.id
await asyncio.sleep(60)
if message.id == snipe_message_id:
snipe_message_author = None
snipe_message_content = None
snipe_message_id = None
@bot.command()
async def snipe(message):
if snipe_message_content is None:
await message.channel.send("Theres nothing to snipe.")
else:
embed = discord.Embed(description=f"{snipe_message_content}")
embed.set_footer(text=f"Asked by {message.author.name}#{message.author.discriminator}", icon_url=message.author.avatar_url)
embed.set_author(name=f"<@{snipe_message_author}>")
await message.channel.send(embed=embed)
return
遗憾的是,您无法在嵌入的 author
部分对用户执行 ping 操作。但是,您可以说出他们的名字,这就是我们要做的。
作者姓名
获取作者姓名就像将 snipe_message_author = message.author.id
(获取 ID)更改为 snipe_message_author = message.author.name
.
一样简单
显示作者姓名
要以良好的格式显示作者姓名,请从显示姓名的位置删除“<@>”。
代码
@client.event
async def on_message_delete(message):
global snipe_message_content
global snipe_message_author
global snipe_message_id
snipe_message_content = message.content
snipe_message_author = message.author.name
snipe_message_id = message.id
await asyncio.sleep(60)
if message.id == snipe_message_id:
snipe_message_author = None
snipe_message_content = None
snipe_message_id = None
@client.command()
async def snipe(message):
if snipe_message_content is None:
await message.channel.send("Theres nothing to snipe.")
else:
embed = discord.Embed(description=f"{snipe_message_content}")
embed.set_footer(text=f"Asked by {message.author.name}#{message.author.discriminator}", icon_url=message.author.avatar_url)
embed.set_author(name=f"{snipe_message_author}")
await message.channel.send(embed=embed)
return
输出
编辑
不小心泄露了我的不和谐名称和鉴别器,将鉴别器更改为随机整数。代码仍然有效。
我做了这个狙击命令(来自 Whosebug 的一个问题)但是当机器人回答时,标题说的是被删除消息的作者的 ID,而不是 ping 或名字。我该如何更改?
代码:
snipe_message_content = None
snipe_message_author = None
snipe_message_id = None
@bot.event
async def on_message_delete(message):
global snipe_message_content
global snipe_message_author
global snipe_message_id
snipe_message_content = message.content
snipe_message_author = message.author.id
snipe_message_id = message.id
await asyncio.sleep(60)
if message.id == snipe_message_id:
snipe_message_author = None
snipe_message_content = None
snipe_message_id = None
@bot.command()
async def snipe(message):
if snipe_message_content is None:
await message.channel.send("Theres nothing to snipe.")
else:
embed = discord.Embed(description=f"{snipe_message_content}")
embed.set_footer(text=f"Asked by {message.author.name}#{message.author.discriminator}", icon_url=message.author.avatar_url)
embed.set_author(name=f"<@{snipe_message_author}>")
await message.channel.send(embed=embed)
return
遗憾的是,您无法在嵌入的 author
部分对用户执行 ping 操作。但是,您可以说出他们的名字,这就是我们要做的。
作者姓名
获取作者姓名就像将 snipe_message_author = message.author.id
(获取 ID)更改为 snipe_message_author = message.author.name
.
显示作者姓名
要以良好的格式显示作者姓名,请从显示姓名的位置删除“<@>”。
代码
@client.event
async def on_message_delete(message):
global snipe_message_content
global snipe_message_author
global snipe_message_id
snipe_message_content = message.content
snipe_message_author = message.author.name
snipe_message_id = message.id
await asyncio.sleep(60)
if message.id == snipe_message_id:
snipe_message_author = None
snipe_message_content = None
snipe_message_id = None
@client.command()
async def snipe(message):
if snipe_message_content is None:
await message.channel.send("Theres nothing to snipe.")
else:
embed = discord.Embed(description=f"{snipe_message_content}")
embed.set_footer(text=f"Asked by {message.author.name}#{message.author.discriminator}", icon_url=message.author.avatar_url)
embed.set_author(name=f"{snipe_message_author}")
await message.channel.send(embed=embed)
return
输出
编辑
不小心泄露了我的不和谐名称和鉴别器,将鉴别器更改为随机整数。代码仍然有效。