代码一次触发 2 个函数 |不和谐.py
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【中文标题】代码一次触发 2 个函数 |不和谐.py【英文标题】:The code is triggering 2 function at once | discord.py 【发布时间】:2020-12-17 03:49:45 【问题描述】:所以我尝试制作一个过滤系统,当警告小于 3 时,它可以正常工作,直到达到 4,这意味着(禁止时间)它会触发静音。这是代码
@commands.Cog.listener()
async def on_message(self, message):
if not message.author.bot:
if not message.author.guild_permissions.manage_messages:
for b in ignored:
if b in message.content.lower():
return
else:
self.cur.execute(f"SELECT text FROM badwords WHERE guild_id = message.guild.id")
result = self.cur.fetchall()
bword = [x[0] for x in result]
if any(b in bword for b in message.content.lower().split(" ")):
self.cur.execute(f"SELECT onoff from mutedet WHERE guild_id = message.guild.id")
result3 = self.cur.fetchone()
if result3 is not None:
self.cur.execute(f"SELECT role FROM mute WHERE guild_id = message.guild.id")
mutedet = self.cur.fetchone()
if mutedet is not None:
self.cur.execute(f"SELECT warn FROM warning WHERE guild_id = message.guild.id and user_id = message.author.id")
result2 = self.cur.fetchone()
if result2 is None:
role = discord.utils.get(message.author.guild.roles, id=int(mutedet[0]))
sql = f"INSERT INTO warning(guild_id, user_id, warn) VALUES(?,?,?)"
val = (message.guild.id, message.author.id, 1)
self.cur.execute(sql, val)
self.db.commit()
await message.channel.purge(limit=1)
await message.author.send(f"Hey message.author.mention you are muted for 10 Minutes, keep doing this will result on getting banned.")
await message.channel.send(f"Hey message.author.mention, That word isn't allowed here!")
await message.author.add_roles(role)
await asyncio.sleep(600)
await message.author.remove_roles(role)
await message.author.send("You are unmuted, If you keep doing this you'll be banned!")
elif result2 is not None:
self.cur.execute(f"SELECT warn FROM warning WHERE guild_id = message.guild.id and user_id = message.author.id")
warnamount = self.cur.fetchone()
role = discord.utils.get(message.author.guild.roles, id=int(mutedet[0]))
wcount = int(warnamount[0])
if wcount >= 3:
self.cur.execute(f"DELETE FROM warning WHERE guild_id = message.guild.id and user_id = message.author.id")
self.db.commit()
await message.author.send("You have been banned, because of 3 Violates in a row!")
await message.author.ban(reason="3 Violates in a row!")
else:
self.cur.execute(f"SELECT warn FROM warning WHERE guild_id = message.guild.id and user_id = message.author.id")
warnings = self.cur.fetchone()
warns = int(warnings[0])
sql = f"UPDATE warning SET warn = ? WHERE guild_id = ? and user_id = ?"
val = (warns + 1, message.guild.id, message.author.id)
self.cur.execute(sql, val)
self.db.commit()
await message.channel.purge(limit=1)
await message.author.send(f"Hey message.author.mention you are muted for 10 Minutes, keep doing this will result on getting banned.")
await message.channel.send(f"Hey message.author.mention, That word isn't allowed here!")
await message.author.add_roles(role)
await asyncio.sleep(600)
await message.author.remove_roles(role)
await message.author.send("You are unmuted, If you keep doing this you'll be banned!")
消息作者被禁止后,不知何故,它再次以 1 的形式添加到 db 中:/ 请问有更好的代码逻辑建议吗?
【问题讨论】:
【参考方案1】:该函数非常大,您应该将其折射为更小更抽象的版本。 就像我认为您正在测试包含坏词的消息一样,您可以为其创建一个函数来检查坏词并返回真或假。 您可以将数据库操作包装在更有意义、更易于编写和推理的函数中。
将您的数据库逻辑与代码分离,将其包装成一个函数
您再次收到静音消息的原因是您在达到 3 后删除警告,并且当他发送回消息作为警告用户已被删除时,结果 2 再次为 None,此循环继续。
一个可能的解决方案是使用包含被禁止成员的表格,一旦警告超出您可以将用户放入该表格。
【讨论】:
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