IndentationError：意外取消缩进

Posted 2023-03-23

【中文标题】IndentationError：意外取消缩进

【英文标题】：IndentationError: unexpected unindent

【发布时间】：2018-02-04 00:22:55

【问题描述】：

不久前开始编码，大约一个月左右。我目前正在为 Discord 编写一个机器人，一切正常，直到我在添加新命令后尝试运行该机器人时收到此错误消息：

Traceback (most recent call last):
  File "C:\Users\Jeriel\Desktop\JerryBot\run.py", line 162, in main
    from musicbot import MusicBot
  File "C:\Users\Jeriel\Desktop\JerryBot\musicbot\__init__.py", line 1, in <modu
le>
    from .bot import MusicBot
  File "C:\Users\Jeriel\Desktop\JerryBot\music\bot.py", line 2094
    if __name__ == "__main__":
                             ^
IndentationError: unexpected unindent

它是在我添加这个之后开始的。我检查了这些之前的每一行，我在任何地方都找不到一个取消缩进：

    async def kick(message,*args):
        """Kicks the specified user from the server"""
        if len(message.mentions) < 1:
            return False

        if message.channel.is_private:
            msg = await client.send_message(message.channel,'Users cannot be kicked/banned from private channels.')
            asyncio.ensure_future(message_timeout(msg, 40))
            return

        if not message.channel.permissions_for(message.server.get_member(client.user.id)).kick_members:
            msg = await client.send_message(message.channel, message.author.mention + ', I do not have permission to kick users.')
            asyncio.ensure_future(message_timeout(msg, 40))
            return

        members = []

        if not message.channel.is_private and message.channel.permissions_for(message.author).kick_members:
            for member in message.mentions:
                if member != message.author:
                    try:
                        await client.kick(member)
                        members.append(member.name)
                    except:
                        pass
                else:
                    msg = await client.send_message(message.channel, message.author.mention + ', You should not kick yourself from a channel, use the leave button instead.')
                    asyncio.ensure_future(message_timeout(msg, 40))
        else:
            msg = await client.send_message(message.channel, message.author.mention + ', I do not have permission to kick users, or this is a private message channel.')
            asyncio.ensure_future(message_timeout(msg, 40))

        msg = await client.send_message(message.channel,'Successfully kicked user(s): ``'.format('`, `'.join(members)))
        asyncio.ensure_future(message_timeout(msg, 60))

    @register('ban','@<mention users>',owner=True)

    async def ban(message,*args):
        """Bans the specified user from the server"""
        if len(message.mentions) < 1:
            return False

        if message.channel.is_private:
            msg = await client.send_message(message.channel,'Users cannot be kicked/banned from private channels.')
            asyncio.ensure_future(message_timeout(msg, 40))
            return

        if not message.channel.permissions_for(message.server.get_member(client.user.id)).ban_members:
            msg = await client.send_message(message.channel, message.author.mention + ', I do not have permission to ban users.')
            asyncio.ensure_future(message_timeout(msg, 40))
            return

        members = []

        if message.channel.permissions_for(message.author).ban_members:
            for member in message.mentions:
                if member != message.author:
                    try:
                        await client.ban(member)
                        members.append(member.name)
                    except:
                        pass
                else:
                    msg = await client.send_message(message.channel, message.author.mention + ', You should not ban yourself from a channel, use the leave button instead.')
                    asyncio.ensure_future(message_timeout(msg, 40))
        else:
            msg = await client.send_message(message.channel, message.author.mention + ', I do not have permission to ban users, or this is a private message channel.')
            asyncio.ensure_future(message_timeout(msg, 40))

        msg = await client.send_message(message.channel,'Successfully banned user(s): ``'.format('`, `'.join(members)))
        asyncio.ensure_future(message_timeout(msg, 30))

    @register('bans',alias='bannedusers')
    @register('bannedusers')


if __name__ == "__main__":
    bot = JerryBot()
    bot.run("---")

【问题讨论】：

错误信息指向if __name__ == "__main__":。为什么该行相对于其余代码没有缩进？

您是复制并粘贴代码，还是自己输入？如果您复制并粘贴它，可能是由于您使用制表符在某些地方缩进，而在其他地方使用空格来缩进的问题。这样做总是很有趣……

@Blorgbeard 我认为 OP 想使用 if __name__ == "__main__": 块作为主要功能，如果是这样，那么它必须不缩进

【参考方案1】：

你的问题在这里：

    @register('bans',alias='bannedusers')
    @register('bannedusers')


if __name__ == "__main__":
    bot = JerryBot()
    bot.run("---")

装饰器语法需要在@ 行下定义函数，该函数必须与@ 处于相同的缩进级别。换句话说，你不能在那里有你的if 声明。您需要一个函数定义。缩进只是它发现的第一个问题。如果你缩进你的if 语句，你会得到一个不同的错误。

我不知道您是否在此处错误地包含了 @ 行，或者您是否省略了您打算放置在那里的函数定义。根据您的目的，编写函数或删除 @ 行。

【讨论】：

谢谢！我就是这样做的，它似乎解决了缩进问题，但是一旦我这样做，就会出现：File "C:\Users\Jeriel\Desktop\JerryBot\musicbot\bot.py", line 2018 async def kick(message,*args): ^ TabError: inconsistent use of tabs and spaces in indentation C:\Users\Jeriel\Desktop\JerryBot>关于如何解决这个问题的任何想法？

【参考方案2】：

您是否可以混合使用制表符和空格？这是python初学者的常见错误。从技术上讲，您可以使用其中任何一种，但不能混合使用。我对此的解决方案（我使用 vim 作为 IDE）是在我的 .vimrc 中将 tab 键设置为实际上是 4 个空格。

您可以尝试在 vim 中通过输入 gg=G 并在命令模式下按 Enter 键重新插入。