如何修复python不一致的错误？

Question

if message.content.lower().startswith('!kick') and (roleLFJob in message.author.roles or roleLFAba in message.author.roles):
    await client.delete_message(message)
    serverchannel = '405090256124248065'
    messageParsed = message.content.split()
    kick = messageParsed[0]
    mention = messageParsed[1]
    msg = messageParsed[2:]
    for member in message.mentions:
        await client.kick(member)
        await client.send_message(discord.Object(id=serverchannel), '{0} was kicked by {1}, with reason:"**'.format(member.mention, message.author.mention) + msg + '**"')

当我不和谐地写下这个命令时：

^{pr2}$

出现此错误：

Ignoring exception in on_message Traceback (most recent call last):  
File "C:\Users\senuk\AppData\Local\Programs\Python\Python35\lib\site-packages\discord\client.py", line 307, in _run_event
    yield from getattr(self, event)(*args, **kwargs)   
File "overmind.py", line 128, in on_message
    await client.send_message(discord.Object(id=serverchannel), '{0} was kicked by {1} with reason:"**'.format(member.mention, message.author.mention) + msg + '**"') 
TypeError: Can't convert 'list' object to str implicitly

Answer 1

既然你打了这个：

!kick @(member.mention) reason, reason and reason

然后当你拆分整个内容时：

^{pr2}$

结果将是：

kick = "!kick"
mention = "@(member.mention)"
msg = ["reason,", "reason", "and", "reason"]

在代码的最后一行，尝试用msg（list）连接两个字符串。这是不可能的。在

您可能希望msg是一个字符串，因此应该使用：

msg = " ".join(messageParsed[2:])

它将msg还原为原始形式：

msg = "reason, reason and reason"