Question

我遇到了SyntaxError，但错误指向没有意义的

我已经尝试过移动代码并在Google中搜索该错误，但没有帮助

@client.command()
@has_permissions(kick_members=True)
async def mute(ctx, member:discord.Member, *, reason=None):
 arg=reason
 author=ctx.author
 guild=ctx.message.guild
 perms=discord.Permissions(connect=False, speak=False, read_text_channels_&_see_voice_channels=False, add_reactions=False, send_messages=False)
 role=discord.utils.get(ctx.guild.roles, name="muted")

 await guild.create_role(name="muted", colour=discord.Colour(0x808080), permissions=perms)
 await member.send(f'You got muted for: ```\n{arg}\n``` Muted by: {author}')
 await member.add_roles(role)
 await ctx.send(f'{member.mention} got muted for: ```\n{arg}\n``` Muted by: {author}!')

错误消息是：SyntaxError：关键字不能是表达式，然后在我放置^的地方以红色突出显示

@client.command()
@has_permissions(kick_members=True)
async def mute(ctx, member:discord.Member, *, reason=None):
 arg=reason
 author=ctx.author
 guild=ctx.message.guild
 perms=discord.Permissions(connect=False, speak=False, read_text_channels_&_see_voice_channels=False, add_reactions=False
```                                                   ^

Answer 1

答案很简单：您在Debug JS Remotely中有一个&，我相信您打算以此作为参数的名称。

由于read_text_channels_&_see_voice_channels不能使用名称，Python将其解析为&，这与将read_text_channels_ & _see_voice_channels运算符（二进制和）应用于&和{{ 1}}。

Python 3.7.3：我收到了SyntaxError：关键字不能是表达式

1 个答案: