[英]discord.py SyntaxError: 'await' outside async function
I got an error in the whole code below.我在下面的整个代码中遇到错误。 I would like to seek help with the error.
我想就错误寻求帮助。 Can I ask for help by looking at the code below?
我可以通过查看下面的代码寻求帮助吗?
async def ban(ctx, member: discord.Member, *, reason: typing.Optional[str] = "사유 없음."):
await ctx.message.delete()
author = ctx.message.author
embed = None
ch = bot.get_channel(id=772349649553850368)
mesge = await ctx.send("차단을 시킬까요?")
await mesge.add_reaction('✅')
await mesge.add_reaction('❌')
def check1(reaction, user):
return user == ctx.message.author and str(reaction.emoji) == "✅"
try:
reaction, user = await bot.wait_for("reaction_add", timeout = 30.0, check = check1)
embed = discord.Embed(title="종합게임 커뮤니티 제재내역 - 차단", description=f'담당자 : {author.mention} \n대상자 : {member.mention} \n제재사유 : {reason} \n\n위와 같은 사유로 인해 제재처리 되었습니다.', color=0xff0000)
embed.set_author(name=f"{str(member)}님을 서버에서 영구적으로 차단했어요.", icon_url=member.avatar_url_as(static_format='png', size=2048))
embed.set_footer(text=f'처리 시각 - {str(now.year)} 년 {str(now.month)} 월 {str(now.day)} 일 | {str(now.hour)} 시 {str(now.minute)} 분 {str(now.second)}초 - 담당자 : {author.display_name}')
await ch.send(embed=embed)
await member.send(embed=embed)
await ctx.guild.ban(member, reason=f'사유 : {reason} - 담당자 : {author.display_name}')
except asyncio.TimeoutError:
print("Timeout")
def check2(reaction, user):
return user == ctx.message.author and str(reaction.emoji) == "❌"
try:
reaction, user = await bot.wait_for("reaction_add", timeout = 30.0, check = check2)
await ctx.send("취소되었다")
except asyncio.TimeoutError:
print("Timeout")
The following error appears in the above code.上面的代码中出现以下错误。
reaction, user = await bot.wait_for("reaction_add", timeout = 30.0, check = check1)
^
SyntaxError: 'await' outside async function
If you know how to fix it, please help.如果您知道如何修复它,请帮助。
I used a translator.我用了翻译器。
Python uses identation to identify code blocks. Python 使用标识来标识代码块。 In your code, you placed the
await
call inside of the non-async function check1
.在您的代码中,您将
await
调用放在非异步函数check1
。 Here is an example of the same problem:以下是同一问题的示例:
async def foo():
def check1():
return True
baz = await bar() # improperly indented and in fact can never
# run because it is after the function `return`
The fix is to move the code outside of check1
.解决方法是将代码移到
check1
之外。 It should align with the "def" statement above.它应该与上面的“def”语句一致。
async def foo():
def check1():
return True
baz = await bar()
Your issue is with indentation after both checks.您的问题在于两次检查后的缩进。
I have added # ---- here ----
So that you know where to end the check
我在
# ---- here ----
添加了# ---- here ----
这样您就知道在哪里结束check
async def ban(ctx, member: discord.Member, *, reason: typing.Optional[str] = "사유 없음."):
await ctx.message.delete()
author = ctx.message.author
embed = None
ch = bot.get_channel(id=772349649553850368)
mesge = await ctx.send("차단을 시킬까요?")
await mesge.add_reaction('✅')
await mesge.add_reaction('❌')
def check1(reaction, user):
return user == ctx.message.author and str(reaction.emoji) == "✅"
# ---- here ----
try:
reaction, user = await bot.wait_for("reaction_add", timeout = 30.0, check = check1)
embed = discord.Embed(title="종합게임 커뮤니티 제재내역 - 차단", description=f'담당자 : {author.mention} \n대상자 : {member.mention} \n제재사유 : {reason} \n\n위와 같은 사유로 인해 제재처리 되었습니다.', color=0xff0000)
embed.set_author(name=f"{str(member)}님을 서버에서 영구적으로 차단했어요.", icon_url=member.avatar_url_as(static_format='png', size=2048))
embed.set_footer(text=f'처리 시각 - {str(now.year)} 년 {str(now.month)} 월 {str(now.day)} 일 | {str(now.hour)} 시 {str(now.minute)} 분 {str(now.second)}초 - 담당자 : {author.display_name}')
await ch.send(embed=embed)
await member.send(embed=embed)
await ctx.guild.ban(member, reason=f'사유 : {reason} - 담당자 : {author.display_name}')
except asyncio.TimeoutError:
print("Timeout")
def check2(reaction, user):
return user == ctx.message.author and str(reaction.emoji) == "❌"
# ---- here ----
try:
reaction, user = await bot.wait_for("reaction_add", timeout = 30.0, check = check2)
await ctx.send("취소되었다")
except asyncio.TimeoutError:
print("Timeout")
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.