如何嵌入每次运行命令时更新的变量？
如何嵌入每次运行命令时更新的变量？
提问于 2022-12-04 11:05:23
我试图让我的不和谐机器人(用python编码)嵌入我设置的字符串变量的内容。我无法弄清楚如何让bot嵌入字符串，以及如何在每次运行命令时进行变量更新。
我为播放列表变量做了一个函数，希望每次调用它时它都能运行，但是我想得到一些关于这个变量是否正确的指导，以及显示它的问题。
def playerlist():
req = Request(url, headers = {'User-Agent': 'Mozilla/5.0'}) #spoopy disguise
webpage = urlopen(req).read()
soup = soup(webpage, "html.parser")
cleansoup = (soup.get_text(strip=True, separator=" "))
x = cleansoup.split("""Play time""")[-1]
x = x.split("""Most Time Played""")[0]
print(x)@commands.hybrid_command(
name="playerlist",
description="This should pull a player list with playtimes",
)
# This will only allow non-blacklisted members to execute the command
@checks.not_blacklisted()
async def playerlist(self, context: Context) -> None:
"""
playerlist from battlemetrics
"""
playerlist()
embed = discord.Embed(
title="Playerlist",
description=x,
color=0x9C84EF
)
await context.send(embed=embed)我收到一条错误消息，表示没有定义"x“。我能理解为什么它会说它没有定义，但它仍然令人困惑，因为它应该是从函数定义的？
非常感谢你抽出时间
回答 1
Stack Overflow用户
发布于 2022-12-04 12:35:12
您可以从播放列表函数返回x。
def playerlist():
req = Request(url, headers = {'User-Agent': 'Mozilla/5.0'}) #spoopy disguise
webpage = urlopen(req).read()
soup = soup(webpage, "html.parser")
cleansoup = (soup.get_text(strip=True, separator=" "))
x = cleansoup.split("""Play time""")[-1]
x = x.split("""Most Time Played""")[0]
return x任何时候都可以回忆
@commands.hybrid_command(
name="playerlist",
description="This should pull a player list with playtimes",
)
# This will only allow non-blacklisted members to execute the command
@checks.not_blacklisted()
async def playerlist(self, context: Context) -> None:
"""
playerlist from battlemetrics
"""
x = playerlist()
embed = discord.Embed(
title="Playerlist",
description=x,
color=0x9C84EF
)
await context.send(embed=embed)页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/74674934
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