我试图使一个机器人,每当它检测到页面状态变化时发送消息,但3-5秒后,它随机发送“服务器在线”,即使页面没有任何变化。
import os
import discord
from dotenv import load_dotenv
import time
import requests
def check(r):
if "online" in r.text:
return True
else:
return False
online = False
load_dotenv()
TOKEN = "#hidden"
GUILD = #hidden
client = discord.Client()
@client.event
async def on_ready():
for guild in client.guilds:
if guild.name == GUILD:
break
print(
f'{client.user} is connected to the following guild:\n'
f'{guild.name}(id: {guild.id})')
channel = client.get_channel(#hidden)
last_status = check(requests.get("#page"))
while True:
if check(requests.get("#page")) == last_status:
continue
else:
if check(requests.get(#page")):
await channel.send("server is online")
last_status = check(requests.get("#page"))
else:
await channel.send("Server is offline")
last_status = check(requests.get("#page"))
client.run(TOKEN)发布于 2020-04-20 19:28:05
这可能是因为您有多个正在运行的on_ready函数实例。
不一致API通常会发送重新连接的指令(特别是在由于流行病而导致负载过载的这段时间内)。
当discord.py收到此指令时,它会重新连接并再次调用on_ready,而不会终止另一个。
解决方案是使用asyncio.ensure_future和client.wait_until_ready来确保只有一个实例
代码:
import os
import discord
from dotenv import load_dotenv
import time
import requests
import asyncio
def check(r):
if "online" in r.text:
return True
else:
return False
online = False
load_dotenv()
TOKEN = "#hidden"
GUILD = #hidden
client = discord.Client()
async def routine():
await client.wait_until_ready()
channel = client.get_channel(#hidden)
last_status = check(requests.get("#page"))
while True:
if check(requests.get("#page")) == last_status:
continue
else:
if check(requests.get(#page")):
await channel.send("server is online")
last_status = check(requests.get("#page"))
else:
await channel.send("Server is offline")
last_status = check(requests.get("#page"))
@client.event
async def on_ready():
for guild in client.guilds:
if guild.name == GUILD:
break
print(
f'{client.user} is connected to the following guild:\n'
f'{guild.name}(id: {guild.id})')
asyncio.ensure_future(routine())
client.run(TOKEN)https://stackoverflow.com/questions/61264878
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