ClientConnectorError:无法连接到主机discordapp.com:443 ssl:默认[连接呼叫失败('162.159.134.233'，443)] [英] ClientConnectorError: Cannot connect to host discordapp.com:443 ssl:default [Connect call failed ('162.159.134.233', 443)]

问题描述

因此，我尝试在Discord中制作一个bot，并尝试在Gitpod上运行我的discord bot，并且该bot能够运行，但是当我尝试在pythonanywhere.com上运行它时，出现此错误:

  aiohttp.client_exceptions.ClientConnectorError:无法连接到主机discordapp.com:443 ssl:默认[连接呼叫失败('162.159.134.233'，443)] 

这是我的代码段:

 导入不和谐从discord.ext导入命令导入json使用open("credentials.json")作为凭据:creds = json.loads(creds.read())令牌=凭证[令牌"]客户端= discord.Client()Bot类(commands.Bot):def __init __():超级(自动，自我).__init __(command_prefix ="$"，case_insensitive = True)self.pool =无机器人= Bot()@ bot.event异步def on_ready():打印('登录为')打印(bot.user)打印(' -  -  - ')# 计算@ bot.command()异步def计算(ctx):等待ctx.send("foo")bot.run(令牌) 

任何帮助将不胜感激.

解决方案

PythonAnywhere上的免费帐户无法使用Discord websockets API，但可以使用其基于HTTP的API.请参阅论坛帖子，其中介绍了如何执行此操作./p>

So I was trying out making a bot in Discord and I tried running my discord bot over Gitpod and it was able to run, but when I tried running it on pythonanywhere.com, I get this error:

aiohttp.client_exceptions.ClientConnectorError: Cannot connect to host discordapp.com:443 ssl:default [Connect call failed ('162.159.134.233', 443)]

Here is a snippet of my code:

import discord
from discord.ext import commands
import json

with open("credentials.json") as creds:
    creds = json.loads(creds.read())
    TOKEN = creds["TOKEN"]

client = discord.Client()

class Bot(commands.Bot):
    def __init__(self):
        super(Bot, self).__init__(command_prefix="$", case_insensitive=True)
        self.pool = None

bot = Bot()

@bot.event
async def on_ready():
    print('Logged in as')
    print(bot.user)
    print('------')

# calculate
@bot.command()
async def calculate(ctx):
    await ctx.send("foo")

bot.run(TOKEN)

Any help would be greatly appreciated.

解决方案

Free accounts on PythonAnywhere can't use the Discord websockets API, but you can use their HTTP-based one. Take a look at the forum post that explains how to do that.

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