用戶在這裏是我的代碼如何檢查是否不和諧機器人是輸入命令
import discord
import asyncio
import sys
import commands
import pickle
import os
import random
import array
#if len(sys.argv) != 2:
# print('Usage: python3 main.py [token]')
commandz=("yes","no","hi","hi")
try:
with open(os.path.join(os.path.dirname(__file__), "token.pickle"), 'rb') as file:
token = pickle.load(file)
if len(token) == 59:
key = int(random.random() * 10000000000000000)
print('Found saved token in stored.py, use phrase tokenreset'+str(key), 'to undo this.')#youll need to add the command
else:
raise
except:
#code tokenreset with admin permissions
token = input('What is your Discord bot token? (found on Discord developer page): ')
with open(os.path.join(os.path.dirname(__file__), "token.pickle"), 'wb') as file:
pickle.dump(token, file)
client = discord.Client()
@client.event
async def on_ready():
print('Logged in as' + '[' + client.user.id + ']' + client.user.name)
print('--------')
@client.event
async def on_message(message):
await handle_command(message)
async def handle_command(message):
print('Noticed: ' + message.content)
if message.content == 'tokenreset'+str(key):
await client.send_message(message.channel, 'code accepted')
i = 0
if str(message.author) == "NOTAKOALAINVENEZUELA#6895":
i = 1
x = 0
await client.send_message(message.channel, message.author)
while i == 0:
if commandz[x] in message.content:
x = x + 1
await client.send_message(message.channel, commandz[x])
i = i + 1
else:
if x == len(commandz) - 2:
i = i + 1
else:
x = x + 2
client.run(token)
結果是我的機器人一遍又一遍,我已經試過各種觸發其自己的命令得到它不承認它的自己的消息,我在我的智慧結束。
我認爲由於某種原因message.author是一些不能用於比較的奇怪類型的變量,我對編程和python超級新,所以我不確定。
您發佈的代碼不會運行。請仔細檢查所有需要它自己的行的東西都有它,並且所有內容都正確縮進。 –
是的當發佈的格式搞砸了編輯 – trisimix