Cron 如何编辑以前发送的嵌入?

Cron 如何编辑以前发送的嵌入?,cron,discord,discord.js,bots,Cron,Discord,Discord.js,Bots,这是我的代码,它不起作用,它说它需要在一个异步函数中,但我不知道把异步放在哪里,我也在使用cron-npm const Discord = require ('discord.js') const client = new Discord.Client() const config = require('./config.json') const cron = require('cron'); let embed = new Discord.MessageEmbed() .setTitle(&q

这是我的代码,它不起作用,它说它需要在一个异步函数中,但我不知道把异步放在哪里,我也在使用cron-npm

const Discord = require ('discord.js')
const client = new Discord.Client()
const config = require('./config.json')
const cron = require('cron');
let embed = new Discord.MessageEmbed()
.setTitle("EMBED TITLE")
.setDescription("EMBED DESCRIPTION")

client.once('ready', () => {
  console.log(`Logged in as ${client.user.tag}`)

})

const welchannel = client.channels.cache.get("CHANNEL_ID") //the id of the channel is going to be sent
const msg =  await welchannel.send(embed)

let sevencron = new cron.CronJob('00 * 7 * * 1', () => {
  const oneembed = new Discord.MessageEmbed()
  .setTitle("EDIT_EMBED_TITLE")
  .setDescription('EDIT_EMBED_DESCRIPTION')
 msg.edit(oneembed)
});

sevencron.start()


client.login(config.token)
就在这里:

                       |
                       V
client.once('ready', async () => {
  console.log(`Logged in as ${client.user.tag}`)

})

ready
事件中,您应该将其放在
()
之前。