0

所以我正在制作一个命令，当用户加入时向频道发送消息，但由于我缓存数据的方式，我无法从主文件访问它。

那么有什么方法可以在使用 Commando 的命令中使用客户端事件，还是我需要找到一种方法来缓存在我的主文件中？

https://pastebin.com/ZmRy1jTC 

const { CommandoClient } = require("discord.js-commando");
const path = require("path");
const mongo = require("./mongo");
const { token, mongoPath } = require("./config.json");
const cache = {};
const welcomeSchema = require('./models/welcomeSchema')

const client = new CommandoClient({
  commandPrefix: ";",
  owner: ["769231300472995840", "701561771529470074"],
  invite: "https://discord.gg/GJRC4cnU4j",
});

client.registry
  .registerDefaultTypes()
  .registerGroups([
    ["moderation", "Moderation Commands"],
    ["general", "General Commands"],
    ["misc", "Misc Commands"],
    ["server-config", "Server Config Commands"],
    ["economy", "Economy Commands"],
    ["testing", "Testing Commands"],
  ])
  .registerDefaultGroups()
  .registerDefaultCommands()
  .registerCommandsIn(path.join(__dirname, "commands"));

client.once("ready", async () => {
  console.log(`Logged in as ${client.user.tag}! (${client.user.id})`);
  const delay = (msec) => new Promise((resolve) => setTimeout(resolve, msec));
  while (0 == 0) {
    await delay(10000);
    client.user
      .setActivity(`${client.guilds.cache.size} servers`, { type: "WATCHING" })
      .catch(console.error);
    await delay(10000);
    client.user.setActivity(`;help`, { type: "WATCHING" }).catch(console.error);
  }
});

const onJoin = async (member) => {
  const { guild } = member;

  let data = cache[guild.id];

  if (!data) {
    console.log("FETCHING FROM DATABASE");

    await mongo().then(async (mongoose) => {
      try {
        const result = await welcomeSchema.findOne({ _id: guild.id });

        cache[guild.id] = data = [result.channelId, result.text];
      } finally {
        mongoose.connection.close();
      }
    });
  }

  const channelId = data[0];
  const text = data[1];

  const channel = guild.channels.cache.get(channelId);
  channel.send(text.replace(/<@>/g, `<@${member.id}>`));
};

client.on('guildMemberAdd', (member) => {
    onJoin(member)
})

client.on("error", console.error);

client.login(token);

https://pastebin.com/cCruVWWw 

const { Command } = require('discord.js-commando');
const mongo = require('../../mongo');
const WelcomeSchema = require('../../models/welcomeSchema');

const cache = {}

module.exports = class WelcomeSetCommand extends Command {
    constructor(client) {
        super(client, {
            name: 'setwelcome',
            group: 'server-config',
            memberName: 'setwelcome',
            description: 'A command that sets a welcome message.',
            userPermissions: ['ADMINISTRATOR'],
            args: [
                {
                    key: 'message',
                    prompt: 'What do want to say when a new user joins?',
                    type: 'string'
                }
            ]
        })
    }

    async run (msg, { message }) {
        const { channel, guild } = msg;

        cache[guild.id] = [channel.id, message]

        await mongo().then(async (mongoose) => {
            try{
                await WelcomeSchema.findOneAndUpdate({
                    _id: guild.id
                }, {
                    _id: guild.id,
                    channelId: channel.id,
                    text: message
                }, {
                    upsert: true
                })
            } finally{
                mongoose.connection.close()
            }
        })
    }
}

4

1 回答 1

0

如果您可以通过，client您可以使用事件。
示例：
index.js

const Client = new Discord.Client();

const GuildAdd = require("./guildadd");

GuildAdd(Client);

Client.login("TOKEN");

guildadd.js

module.exports = (Client) => {
    Client.on("guildMemberAdd", member => {
        //Do the rest here
    });
}
于 2021-05-11T05:40:56.870 回答