当有人加入特定语音频道时向文本频道发送消息

Question

我正在尝试让我的机器人在有人进入语音支持等候室时通过特定的文本频道通知我的服务器工作人员。

不幸的是，我只找到了 Discord 版本 12 的一些示例，并且在版本 13 中它不适用于我。我不知道该代码中要更改什么。

没有错误信息什么的，只是在有人加入语音频道时不发送信息。

client.on('voiceStateUpdate', (oldState, newState) => {
  const newUserChannel = newState.channelID;
  const textChannel = client.channels.cache.get('931223643362703521');

  if (newUserChannel === '931156705303317013') {
    textChannel.send('Somebody joined');
  }
  console.error();
});

Answer 1

首先，确保添加 GUILD_VOICE_STATES 意图。

fetch 频道可能会更好，而不是依赖缓存。如果通道已经被缓存，fetch 将使用它并且不会向 discord API 发出请求，但如果它没有被缓存，channels.cache.get 将找不到它。

我做了一些更改，并在下面的代码中添加了一些 cmets：

const { Client, Intents } = require('discord.js');

const client = new Client({
  intents: [
    Intents.FLAGS.GUILDS,
    Intents.FLAGS.GUILD_MESSAGES,
    Intents.FLAGS.GUILD_VOICE_STATES,
  ],
});

client.on('voiceStateUpdate', async (oldState, newState) => {
  const VOICE_SUPPORT_ID = '931156705303317013';
  const LOG_CHANNEL_ID = '931223643362703521';
  // if there is no newState channel, the user has just left a channel
  const USER_LEFT = !newState.channel;
  // if there is no oldState channel, the user has just joined a channel
  const USER_JOINED = !oldState.channel;
  // if there are oldState and newState channels, but the IDs are different,
  // user has just switched voice channels
  const USER_SWITCHED = newState.channel?.id !== oldState.channel?.id;

  // if a user has just left a channel, stop executing the code 
  if (USER_LEFT)
    return;

  if (
    // if a user has just joined or switched to a voice channel
    (USER_JOINED || USER_SWITCHED) &&
    // and the new voice channel is the same as the support channel
    newState.channel.id === VOICE_SUPPORT_ID
  ) {
    try {
      let logChannel = await client.channels.fetch(LOG_CHANNEL_ID);

      logChannel.send(
        `${newState.member.displayName} joined the support channel`,
      );
    } catch (err) {
      console.error('❌ Error finding the log channel, check the error below');
      console.error(err);
      return;
    }
  }
});