需要帮助向用户添加角色。 (discord.js)

时间:2021-04-01 15:46:40

标签: javascript node.js discord.js

当我在线时,机器人会给我一个角色,一旦我下线,机器人就会从我身上删除该角色。

当它删除角色时,我希望机器人将角色授予特定用户。我该怎么做?

我的当前代码如下:

client.on('presenceUpdate', (oldPresence, newPresence) => {
  const member = newPresence.member;
  if (member.id === 'user.id') {
    if (oldPresence.status !== newPresence.status) {
      var gen = client.channels.cache.get('channel.id');
      if (
        newPresence.status == 'idle' ||
        newPresence.status == 'online' ||
        newPresence.status == 'dnd'
      ) {
        gen.send('online');
        member.roles.add('role.id');
      } else if (newPresence.status === 'offline') {
        gen.send('offline');
        member.roles.remove('role.id');
      }
    }
  }
});

1 个答案:

答案 0 :(得分:0)

您可以通过其 ID 获取其他成员。 newPresence 有一个 guild 属性,它有一个 members 属性;通过使用它的 .fetch() 方法,您可以获得要分配角色的成员。拥有此成员后,您可以再次使用 .toles.add()。检查下面的代码:

// use an async function so we don't have to deal with then() methods
client.on('presenceUpdate', async (oldPresence, newPresence) => {
  // move all the variables to the top, it's just easier to maintain
  const channelID = '81023493....0437';
  const roleID = '85193451....5834';
  const mainMemberID = '80412945....3019';
  const secondaryMemberID = '82019504....8541';
  const onlineStatuses = ['idle', 'online', 'dnd'];
  const offlineStatus = 'offline';
  const { member } = newPresence;

  if (member.id !== mainMemberID || oldPresence.status === newPresence.status)
    return;

  try {
    const channel = await client.channels.fetch(channelID);

    if (!channel) return console.log('Channel not found');

    // grab the other member
    const secondaryMember = await newPresence.guild.members.fetch(secondaryMemberID);

    if (onlineStatuses.includes(newPresence.status)) {
      member.roles.add(roleID);
      secondaryMember.roles.remove(roleID);
      channel.send('online');
    }

    if (newPresence.status === offlineStatus) {
      member.roles.remove(roleID);
      secondaryMember.roles.add(roleID);
      channel.send('offline');
    }
  } catch (error) {
    console.log(error);
  }
});