Question

我当时正在开发一款不和谐的漫游器以及一个验证频道。我希望用户仅键入/verify命令：他们键入的除/verify以外的每条消息或命令都应自动删除。我该怎么办？
当前代码：

if (command === "verify") {
  if (message.channel.id !== "ChannelID") return;
  let role = message.guild.roles.find(rol => rol.name === 'Member')

  const reactmessage = await message.channel.send('React with  to verify yourself!');
  await reactmessage.react('');
  const filter = (reaction, user) => reaction.emoji.name === '' && !user.bot;
  const collector = reactmessage.createReactionCollector(filter, {
    time: 15000
  });
  collector.on('collect', async reaction => {
    const user = reaction.users.last();
    const guild = reaction.message.guild;
    const member = guild.member(user) || await guild.fetchMember(user);

    member.addRole(role);
    message.channel.send(`Verification Complete.. ${member.displayName}. You have got access to server. `)

  });
  message.delete();
}

Answer 1

您可以在client.on('message')侦听器的顶部添加一个检查：

client.on('message', message => {
  let verified = !!message.member.roles.find(role => role.name == 'Member');
  // ... command parsing ect...
  if (!verified && command == 'verify') {...}
  else if (verified) {
    // other commands...
  }
});

如何删除特定通道中除一条命令外的所有用户输入？

1 个答案: