Question

我正在尝试编写代码，以检查通道中的所有消息是否包含某些单词，如果包含某些单词，则将其删除。像这样：

if(msg.content.startsWith(prefix+'clean') {
  let check = msg.content.split(prefix+'clean')[1]; // Condition, in this case if it containts a certain string
  msg.channel.fetchMessages().then(msgs => { // Get messages to check
    let msglog = msgs.array() // Make an array with all the messages fetched
    for(var i = 0; i < msglog.size; i++) { // Loop to check all messages in array
      if (check in msglog[i]) {
        // Code to delete that message
      };
    };
  });
};

我知道这不会检查整个频道，而只会检查最近的50条消息，但是我不知道如何检查整个频道，因此在我确定如何做之前将如此。

但是什么代码将删除通过检查的消息？还是我可以采用其他方式？

编辑：

似乎我还不够清楚，所以假设某个频道进行了以下对话：

人员A：嗨，大家好！

乙：嗨

人员C：再见

假设我要通过我的机器人删除其中所有带有“ Hi”的消息，该怎么办？注意：我不希望在邮件发送后立即删除它，我只想在要删除时删除它。

Answer 1

好吧，这是我意识到fetchMessages（）的2周限制后解决问题的方法

  else if(msg.content.startsWith(`${prefix}clean`}) { // Check for command
    let check = msg.content.split(`${prefix}clean`)[1] // Defines a check
    msg.channel.fetchMessages({ limit: 100 }).then(msgs => { // Fetches the last 100 messages of the channel were the command was given
      const msgstodelete = msgs.filter(del => del.content.includes(check)) // Filters the messages according to the check
      msg.delete() // Deletes the original message with the command
      for (var i = 0; i<Array.from(msgstodelete.keys()).length; i++) { 
        msg.channel.fetchMessage(Array.from(msgstodelete.keys())[i]).then(deldel => deldel.delete())
      } // Loop to delete all messages that passed the filter
    })
  }

Answer 2

bulkDelete函数会删除给定的消息，这些消息要早于两周。

if(msg.content.startsWith(prefix+'clean') {
  let check = msg.content.split(prefix+'clean')[1]; // Condition, in this case if it containts a certain string
  msg.channel.fetchMessages().then(msgs => { // Get messages to check
    let msgDel = msgs.filter(msgss => msgss.content.includes(check)) // Finds all messages with 'check'
    msg.channel.bulkDelete(msgDel) // Deletes all messages that got found
  });
};

要删除超过2周的邮件，您必须手动遍历邮件以将其删除：

async function deleteReturnLast(chan, option, prevMsg, cond) {
  return chan.fetchMessages(option)
    .then(async msgs => {
      if (msgs.size === 0){
    if (cond(prevMsg)) {
      prevMsg.delete()
        .then(d => console.log('last message deleted: ' + d.content))
        .catch(err => console.log('ERR>>', err, prevMsg.content, option.before));   }
    return prevMsg;
      };
      let last = msgs.last();
      for (const[id, msg] of msgs) {
    let tmp = (id === last.id) ? prevMsg : msg;
    if (cond(tmp)) {
      tmp.delete()
        .then(d => console.log('Message deleted: ' + d.content))
        .catch(err => console.log('ERR>>', err));
    }
      };
      return last;
    })
    .catch(err => console.log('ERR>>', err));
}

function cond(msg) {
  return !msg.content.includes('a');
}

client.on('message', async function(msg) {
  let chan = msg.channel;
  let last = chan.lastMessage;
  while (last !== (last = await deleteReturnLast(chan, {limit: 2, before: last.id}, last, cond))){
  };
});

在给定的不和谐频道中搜索所有满足条件的消息，然后删除

2 个答案: