我收到此错误:“ TypeError:无法读取未定义的属性'channel'”
如果表格“ steamid”中有新条目,则漫游器应自动在不和谐频道中发布消息
如果有人可以帮助我,那将很酷:)
我的代码:
const mysql = require('mysql');
let lastId = 0;
let id;
let lastBan;
let conn = mysql.createConnection({
host: '',
user: '',
password: '',
database: ''
});
conn.connect();
client.on('message', (msg) => {
if (msg.channel.id === '') {
msg.reply(`Der letzte Ban ist: ${lastBan}`);
}
});
const checkEntries = () => {
conn.query(`SELECT id, steamid FROM banned ORDER BY id DESC LIMIT 1`, (err, res, fields) => {
if(err) throw err;
id = res[0].id;
if(id > lastId){
lastBan.channel.send(res);
lastBan = id;
}
lastId = id;
});
}
conn.on('error', function(err) {
console.log("[MySQL Fehler] ",err);
});
setInterval(checkEntries, 500);
最佳答案
仅声明lastBan
。您没有在lastBan
上设置任何频道属性
编辑:
也许您是说client.channel.send(res);
吗?
编辑2:client.channels.get(/*channelid*/).send(res)
吗?
client.on('ready', function(){
setInterval(checkEntries, 500);
});
关于javascript - 变量不是 channel 对象,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/59493135/