我已经使用了初学者指南,开始为我们的公会不和谐而使用一个简单的机器人。但是,出于对我的爱,我无法找到一种方法来让bot引用服务器特定的昵称而不是用户名。
代码看起来像这样:
bot.on('message', function (user, userID, channelID, message, evt) {
if (message.substring(0, 1) == '!') {
var args = message.substring(1).split(' ');
var cmd = args[0];
args = args.splice(1);
switch(cmd) {
case 'ping':
bot.sendMessage({
to: channelID,
message: 'Pong! was returned to user with nickname ' + ????????????.nickname
});
break;
}
}
}
有人知道命令/变量是什么来检索昵称而不是用户名吗?
非常感谢您:)
答案 0 :(得分:0)
我不太理解您的问题,但是如果我理解正确的话。您需要在回复消息中获取member.nickname。
在第一个message.channel.send('text')
事件中,仅返回message
对象。
要在同一频道上回复,可以使用bot.on('message', message => {
if (message.substring(0, 1) == '!') {
var args = message.substring(1).split(' ');
var cmd = args[0];
args = args.splice(1);
switch(cmd) {
case 'ping':
message.channel.send(`Pong! was returned to user with nickname ${message.member.displayName}`)
break;
}
}
})
如果需要获取成员昵称,则可以使用member.displayName
,它将返回成员服务器的昵称,如果成员没有昵称,则返回成员用户名+标记
private fun sendNotification(messageBody: String) {
val intent = Intent(this, MainActivity::class.java)
intent.addFlags(Intent.FLAG_ACTIVITY_CLEAR_TOP)
val pendingIntent = PendingIntent.getActivity(this, 0 /* Request code */, intent,
PendingIntent.FLAG_ONE_SHOT)
val channelId = "" // TODO CHANNEL ID NEEDED
val defaultSoundUri = RingtoneManager.getDefaultUri(RingtoneManager.TYPE_NOTIFICATION)
val notificationBuilder = NotificationCompat.Builder(this, channelId)
.setContentText(messageBody)
.setAutoCancel(true)
.setSound(defaultSoundUri)
.setContentIntent(pendingIntent)
val notificationManager = getSystemService(Context.NOTIFICATION_SERVICE) as NotificationManager
// Since android Oreo notification channel is needed.
if (Build.VERSION.SDK_INT >= Build.VERSION_CODES.O) {
val channel = NotificationChannel(channelId,
"Channel human readable title",
NotificationManager.IMPORTANCE_DEFAULT)
notificationManager.createNotificationChannel(channel)
}
notificationManager.notify(0 /* ID of notification */, notificationBuilder.build())
}