我将两条消息发回用户作为回复,如下所示, static Timer t = new Timer(new TimerCallback(TimerEvent));static Timer t1 = new Timer(new TimerCallback(TimerEventInActivity));static int timeOut = Convert.ToInt32(ConfigurationManage
static Timer t = new Timer(new TimerCallback(TimerEvent));
static Timer t1 = new Timer(new TimerCallback(TimerEventInActivity));
static int timeOut = Convert.ToInt32(ConfigurationManager.AppSettings["disableEndConversationTimer"]); //3600000
public static void CallTimer(int due) {
t.Change(due, Timeout.Infinite);
}
public static void CallTimerInActivity(int due) {
t1.Change(due, Timeout.Infinite);
}
public async static Task PostAsyncWithDelay(this IDialogContext ob, string text) {
try {
var message = ob.MakeMessage();
message.Type = Microsoft.Bot.Connector.ActivityTypes.Message;
message.Text = text;
await PostAsyncWithDelay(ob, message);
CallTimer(300000);
if ("true".Equals(ConfigurationManager.AppSettings["disableEndConversation"])) {
CallTimerInActivity(timeOut);
}
} catch (Exception ex) {
Trace.TraceInformation(ex.Message);
}
}
await context.PostAsyncWithDelay("Great!");
await context.PostAsyncWithDelay("I can help you with that.");
但是,收到它们之间没有延迟.这两条消息都是一次性收到的.
如何延迟第二条消息一段时间?
在根对话框中要延迟消息,可以使用Task.Delay方法.将PostAsyncWithDelay更改为:
public async static Task PostAsyncWithDelay(IDialogContext context, string text)
{
await Task.Delay(4000).ContinueWith(t =>
{
var message = context.MakeMessage();
message.Text = text;
using (var scope = DialogModule.BeginLifetimeScope(Conversation.Container, message))
{
var client = scope.Resolve<IConnectorClient>();
client.Conversations.ReplyToActivityAsync((Activity)message);
}
});
}
如果要延迟消息,可以调用PostAsyncWithDelay方法,否则使用context.PostAsync方法发送消息.
private async Task MessageReceivedAsync(IDialogContext context, IAwaitable<object> result)
{
//Sending a message nomally
await context.PostAsync("Hi");
//Notify the user that the bot is typing
var typing = context.MakeMessage();
typing.Type = ActivityTypes.Typing;
await context.PostAsync(typing);
//The message you want to delay.
//NOTE: Do not use context.PostAsyncWithDelay instead simply call the method.
await PostAsyncWithDelay(context, "2nd Hi");
}
OUTPUT
