Botframework 以编程方式指定LUIS对话框拼写检查插槽
到目前为止,通过执行以下操作,我可以避免对我的LUIS appId和key进行硬编码:Botframework 以编程方式指定LUIS对话框拼写检查插槽,botframework,Botframework,到目前为止,通过执行以下操作,我可以避免对我的LUIS appId和key进行硬编码: var luisService = new LuisService(new LuisModelAttribute(ConfigurationManager.AppSettings["LuisAppId"], ConfigurationManager.AppSettings["LuisAppKey"])); context.Call(new LuisDialog(luisService), ResumeAfte
var luisService = new LuisService(new LuisModelAttribute(ConfigurationManager.AppSettings["LuisAppId"], ConfigurationManager.AppSettings["LuisAppKey"]));
context.Call(new LuisDialog(luisService), ResumeAfterDialog);
然后将我的LUIS对话框声明为:
[Serializable]
public class LuisDialog : LuisDialog<object>
{
public LuisDialog(ILuisService ls) : base(ls)
{
}
....
}
[可序列化]
公共类LuisDialog:LuisDialog
{
公共LuisDialog(ILuisService ls):基本(ls)
{
}
....
}
}
但我也希望能够以编程方式设置LuisModel属性中可用的SpellCheck=true、Log、Verbose和其他参数,有没有办法做到这一点
谢谢我找到了答案,在创建LuissService之前,我只需要在代码中设置LuisModelAttribute属性:
var luisSettings = new LuisModelAttribute(ConfigurationManager.AppSettings["LuisAppId"], ConfigurationManager.AppSettings["LuisAppKey"]);
luisSettings.Log = true;
luisSettings.SpellCheck = true;
luisSettings.Log = true;
var luisService = new LuisService(luisSettings);
context.Call(new LuisDialog(luisService), ResumeAfterDialog);