Question

我试图使用Azure Bot框架创建一个机器人，它将要求一个未知数量的复杂对象（每个都需要三个响应）。但我不知道如何为根形式中的每个复杂对象创建表单。见http://docs.botframework.com/sdkreference/csharp/forms.html。它指出：＆＃34;为了处理复杂对象列表，您需要为顶级C＃类创建一个表单，并为复杂对象创建一个表单。您可以使用Dialogs系统一起组合表单。＆＃34; 这就是我不知道该怎么做。

public enum SystemSelection { SharePoint, BizTalk, Azure, Office365 };

public enum RequestType { Bug, SupportRequest, Question };

public enum Importance { Blocking, High, Medium, Low };

[Serializable]
class Declaration
{
    public string Type;
    public string Amount;
    public string Date;

    public static IForm<Declaration> BuildForm()
    {
        return new FormBuilder<Declaration>()
                .Message("Add a declaration")
                .Build();
    }
}

[Serializable]
class SupportRequest
{

    public SystemSelection? SystemSelection;
    public RequestType? RequestType;
    public Importance? Importance;

    public List<Declaration> Declarations;

    public static IForm<SupportRequest> BuildForm()
    {
        IForm<Declaration> aForm = new FormBuilder<Declaration>().Message("Add declaration").Build();

        // now what?

        return new FormBuilder<SupportRequest>()
                .Message("Welcome to the simple support bot!")
                .Build();
    }
}

控制器：

  [BotAuthentication]
  public class MessagesController : ApiController
{
    internal static IDialog<SupportRequest> MakeRootDialog()
    {
        // change something here??
        return Chain.From(() => FormDialog.FromForm(SupportRequest.BuildForm));
    }

    public async Task<Message> Post([FromBody]Message message)
    {
        if (message.Type == "Message")
        {
            return await Conversation.SendAsync(message, MakeRootDialog);
        }
        else
        {
            return HandleSystemMessage(message);
        }
    }

Answer 1

您可以使用Chain课程中的方法撰写对话框。由于这些方法也支持LINQ语法，因此您可以在MakeRootDialog中编写类似的内容，以按顺序执行SupportRequest对话框和Decalaration对话框：

    internal static IDialog<SupportRequest> MakeRootDialog()
    {
        var dlg = from x in FormDialog.FromForm(SupportRequest.BuildForm)
                  from y in FormDialog.FromForm(Declaration.BuildForm)
                  select x;
        return dlg;
        // return Chain.From(() => FormDialog.FromForm(SupportRequest.BuildForm));
    }

您也可以手动链接对话框：

        var dlg = Chain.From(() => FormDialog.FromForm(SupportRequest.BuildForm))
            .ContinueWith<SupportRequest,Declaration>(async (ctx, sr) =>
            {
                var res = await sr;
                return FormDialog.FromForm(Declaration.BuildForm);
            });

Azure Bot框架FormFlow复杂形式

1 个答案: