EWS托管API 2.0 ID格式错误

Question

我正在使用EWS Managed API 2.0发送会议邀请并捕获用户响应。

我遵循了官方网站https://msdn.microsoft.com/zh-cn/library/office/jj220499(v=exchg.80).aspx和https://msdn.microsoft.com/zh-CN的参考/library/office/dd633661(v=exchg.80).aspx 。 我，由于ID格式错误，出现错误。

我在官方网站上使用了确切的代码。 我进行了一些谷歌搜索，发现如果id具有特殊字符，则可能会出现此错误。 但是我提供的ID没有特殊字符，只是一个简单字符。

这是代码

ServicePointManager.ServerCertificateValidationCallback = CertificateValidationCallBack;
            ExchangeService service = new ExchangeService(ExchangeVersion.Exchange2007_SP1);
            service.Credentials = new WebCredentials("********@live.com", "************");

            service.TraceEnabled = true;
            service.TraceFlags = TraceFlags.All;

            service.AutodiscoverUrl("***********@live.com", RedirectionUrlValidationCallback);

            // Create the appointment.

            Appointment appointment = Appointment.Bind(service, "AAMkA=");


            // Set properties on the appointment. Add two required attendees and one optional attendee.
            appointment.Subject = "Status Meeting";
            appointment.Body = "The purpose of this meeting is to discuss status.";
            appointment.Start = new DateTime(2009, 3, 1, 9, 0, 0);
            appointment.End = appointment.Start.AddHours(2);
            appointment.Location = "Conf Room";
            appointment.RequiredAttendees.Add("***********@live.com");

            // Send the meeting request to all attendees and save a copy in the Sent Items folder.
            appointment.Save(SendInvitationsMode.SendToAllAndSaveCopy);

            foreach (Attendee t in appointment.RequiredAttendees)
            {
                if (t.ResponseType != null)
                    Console.WriteLine("Required attendee - " + t.Address + ": " + t.ResponseType.Value.ToString());
            }

在Appointment appointment = Appointment.Bind(service, "AAMkA=");遇到错误Appointment appointment = Appointment.Bind(service, "AAMkA="); ID是故障。

Answer 1

Exchange服务器将在创建约会时提供唯一的ID。 因此，需要在获取用户响应时保存ID并传递相同的ID。

var Id = appointment.Id;

因此，传递ID来获取响应。