topic Re: Automatically OpenAndActivate a cloud model on Revit startup in Revit API Forum

Automatically OpenAndActivate a cloud model on Revit startup

Rockit_for_Revit — Sun, 02 Feb 2025 12:29:36 GMT

Hello,

I'm trying to automatically OpenAndActivate a cloud model. Given I need to access UIApplication I am doing it via UIControlledApplication.Idling event.

The command works fine when I open a local Revit model. However when I open a cloud model I am left with the following dialog 'Open Model' indefinitely and I have no option but to kill the Revit session.

I am opening the cloud model using the following:

Dim modelPath As ModelPath = ModelPathUtils.ConvertCloudGUIDsToCloudPath(If(region, ModelPathUtils.CloudRegionUS), ' Default to US region if not provided New Guid(projectGUID), New Guid(modelGUID) ) Dim openOptions As New OpenOptions() Dim objUIdocument As UIDocument = uiApp.OpenAndActivateDocument(modelPath, openOptions, False)

Any ideas?

Re: Automatically OpenAndActivate a cloud model on Revit startup

ricaun — Sun, 02 Feb 2025 16:38:01 GMT

Maybe it's Idling, it's usually not safe to run long processes in there.

You could use ExternalEvent and verify if that works.

Some sample code like this:

using Autodesk.Revit.DB; using Autodesk.Revit.UI; using System; namespace RevitAddin.Forum { public class AppCloud : IExternalApplication { public Result OnStartup(UIControlledApplication application) { var region = ModelPathUtils.CloudRegionUS; var projectGuid = Guid.Parse("00000000-0000-0000-0000-000000000000"); var modelGuid = Guid.Parse("00000000-0000-0000-0000-000000000000"); // Create ExternalEvent and Raise to execute in the next idle cycle ExternalEvent.Create(new CloudOpenExternalEvent(region, projectGuid, modelGuid)).Raise(); return Result.Succeeded; } public Result OnShutdown(UIControlledApplication application) { return Result.Succeeded; } } public class CloudOpenExternalEvent : IExternalEventHandler { private string Region; private Guid ProjectGuid; private Guid ModelGuid; public CloudOpenExternalEvent(string region, Guid projectGuid, Guid modelGuid) { Region = region; ProjectGuid = projectGuid; ModelGuid = modelGuid; } public void Execute(UIApplication uiapp) { try { var modelPath = ModelPathUtils.ConvertCloudGUIDsToCloudPath(Region, ProjectGuid, ModelGuid); var options = new OpenOptions(); uiapp.OpenAndActivateDocument(modelPath, options, false); } catch (Exception ex) { TaskDialog.Show(GetName(), ex.ToString()); } } public string GetName() { return nameof(CloudOpenExternalEvent); } } }

Re: Automatically OpenAndActivate a cloud model on Revit startup

Rockit_for_Revit — Mon, 03 Feb 2025 04:06:11 GMT

Perfect thank you so much. I needed to seperate it out of OnStartup, however its essentally the same:

Private Shared _externalEvent As ExternalEvent Public Sub CreateAndRaiseExternalEvent(filePath As String) _externalEvent = ExternalEvent.Create(New ExternalEvent(filePath)) _externalEvent.Raise() End Sub

Re: Automatically OpenAndActivate a cloud model on Revit startup

jeremy_tammik — Mon, 03 Feb 2025 05:34:48 GMT

OnStartup is useful to register event handlers and set up add-ins and stuff, but it is not the optimal time and place to start executing code. ApplicationInitialized is a later and probably more suitable choice:

https://www.revitapidocs.com/2024/f35ba9fc-0b6b-4284-60eb-91788761127c.htm

Subscribe to this event to get notified after the Revit application has been initialized.