topic Room Separation Line - Associated Level in Revit API Forum

Room Separation Line - Associated Level

gprofile — Fri, 08 Dec 2017 10:16:31 GMT

Is it possible to get the associated level of a Room separation Line by creating a simple Macro, as it isn't available in the properties.

I checked in the Revit 2017.1 API documentation but couldnt find anything related to Room Separation Line.

Re: Room Separation Line - Associated Level

Anonymous — Fri, 08 Dec 2017 10:45:28 GMT

You should be able to get the owner view, cast it as ViewPlan and get the Level from the view plan.

 var viewPlan = (ViewPlan) Document.GetElement(modelCurve.OwnerViewId);
 var level = (Level) Document.GetElement(viewPlan.LevelId);

Re: Room Separation Line - Associated Level

FAIR59 — Fri, 08 Dec 2017 15:23:11 GMT

the room separation line can be visible in floorplans of a different level, or even in 3D views.

this will sound a bit hacky, but the name of the sketchplane of the separationline is equal to the name of the level.

                SketchPlane sk = modelcurve.SketchPlane;
                IEnumerable<Level> levels = new FilteredElementCollector(doc)
                	.OfClass(typeof(Level))
                	.Cast<Level>()
                	.Where( l => l.Name.Equals(sk.Name));
                Level LevelofLine = levels.FirstOrDefault();

another approach would be to find the level by the height from the sketchplane.Plane [ height = sk.GetPlane().Origin.Z ]

Re: Room Separation Line - Associated Level

gprofile — Fri, 08 Dec 2017 15:58:34 GMT

Excuse me for my ignorance in C# i tried the code below but couldn't solve the error

Error CS0039: The Autodesk.Revit.DB.GeometryObject type cannot be converted to Autodesk.Revit.DB.ModelCurve with a reference conversion, boxing conversion, unboxing conversion, wrapping conversion, or NULL type conversion.

UIDocument uiDoc = this.ActiveUIDocument;
Document doc = uiDoc.Document;
			
Reference myRef = uiDoc.Selection.PickObject(ObjectType.Element);
Element ele = doc.GetElement(myRef);
			
GeometryObject geomObj = ele.GetGeometryObjectFromReference(myRef);
ModelCurve modelcurve = geomObj as ModelCurve;
			
SketchPlane sk = modelcurve.SketchPlane;
IEnumerable<Level> levels = new FilteredElementCollector(doc).OfClass(typeof(Level)).Cast<Level>().Where( l => l.Name.Equals(sk.Name));
Level levelOfLine = levels.FirstOrDefault();

Please guide me.

Re: Room Separation Line - Associated Level

FAIR59 — Fri, 08 Dec 2017 16:09:08 GMT

The GeometryObject is not needed.

you can get the modelcurve directly:

ModelCurve modelcurve = doc.GetElement(myRef) as ModelCurve;

Re: Room Separation Line - Associated Level

gprofile — Fri, 08 Dec 2017 16:31:51 GMT

I cannot understand why this code throws a null Exception, when I select a Line

UIDocument uiDoc = this.ActiveUIDocument;
			Document doc = uiDoc.Document;
			
			Reference myRef = uiDoc.Selection.PickObject(ObjectType.Element);
			Element e = doc.GetElement(myRef);
			ModelCurve modelcurve = e as ModelCurve;
			
			var viewPlan = (ViewPlan) doc.GetElement(modelcurve.OwnerViewId);
 			var level = (Level) doc.GetElement(viewPlan.LevelId);
 			
 			TaskDialog.Show("Level is : ",e.Name + level.Name);

Re: Room Separation Line - Associated Level

FAIR59 — Sun, 10 Dec 2017 14:59:41 GMT

a room separation line is a model element, and as such it doesn't have a OwnerView. ( Only annotation elements have a Ownerview.)

modelcurve.OwnerViewId = ElementId.InvalidElementId, resulting in doc.GetElement(modelcurve.OwnerViewId) = null