topic Enclosed Region using Room Separator in Revit API Forum

Enclosed Region using Room Separator

Anonymous — Tue, 04 Oct 2016 12:29:16 GMT

How can I programmatically create an enclosed space using Room Separators and assign Room object to it. Can somebody provide a sample code to work with Room Separator.

Re: Enclosed Region using Room Separator

atiefenbach — Thu, 06 Oct 2016 11:58:41 GMT

This should get you going:

public void CreateRoomAndSeperators()
{
	Document doc = this.ActiveUIDocument.Document;
	View activeView = doc.ActiveView; //You should make sure you're in a Plan View
	
	XYZ pt1 = new XYZ(0, 0, 0);
	XYZ pt2 = new XYZ(0, 10, 0);
	XYZ pt3 = new XYZ(10, 10, 0);
	XYZ pt4 = new XYZ(10, 0, 0);
	
	UV pt5 = new UV(5, 5);
	
	CurveArray curveArray = new CurveArray();
	curveArray.Append(Line.CreateBound(pt1, pt2));
	curveArray.Append(Line.CreateBound(pt2, pt3));
	curveArray.Append(Line.CreateBound(pt3, pt4));
	curveArray.Append(Line.CreateBound(pt4, pt1));
	
	using (Transaction trans = new Transaction(doc))
	{
		trans.Start("Create Room and Boundaries");
		ModelCurveArray lines = doc.Create.NewRoomBoundaryLines(activeView.SketchPlane, curveArray, activeView);
		Room room = doc.Create.NewRoom(activeView.GenLevel, pt5);
		//now modify the "room" Element's Name, Number, etc.
		trans.Commit();
	}
}

Re: Enclosed Region using Room Separator

JimJia — Tue, 11 Oct 2016 08:30:51 GMT

Yes, NewRoomBoundaryLines() is the answer!

Thanks Anthony's code!

Re: Enclosed Region using Room Separator

MiguelGT17 — Thu, 19 Jan 2023 16:32:48 GMT

Some how the code doesnt work if I set the current view as a 3D view. Indeed, the following issue pops up:

Re: Enclosed Region using Room Separator

Tabeer_Tariq — Mon, 22 Jul 2024 17:32:00 GMT

Can this be done using dynamo only?

Re: Enclosed Region using Room Separator

AnthonyTiefenbach — Tue, 23 Jul 2024 14:17:04 GMT

I’m sure it’s possible, but you’d probably get a better answer in a Dynamo forum.