topic Re: Create dimensions in a room in Revit API Forum

Create dimensions in a room

Anonymous — Wed, 27 Mar 2019 17:33:20 GMT

Hi everyone!

I am trying to get the dimensions in a room, but I have a mistake in the code. Because when I built it, I get every dimension with the value 0.0, like you could see in the next picture.

I don't attach the completely code because is so long, there are a lot of verifications...

Therefore I am going to attach the parts that I think that is important,and where maybe the mistake is...

I first created a loop to run through the walls of the room.

With this I create the point, the start of the wall and the end of the wall.

var startVertice1 = boundSeg1.GetCurve().GetEndPoint(0);
var endVertice1 = boundSeg1.GetCurve().GetEndPoint(1);

Line dimensionLine1 = Line.CreateBound(startVertice1, endVertice1);

Then create a sketch and the ModelCurve

SketchPlane sketch = SketchPlane.Create(doc, geomPlane);
modelCurve1 = doc.Create
.NewModelCurve(dimensionLine1, sketch);

The same for the other line!

Then I add the geometry details references to an array:

references.Append(modelCurve1.GeometryCurve.Reference);
references.Append(modelCurve2.GeometryCurve.Reference);

I prove that the two lines are paralell and also that they are different:

var paralell = Math.Abs(Vector3D.DotProduct(vVector3D1, vVector3D2));

if (boundSeg2.ElementId != boundSeg1.ElementId)
{
 //comprobamos si es paralelo 
         if (paralell > 1E-1)
          {....

If the conditions are true, I create a line for the dimension:

 XYZ dimensionLineStart = new XYZ(startVerticeXyz1.X + Cota,startVerticeXyz1.Y, 0);
 XYZ dimensionLineEnd = new XYZ(endVertice2.X + Cota,endVertice2.Y,0);
 dimensionLine = Line.CreateBound(dimensionLineStart,dimensionLineEnd);

And finally I use NewDimension with the view

using (Transaction tCota = new Transaction(doc, "dimensions"))
 {
  tCota.Start();
  var comprodDimension =doc.Create.NewDimension(doc.ActiveView, dimensionLine, references);
  tCota.Commit();
 }

Re: Create dimensions in a room

jeremytammik — Thu, 28 Mar 2019 07:21:32 GMT

Does this help?

https://thebuildingcoder.typepad.com/blog/about-the-author.html#5.45

Good luck!

Cheers,

Jeremy

Re: Create dimensions in a room

naveen.kumar.t — Thu, 28 Mar 2019 10:07:57 GMT

Hi @Anonymous ,

So you want to create dimension in a room?

Try using the below code

Select the Room and try to run the below code

if(Room!=null)
                {
                    SpatialElementBoundaryOptions op = new SpatialElementBoundaryOptions();
                    op.SpatialElementBoundaryLocation = SpatialElementBoundaryLocation.Center;
                    IList<IList<BoundarySegment>> segmensts = Room.GetBoundarySegments(op) as IList<IList<BoundarySegment>>;
                    foreach ( IList<BoundarySegment> BSS in segmensts)
                    {
                        foreach(BoundarySegment BS in BSS)
                        {

                            XYZ p1 = new XYZ(BS.GetCurve().GetEndPoint(0).X, BS.GetCurve().GetEndPoint(0).Y, BS.GetCurve().GetEndPoint(0).Z);
                            XYZ p2= new XYZ(BS.GetCurve().GetEndPoint(1).X, BS.GetCurve().GetEndPoint(1).Y, BS.GetCurve().GetEndPoint(1).Z);

                            Line l1 = Line.CreateBound(p1, p2);
                            ModelCurve mc = doc.Create.NewModelCurve(l1, SketchPlane.Create(doc, r.LevelId));
                            if(l1!=null)
                            {
                                ReferenceArray RA = new ReferenceArray();
                                RA.Append(mc.GeometryCurve.GetEndPointReference(0));
                                RA.Append(mc.GeometryCurve.GetEndPointReference(1));
                                doc.Create.NewDimension(doc.ActiveView, l1, RA, newdimesionType);
                            }
                        }                       
                    }
                }

and

while setting the op.SpatialElementBoundaryLocation, make sure you are giving the proper boundary options for the spatial element geometry calculation

Also refer the below link

SpatialElementBoundaryLocation

If this helped solve your problem please mark it as solution, so other users can get this solutions as well

Re: Create dimensions in a room

jeremytammik — Thu, 28 Mar 2019 10:54:53 GMT

Dear Naveen,

Thank you very much for a nice approach!

One little note on the code: it would be more readable if you just called BS.GetCurve() once (not six times!) and stored the result in a variable, e.g. `c`.

Next, why don't you simply say:

  XYZ p1 = c.GetEndPoint(0);
  XYZ p2 = c.GetEndPoint(1);

Finally, a more serious and non-trivial problem:

This solution creates a model curve.

If the room changes, the model curve will not, so the dimension will not adapt.

The association is lost.

You could implement the association yourself using DMU to update the model curve when the room changes.

It would be simpler if the dimension could use the built-in Revit associativity functionality to adapt when the room changes.

After all, that is what BIM is all about...

Can you retrieve references from the room? Maybe, to achieve that, you have to retrieve references from the room bounding walls?

Cheers,

Jeremy

Re: Create dimensions in a room

Revitalizer — Thu, 28 Mar 2019 11:05:57 GMT

Hi,

I don't consider an Updater as a good idea because the NewDimension method takes a view argument.

A Room may be changed while a ViewSection being the active view, for example.

So you would need to store the view Id, but the view could be deleted meanwhile etc.

Revitalizer

Re: Create dimensions in a room

Anonymous — Fri, 29 Mar 2019 16:33:34 GMT

Hi Naveen!

Thank you so much for your code. Thanks to your code I have realized that I had added in the array References the geometry of the two differentlines, but as you had written in the answer I have to add two points like references (of the same line) :

RA.Append(mc.GeometryCurve.GetEndPointReference(1))

After fixed this, my code works perfectly.

Thanks so much!

Lorena

Re: Create dimensions in a room

Anonymous — Fri, 27 Mar 2020 15:00:31 GMT

Where did you write this code?

I am using python node inside dynamo, so I noticed that this syntax is not written in python.

Kindly advice

Thank you

Re: Create dimensions in a room

Anonymous — Fri, 27 Mar 2020 15:02:30 GMT

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