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Get coordinate of polyline by VBA


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I've just started learning VBA in CAD.I wanna select polyline onscreen and then get its coordinate(of every vertex).But I don't know how to do it.

Thanks for any help.


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Something like this?


Sub Example_Coordinates()

Dim Selection As AcadSelectionSet
Dim Poly As AcadLWPolyline
Dim Obj As AcadEntity
Dim Bound As Double

'Makes a selectionset.
'On Error Resume Next
   Set Selection = ThisDrawing.SelectionSets.Item("Select polyline.")
If Err Then
   Set Selection = ThisDrawing.SelectionSets.Add("Select polyline.")
End If

'Select the polyline.

For Each Obj In Selection

   If Obj.ObjectName = "AcDbPolyline" Then
           Set Poly = Obj
           On Error Resume Next
           Bound = UBound(Poly.Coordinates)
           x = 0
           y = 1
           For i = 0 To Bound / 2
               MsgBox "X= " & Poly.Coordinates(x) & "  Y= " & Poly.Coordinates(y)
               If Err Then Err.Clear
               x = x + 2
               y = y + 2
   End If

Next Obj

End Sub


I hope this helps you!



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  • 5 years later...
Hello, someone please,


Sub Example_Coordinates()

without the sub xx()


how to run this code?


Try another one instead,type vbarun in the command line,

select "demo" sub


Option Explicit
Function PolyCoords(oEnt As AcadEntity) As Variant
    Dim cnt As Integer
    Dim i As Integer
    Dim j As Integer
    Dim iStep As Integer
    Dim varPt As Variant
    Dim dblCoords() As Double
    Dim dblVert() As Double

    If TypeOf oEnt Is AcadLWPolyline Then
         iStep = 2
    ElseIf TypeOf oEnt Is Acad3DPolyline Or _
           TypeOf oEnt Is AcadPolyline Then
         iStep = 3
    End If
    dblCoords = oEnt.Coordinates

    ReDim ptsArr(0 To (UBound(dblCoords) + 1) \ iStep - 1, 0 To iStep - 1) As Double
    For i = 0 To (UBound(dblCoords) + 1) \ iStep - 1
         For j = 0 To iStep - 1
              ptsArr(i, j) = dblCoords(cnt)
              Debug.Print ptsArr(i, j)
              cnt = cnt + 1
    PolyCoords = ptsArr
End Function

Sub demo()
    Dim pts As Variant
    Dim varPt As Variant
    Dim oEnt As AcadEntity
    ThisDrawing.Utility.GetEntity oEnt, varPt, vbCr & "Select polyline"
    If Not TypeOf oEnt Is AcadLWPolyline And _
       Not TypeOf oEnt Is Acad3DPolyline And _
       Not TypeOf oEnt Is AcadPolyline Then
         MsgBox "Method is not applicable for this entity type"
         Exit Sub
    End If
    pts = PolyCoords(oEnt)
End Sub


Next tyme post your question on VBA branch, please:




Edited by fixo
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  • 8 months later...

I’m a bit discouraged, I started to learn acad-vba since a few days too & I wanted to understand Steven Bastiaanse’s example which seems simple.

Why #40 & #41 , what does it mean.

Set Selection = ThisDrawing.SelectionSets.Item (”Select polyline” & #41; ???

(ThisDrawing.selectionsets) expected object or method isn’t it?

Thanks for helping me

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  • 1 year later...

Hello Sir

Please hlep me with this

I need to add a polyline based of part of coodonates of first polyline

So i need to copy vertex example from 0 to 5 of first polyline.


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Here is a VL example you will end up with a variable co-ordsxy which contains all the pts for a 2d pline as 2 values (x1 y1 x2 y2 x3 y3..... you can then get each pair from the list using nth x & nth (+ x 1)


; pline co-ords example
; By Alan H
(defun getcoords (ent)
   (vlax-ename->vla-object ent)

(defun co-ords2xy ()
; convert now to a list of xy as co-ords are x y x y x y if 3d x y z x y z
(setq len (length co-ords))
(setq numb (/ len 2)) ; even and odd check required
(setq I 0)
(repeat numb
(setq xy (list (nth i co-ords)(nth (+ I 1) co-ords) ))
; odd (setq xy (list (nth i co-ords)(nth (+ I 1) co-ords)(nth (+ I 2) co-ords) ))
(setq co-ordsxy (cons xy co-ordsxy))
(setq I (+ I 2))
; program starts here
(setq co-ords (getcoords (car (entsel "\nplease pick pline"))))
(co-ords2xy) ; list of 2d points making pline

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