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本帖最后由 woaishuijia 于 2011-7-21 05:43 编辑 0 h- g2 p7 x' [, L# j( {6 n, k
+ C; U6 ]; Y. a用VBa做了个简单的小程序,居然提示“无效的过程调用或参数”
9 z) C7 D6 |" f8 b/ \2 p程序:
* f$ s4 a' r* f- Public Sub HTT()
2 z3 r" J: s* g: ?- B! K0 p7 q - Dim Pt0, PT00 As Variant
. X1 k) D+ I% P+ i" W3 | - Dim PT1(0 To 2), PT2(0 To 2), PT3(0 To 2) As Double
, E, M% D2 W5 O8 N% v4 g! k7 W - Dim L0, L1 As Double
* l, s7 r8 f: R+ {! T - Dim i, m, n As Integer
# y' x8 ]5 u0 _$ m3 D6 _3 ^. } - Dim ALine As AcadLine
0 W" j! S/ l) C0 M7 l - Pt0 = ThisDrawing.Utility.GetPoint(, "基点:")
) D# L5 J7 R& N' g9 U5 c- e6 X - X1 = Pt0(0)
' r2 _7 w5 [! j- v- h( | - Y1 = Pt0(1)
2 M: M( D' }" [) ?% m* o - L0 = ThisDrawing.Utility.GetDistance(, "单节筒节宽度:")
) O& k8 c6 f8 U - L1 = ThisDrawing.Utility.GetDistance(, "筒节直径:")# N1 \& W! D) h+ u- @
- PT1(0) = Pt0(0) + L1
5 v5 H2 Y% x" o) [0 N3 N - PT1(1) = Pt0(1)' J$ x+ V" [; a' u% h" r
- PT1(2) = Pt0(2)# L0 m( Z9 L( n
- PT2(0) = Pt0(0) + L1
! j! y# x" ~. [ - PT2(1) = Pt0(1) - l2 ~& X9 N( D# n% i
- PT2(2) = Pt0(2)& e7 B ^! y& l y! w: U* k
- PT3(0) = Pt0(0)3 Z4 `* Y# L! f) s8 |
- PT3(1) = Pt0(1) - l2: K' K9 V; R6 E# w- y7 Z- G- G' G+ j
- PT3(2) = Pt0(2)% j+ ?, n/ Z9 K0 C% |6 \& {
- ( z6 v3 ]! s r- L6 c8 b, Y3 L
- Set ALine = ThisDrawing.ModelSpace.AddLine(Pt0, PT1)& _( P+ Y( o! C% `
- Set ALine = ThisDrawing.ModelSpace.AddLine(PT1, PT2) U) k( Y7 D7 ]! p
- Set ALine = ThisDrawing.ModelSpace.AddLine(PT2, PT3)
$ O3 M8 e: Q6 E& x8 X1 _" _$ W - Set ALine = ThisDrawing.ModelSpace.AddLine(PT3, Pt0)( { v: E% y0 D7 _
- ZoomAll* C' [# p! |' C" a
- End Sub
复制代码 $ K: V L( _( ^' L( T
错误提示:
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) H G1 F8 Y' g" _& c% ]有知道如何解决的吗? |
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