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楼主 |
发表于 2016-3-11 15:51:37
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来自: 中国浙江嘉兴
) h) B' N; I( ~7 Q至于用宏的方法,那里可以找到相关资料? 之前已分享不少了,若是不能吸收理解,就僅能抄襲無法廣用., ^- y% z) ]. y( p$ z
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如下參考$ y" @ J+ P1 w1 U
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! \3 }) k v( \' ~' ?' ^' ******************************************************************************+ N( |& {" K d% I5 f. f* u
' macro recorded on 03/04/16 by scliang
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Dim swApp As Object
9 s- h+ E5 H; wDim Part As Object
- V! p4 L; t3 _! _+ UDim boolstatus As Boolean# G* o: L$ @. x: H- d
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Sub main()9 i1 n _$ o$ k5 Q9 o+ g
Set swApp = Application.SldWorks
+ d4 ], \/ R& o2 { K+ ?- ISet Part = swApp.ActiveDoc
: H, ]" B$ j, s) X' X/ FSet myModelView = Part.ActiveView% C2 ]) K% h$ u& [* O6 q/ i* {4 d
Dim myDimension As Object
' {1 G) A% Q, e8 l6 |pi = Atn(1) * 4' C$ d/ `2 [1 ?8 t& O4 W1 u
Set myDimension = Part.Parameter("D15@草圖1") '對應控制角度, `" q' |1 M0 i0 z. h a* A
For i = 0.1 To 90.1 Step 5
% _9 [/ ]) ?+ |! j; A) F+ d5 FA = i * pi / 180) y1 s, A: c& m+ T3 U
myDimension.SystemValue = A3 K& X4 c& p K/ G' i* ~/ V
boolstatus = Part.EditRebuild3()
) m! C( r1 K$ {2 N7 P* n. omyModelView.RotateAboutCenter 0, 0
% u1 P" o0 V- l0 C5 @. sNext; a2 s7 M8 }- ~) R- g+ }
For i = 85.1 To 0.05 Step -5
- d* t: x# ^' [" F1 @$ R% |4 kA = i * pi / 180
/ F$ a* M0 ?5 L: w+ X: CmyDimension.SystemValue = A
: F/ ]$ V+ M; H0 q; f1 Hboolstatus = Part.EditRebuild3()
: ^/ ]$ U# q/ c- t) ? b- ^myModelView.RotateAboutCenter 0, 0; ?8 |( Z) L" l4 J) G- f, i
Next* `( ~$ h7 E5 R, l( u
End Sub+ N# C8 r9 I$ ~0 V3 `
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Macro1.rar
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