_fillt圆角多段线，如何获得多段线的点
<p>感谢大家了！上次的问题，我已经明白了，感谢大家的帮忙！这次又有一个问题！</p><p>我想使用sendcommand 圆角一个直线和一个多段线</p><p>这个直线交在多段线的中间某段</p><p>圆角的时候要获得这两个对象的拾取点，直线的可以line.endpoint，多段线的点该怎么获得啊？？？！</p><p>我试着用几何算法，算了一个点，</p><p>但是，圆角的时候图全乱了，说是 <font color="#f73809">命令: (list(handent"3A14")(list 305.32 .15 0)) ; 错误: 输入中的点位置不正确</font></p><p><font color="#000000">小弟不明白这种问题怎么处理啊？？？请大家帮忙啊！急！！！！</font></p> <p>我现在已经知道怎么获得多段线上的点通过oblect.Coordinates(index)</p><p>我自己做了个小例子：</p><p>Rem 倒圆角是对双元表时<br/>Public Function GetDoubleEntTable(entObj As AcadEntity, Pnt As Variant) As String<br/> Dim entHandle As String<br/> entHandle = entObj.Handle<br/> GetDoubleEntTable = "(list(handent" & Chr(34) & entHandle & Chr(34) & ")(list" & Str(Pnt(0)) & Str(Pnt(1)) & Str(Pnt(2)) & "))"<br/> End Function<br/>Rem 倒圆角实体<br/>Public Function axEnt2lspEnt(entObj As AcadEntity) As String<br/>Dim entHandle As String<br/>entHandle = entObj.Handle<br/>axEnt2lspEnt = "(handent" & Chr(34) & entHandle & Chr(34) & ")"<br/>End Function</p><p><br/> <br/>Public Function TdnPLC(LWPLN1 As AcadLine, LWPLN2 As AcadLWPolyline, PikPt As Variant, FiltR As Double)</p><p>Dim Comd As String<br/>Dim EHadl As String<br/>Dim LHadl As String</p><p>Rem 倒圆角使用command<br/>Comd = "_fillet r " & FiltR & " "</p><p>ThisDrawing.SendCommand Comd</p><p>EHadl = axEnt2lspEnt(LWPLN1)<br/>LHadl = GetDoubleEntTable(LWPLN2, PikPt)</p><p>'Comd = "(handent" & Chr(34) & Trim(LWPLN1.Handle) & Chr(34) & ") " & "(handent" & Chr(34) & Trim(LWPLN2.Handle) & Chr(34) & ") "<br/>Comd = EHadl & " " & LHadl & " "</p><p>ThisDrawing.SendCommand Comd</p><p>ThisDrawing.Application.ZoomExtents</p><p>End Function<br/> <br/> <br/>Public Sub sy()</p><p>Dim myline1 As AcadLine<br/>Dim myline3 As AcadLWPolyline<br/>Dim myline2 As AcadLine<br/>Dim pt1(1 To 3) As Double<br/>Dim pt2(1 To 3) As Double<br/>Dim pt3(1 To 12) As Double<br/>Dim pt13(0 To 2) As Double<br/>Dim pt23(0 To 2) As Double</p><p>Dim r As Double<br/>Dim i As Integer</p><p>pt1(1) = 0<br/>pt1(2) = 60<br/>pt1(3) = 0</p><p>pt2(1) = 10<br/>pt2(2) = 30<br/>pt2(3) = 0</p><p>Set myline1 = ThisDrawing.ModelSpace.AddLine(pt1, pt2)</p><p>pt1(1) = 40<br/>pt1(2) = 30<br/>pt1(3) = 0</p><p>pt2(1) = 50<br/>pt2(2) = 60<br/>pt2(3) = 0</p><p>Set myline2 = ThisDrawing.ModelSpace.AddLine(pt1, pt2)</p><p>r = 3</p><p>pt3(1) = 0<br/>pt3(2) = 0<br/>pt3(3) = 10<br/>pt3(4) = 30<br/>pt3(5) = 20<br/>pt3(6) = 40<br/>pt3(7) = 30<br/>pt3(8) = 40<br/>pt3(9) = 40<br/>pt3(10) = 30<br/>pt3(11) = 50<br/>pt3(12) = 0</p><p>Set myline3 = ThisDrawing.ModelSpace.AddLightWeightPolyline(pt3)</p><p>pt13(0) = myline3.Coordinates(4)<br/>pt13(1) = myline3.Coordinates(5)<br/>pt13(2) = 0</p><p>pt23(0) = myline3.Coordinates(6)<br/>pt23(1) = myline3.Coordinates(7)<br/>pt23(2) = 0</p><p>Call TdnPLC(myline1, myline3, pt13, r)<br/>Call TdnPLC(myline2, myline3, pt23, r)</p><p>End Sub</p><p>完全可以执行！</p><p>但是，我真正用的时候，就不行了！总是提示半径太大！但是不大！因为图形要用到数据库的东西，代码不好发~不好意思！</p><p>我用list查看生成的信息，发现了问题！</p><p>就是小例子的代码执行时：选择第二个对象: (list(handent"4B61")(list 30 40 0)) (<图元名: 7efe7a48> (30 40 <br/>0))</p><p>真正代码执行时：选择第二个对象: (list(handent"4B5C")(list 163.320007324219-2.96499999403954 0)) <br/>(<图元名: 7efe7a20> (nil 0))</p><p>大家注意看多了一个(nil 0)！！！！什么意思？是不是提示没有选中的意思！</p><p>请大家帮忙啊！我都搞了好几天了！实在不懂怎么回事了！</p><p>就像我以前说的，周围就我一个人，连请教的人都没有~就靠大家了！</p><p>我在这里谢过了！</p><p></p> GetDoubleEntTable = "(list(handent" & Chr(34) & entHandle & Chr(34) & ")(list " & Str(Pnt(0)) & " " & Str(Pnt(1)) & " " & Str(Pnt(2)) & "))"
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