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我想在cad上画矩形框后,怎么在vba中得到矩形的左下角和右上角坐标?

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发表于 2002-7-18 19:31 | 显示全部楼层 |阅读模式
我是刚学vba,我想实现的功能是这样的:
运行一个宏,然后画一个临时的框进行选择,然后能在vba中得到临时框的坐标,怎么去实现哪?
我都弄了一天了,也没实现,没什么思路,请大家帮帮忙吧,谢谢了!!!
发表于 2002-7-18 19:54 | 显示全部楼层

以下程序不知是否合用

Sub gcorner()
Dim FstPnt As Variant
Dim SndPnt As Variant
FstPnt = ThisDrawing.Utility.GetPoint(, "选取矩形第一角点:")
SndPnt = ThisDrawing.Utility.GetCorner(FstPnt, "选取矩形对角点:")
MsgBox "矩形框的坐标如下:" & vbCrLf & _
    "第一角点的坐标为" & FstPnt(0) & ", " & FstPnt(1) & ", " & FstPnt(2) & vbCrLf & _
    "第二角点的坐标为" & SndPnt(0) & ", " & SndPnt(1) & ", " & SndPnt(2), , "矩形框坐标"
End Sub
 楼主| 发表于 2002-7-19 12:39 | 显示全部楼层

怎么去画多边形,并取得点的坐标?

版主,太谢谢了,就是这个意思,我还问一下,怎么去画多边形,并取得点的坐标?
发表于 2002-7-19 21:51 | 显示全部楼层

再给个例子吧

Sub polygon()
'以下语句绘制正多边形
    Dim num As Integer
    Dim pnt As Variant
    Dim lpnt As Variant
    num = ThisDrawing.Utility.GetInteger("请选择正多边形的边数:")
    Dim fpnt As Variant
    fpnt = ThisDrawing.Utility.GetPoint(, "请选择正多边形的起点:")
    Dim leng As Double
    leng = ThisDrawing.Utility.GetDistance(fpnt, "请选择正多边形的边长:")
    ReDim lpnt(0 To num * 2 - 1) As Double
    pnt = fpnt
    lpnt(0) = pnt(0)
    lpnt(1) = pnt(1)
    Dim st As Integer
    For st = 1 To num - 1
        pnt = ThisDrawing.Utility.PolarPoint(pnt, (3.14159265 * 2 / num) * (st - 1), leng)
        lpnt(st * 2) = pnt(0)
        lpnt(st * 2 + 1) = pnt(1)
    Next st
    Dim pgon As AcadLWPolyline
    Set pgon = ThisDrawing.ModelSpace.AddLightWeightPolyline(lpnt)
    pgon.Closed = True
    ThisDrawing.Regen (True)
'以下语句获取多边形的顶点
    Dim gpnt As Variant
    gpnt = pgon.Coordinates
    Dim pntcnt As Integer
    pntcnt = UBound(gpnt)
    Dim disptxt As String
    disptxt = "多边形共有" & (pntcnt + 1) / 2 & "个顶点" & vbCrLf
    Dim i As Integer
    For i = 0 To pntcnt - 1 Step 2
        disptxt = disptxt & "第" & i / 2 + 1 & "个顶点的坐标为:" & _
                gpnt(i) & "," & gpnt(i + 1) & vbCrLf
    Next i
    disptxt = disptxt & "明经通道VBA示例 http://www.mjtd.com"
    MsgBox disptxt, , "多边形的坐标显示"
End Sub
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