我正在尝试在Excel中裁剪选定的图片,并尝试使用此代码:
Sub CropPicture()
Dim shpCrop As Shape
Dim sngMemoLeft As Single
Dim sngMemoTop As Single
Set shpCrop = Selection
With shpCrop
sngMemoLeft = .Left
sngMemoTop = .Top
With .PictureFormat
.CropLeft = 10
.CropTop = 10
.CropBottom = 10
.CropRight = 10
End With
.Left = sngMemoLeft
.Top = sngMemoTop
End With
End Sub
但是,这给了我一个类型不匹配的运行时错误。问题是这部分代码: 设置shp =选择
我该如何解决这个问题?
答案 0 :(得分:1)
尝试
Set shpCrop = ActiveSheet.Shapes(Selection.Name)
如果您在“立即”窗格中尝试? typename(selection)
,则会获得“#34;图片"”,但您可以通过其名称属性获取形状。