Thread: Shape Range
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Jon Peltier[_2_] Jon Peltier[_2_] is offline
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Default Shape Range

The next thing I would try is looping through the shapes in the shape
range, and apply the formatting to one shape at a time.

- Jon
-------
Jon Peltier
Peltier Technical Services, Inc.
http://peltiertech.com/


On 4/22/2010 9:01 AM, kuhrty wrote:
Luke,

Thank you for your input, when I attempt to qualify that way, the code
does run but the shapes don't fill with the color passed. I would
have assumed it is the correct way to code it but with an unexpected
result.

Thanks again for your response.

Mark

On Apr 21, 4:37 pm, "Luke wrote:
'Does this not work?

With shtMap.Shapes(strShapeNum).ShapeRange
.Fill.ForeColor.RGB = dColor
.Fill.Visible = msoTrue
.Fill.Solid
End With

--
Best Regards,

Luke wrote in message

...



Below is a subroutine that passes in 2 variables. I am concerned with
the word "Selection" used for ShapeRange and would prefer to eliminate
it. All my attemptes have resulted in errors. I want qualify the
code to prevent error.


Is selection a keyword for ShapeRange or is there a way to qualify it?


Thanks in advance.


Public Sub CountryColor(ByVal strCountry As String, _
ByVal dColor As Double)


Dim shtMap As Worksheet, rgData As Range, strShapeNum As String
Dim i As Integer, dColor1 As Double


Set shtMap =
ThisWorkbook.Application.ThisWorkbook.Worksheets(" Map")
Set rgData =
ThisWorkbook.Application.ThisWorkbook.Worksheets(" Data").Range("RegionCount*ryData")


On Error Resume Next
shtMap.Select


For i = 1 To rgData.Rows.Count


strShapeNum = rgData.Cells(i, 3).Value


shtMap.Shapes(strShapeNum).Select


'This is my issue"
With Selection.ShapeRange
.Fill.ForeColor.RGB = dColor
.Fill.Visible = msoTrue
.Fill.Solid
End With


Next i


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