Hi - I have a workbook with multiple sheets - each having a few shapes
on it. I want to change the color of the fill and line for each one.
My code works ok if I run it on just one sheet, but if I try and run it
on one sheet right after the other (with a subroutine calling this sub
twice), it gives me the "Object doesn't support the property or method"
error. Sometimes this even happens if I run the macro twice in a row
manually, sometimes it doesn't. I am seriously at my wits end....can
someone please help?

Thanks!

Here is the code I am bombing out
on...."Selection.ShapeRange.Fill.ForeColor.SchemeC olor = 16" (or
whichever case it is on)


For Each sh In myDocument.Shapes
sh.Select
If sh.Type = 2 Then
Select Case colorscheme

Case "OcOl"
Selection.ShapeRange.Fill.ForeColor.SchemeColor = 16
Selection.ShapeRange.Line.ForeColor.SchemeColor = 16
With Selection.Characters.Font
.Name = "Arial"
.FontStyle = "Regular"
.Size = 10
.Strikethrough = False
.Superscript = False
.Subscript = False
.OutlineFont = False
.Shadow = False
.Underline = xlUnderlineStyleNone
.ColorIndex = fontcolor
End With

Case "BoTe"
Selection.ShapeRange.Fill.ForeColor.SchemeColor = 34
Selection.ShapeRange.Line.ForeColor.SchemeColor = 34
With Selection.Characters.Font
.Name = "Arial"
.FontStyle = "Regular"
.Size = 10
.Strikethrough = False
.Superscript = False
.Subscript = False
.OutlineFont = False
.Shadow = False
.Underline = xlUnderlineStyleNone
.ColorIndex = fontcolor
End With
Case "EaTe"
Selection.ShapeRange.Fill.ForeColor.SchemeColor = 34
Selection.ShapeRange.Line.ForeColor.SchemeColor = 34
With Selection.Characters.Font
.Name = "Arial"
.FontStyle = "Regular"
.Size = 10
.Strikethrough = False
.Superscript = False
.Subscript = False
.OutlineFont = False
.Shadow = False
.Underline = xlUnderlineStyleNone
.ColorIndex = fontcolor
End With
Case "BoEa"
Selection.ShapeRange.Fill.ForeColor.SchemeColor = 35
Selection.ShapeRange.Line.ForeColor.SchemeColor = 35
With Selection.Characters.Font
.Name = "Arial"
.FontStyle = "Regular"
.Size = 10
.Strikethrough = False
.Superscript = False
.Subscript = False
.OutlineFont = False
.Shadow = False
.Underline = xlUnderlineStyleNone
.ColorIndex = fontcolor
End With
Case Else
'Olive-Ocean is default
Selection.ShapeRange.Fill.ForeColor.SchemeColor = 1
Selection.ShapeRange.Line.ForeColor.SchemeColor = 1
With Selection.Characters.Font
.Name = "Arial"
.FontStyle = "Regular"
.Size = 10
.Strikethrough = False
.Superscript = False
.Subscript = False
.OutlineFont = False
.Shadow = False
.Underline = xlUnderlineStyleNone
.ColorIndex = fontcolor
End With

End Select


End If
Next

Looks like you are trying to format all callouts on the assumption they
contain text.

See this question from earlier today
Subject Font.Color - syntax error ?

In passing, it also looks like you change these throughout the entire
workbook according to your own defined colour scheme. Instead of changing
all those formats you could customize a palette colour.

Eg format everything with (say) colorindex 31 or 31+7 = schemecolor 38
(bottom left in the dropdown palette)

first from the intermediate window (ctrl-g)
?activeworkbook.Colors(16-7)
16711935
?activeworkbook.Colors(35-7)
16776960
?activeworkbook.Colors(34-7)
65535
?vbwhite
16777215

Dim newcolor as long

Select case colorscheme
case "OcOl": newcolor = 16711935
case "BoTe": newcolor = 16776960
Case "EaTe":
Case Else
'Olive-Ocean is default
Selection.ShapeRange.Fill.ForeColor.SchemeColor = 1

End select

Activeworkbook.colors(31) = newcolor

But I don't understand how 'Olive-Ocean goes to SchemeColor = 1, for me
it's vbWhite

In addition, you could set the default colour for all new shapes to be
colorindex 31 (shemecolor38), then you don't need to worry about running a
macro unless you change your colorsheme.

Just a thought

Regards,
Peter


wrote in message
oups.com...
Hi - I have a workbook with multiple sheets - each having a few shapes
on it. I want to change the color of the fill and line for each one.
My code works ok if I run it on just one sheet, but if I try and run it
on one sheet right after the other (with a subroutine calling this sub
twice), it gives me the "Object doesn't support the property or method"
error. Sometimes this even happens if I run the macro twice in a row
manually, sometimes it doesn't. I am seriously at my wits end....can
someone please help?

Thanks!

Here is the code I am bombing out
on...."Selection.ShapeRange.Fill.ForeColor.SchemeC olor = 16" (or
whichever case it is on)


For Each sh In myDocument.Shapes
sh.Select
If sh.Type = 2 Then
Select Case colorscheme

Case "OcOl"
Selection.ShapeRange.Fill.ForeColor.SchemeColor = 16
Selection.ShapeRange.Line.ForeColor.SchemeColor = 16
With Selection.Characters.Font
.Name = "Arial"
.FontStyle = "Regular"
.Size = 10
.Strikethrough = False
.Superscript = False
.Subscript = False
.OutlineFont = False
.Shadow = False
.Underline = xlUnderlineStyleNone
.ColorIndex = fontcolor
End With

Case "BoTe"
Selection.ShapeRange.Fill.ForeColor.SchemeColor = 34
Selection.ShapeRange.Line.ForeColor.SchemeColor = 34
With Selection.Characters.Font
.Name = "Arial"
.FontStyle = "Regular"
.Size = 10
.Strikethrough = False
.Superscript = False
.Subscript = False
.OutlineFont = False
.Shadow = False
.Underline = xlUnderlineStyleNone
.ColorIndex = fontcolor
End With
Case "EaTe"
Selection.ShapeRange.Fill.ForeColor.SchemeColor = 34
Selection.ShapeRange.Line.ForeColor.SchemeColor = 34
With Selection.Characters.Font
.Name = "Arial"
.FontStyle = "Regular"
.Size = 10
.Strikethrough = False
.Superscript = False
.Subscript = False
.OutlineFont = False
.Shadow = False
.Underline = xlUnderlineStyleNone
.ColorIndex = fontcolor
End With
Case "BoEa"
Selection.ShapeRange.Fill.ForeColor.SchemeColor = 35
Selection.ShapeRange.Line.ForeColor.SchemeColor = 35
With Selection.Characters.Font
.Name = "Arial"
.FontStyle = "Regular"
.Size = 10
.Strikethrough = False
.Superscript = False
.Subscript = False
.OutlineFont = False
.Shadow = False
.Underline = xlUnderlineStyleNone
.ColorIndex = fontcolor
End With
Case Else
'Olive-Ocean is default
Selection.ShapeRange.Fill.ForeColor.SchemeColor = 1
Selection.ShapeRange.Line.ForeColor.SchemeColor = 1
With Selection.Characters.Font
.Name = "Arial"
.FontStyle = "Regular"
.Size = 10
.Strikethrough = False
.Superscript = False
.Subscript = False
.OutlineFont = False
.Shadow = False
.Underline = xlUnderlineStyleNone
.ColorIndex = fontcolor
End With

End Select


End If
Next