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Chip Pearson Chip Pearson is offline
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Default Assigning name to button

Grant,

If you have only one button on the sheet, use code like

ActiveSheet.Buttons(1).Name = "TheButton"

You can then delete that shape with code like

ActiveSheet.Buttons("TheButton").Delete

Alternatively, in your recorded code, delete the line

ActiveSheet.Shapes("Button 1").Select

and replace it with

Selection.Name = "TheButton"




--
Cordially,
Chip Pearson
Microsoft MVP - Excel
Pearson Software Consulting, LLC
www.cpearson.com



"Grant Reid" wrote in message
...
Hi

I have recorded a macro that creates a button, assigns a macro

and positions
the button. In my spreadsheet application it is often necessary

to destroy
objects and if I run the macro again to recreate the button, I

get a
Run_time error saying that the item with the specified name

wasn't found. I
understand why it is happening, but don't know what the

workaround can be,
bearing in mind that I will need to frequently destroy the

button and
recreate it. Can one force the button to be created with a

specific name?
Can anyone help? My code is below;

Sub BuildButton()
ActiveSheet.Buttons.Add(4.5, 3, 72, 72).Select
Selection.OnAction = "Connect2Servers"
ActiveSheet.Shapes("Button 1").Select
Selection.Characters.Text = "Connect to Servers"
With Selection.Characters(Start:=1, Length:=18).Font
.Name = "Arial"
.FontStyle = "Regular"
.Size = 10
.Strikethrough = False
.Superscript = False
.Subscript = False
.OutlineFont = False
.Shadow = False
.Underline = xlUnderlineStyleNone
.ColorIndex = xlAutomatic
End With
Selection.ShapeRange.ScaleWidth 2.19, msoFalse,

msoScaleFromTopLeft
Selection.ShapeRange.IncrementTop -0.75
Selection.ShapeRange.IncrementTop -0.75
Selection.ShapeRange.IncrementTop -0.75
Selection.ShapeRange.IncrementTop -0.75
Selection.ShapeRange.IncrementLeft -0.75
Selection.ShapeRange.IncrementLeft -0.75
Selection.ShapeRange.IncrementLeft -0.75
Selection.ShapeRange.IncrementLeft -0.75
Selection.ShapeRange.IncrementLeft -0.75
Selection.ShapeRange.IncrementLeft -0.75
Selection.ShapeRange.IncrementLeft -0.75
Selection.ShapeRange.IncrementTop -0.75
End Sub

Many Thanks - Grant