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Default Combo Change works in 2K not 2003

Hi All

The following code works well in XL 2K but not 2003. I have a
combobox with 7 categories. Test 1 –5, “All” and “Show 123”. Does
anyone have any idea how I might produce the same code in XL 2003.

Thanks in advance.
Chad

Private Sub ComboBox1_Change()
Dim lw As Long

Application.ScreenUpdating = False
Range("A7:A100").EntireRow.Hidden = False
lw = Range("A" & Rows.Count).End(xlUp).Row

' If Range(
For i = lw To 2 Step -1
If Range("A" & i).Value = ComboBox1.Value Then
Range("A" & i).EntireRow.Hidden = True
ElseIf ComboBox1.Value = "All" Then
Range("A7:A100").EntireRow.Hidden = False

ElseIf ComboBox1.Value = "Show 123" Then
If Range("A" & i).Value = "Test4" Or _
Range("A" & i).Value = "Test5" Then
Range("A" & i).EntireRow.Hidden = True
End If
End If

Next i
Calculate
Application.ScreenUpdating = True

End Sub
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Default Combo Change works in 2K not 2003

I don't see any reason why it would not work in xl2003. What indications are
you getting? Error message?

"Chad" wrote:

Hi All

The following code works well in XL 2K but not 2003. I have a
combobox with 7 categories. Test 1 €“5, €śAll€ť and €śShow 123€ť. Does
anyone have any idea how I might produce the same code in XL 2003.

Thanks in advance.
Chad

Private Sub ComboBox1_Change()
Dim lw As Long

Application.ScreenUpdating = False
Range("A7:A100").EntireRow.Hidden = False
lw = Range("A" & Rows.Count).End(xlUp).Row

' If Range(
For i = lw To 2 Step -1
If Range("A" & i).Value = ComboBox1.Value Then
Range("A" & i).EntireRow.Hidden = True
ElseIf ComboBox1.Value = "All" Then
Range("A7:A100").EntireRow.Hidden = False

ElseIf ComboBox1.Value = "Show 123" Then
If Range("A" & i).Value = "Test4" Or _
Range("A" & i).Value = "Test5" Then
Range("A" & i).EntireRow.Hidden = True
End If
End If

Next i
Calculate
Application.ScreenUpdating = True

End Sub

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Posts: 49
Default Combo Change works in 2K not 2003

Hi Thanks for the response. This line

Range("A7:A100").EntireRow.Hidden = False

If I click on the yellow error line for a definition I get the
perfectly helpful message "Identifier under curser not recognised".
It has confused me no end as my code works wonderfully in 2k, but I
put it on the users machine and it fails in 03.

Chad



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