Yes, I remember that one!
Regards.
"Tom Ogilvy" wrote in message
...
Just to add to J.E.'s advice,
Testing won't help if you never get past the set statement. When cells
matching the criteria are not found, that raises a trappable error so it
would be
On Error Resume Next
set rng = .Columns(6).SpecialCells(xlConstants, xlTextValues)
On Error goto 0
if not rng is nothing then
'cells were found
else
' cells were not found
End if
Don't get carried away with the On Error Resume next - nest it around the
command as shown - that is where it is needed. You have been down that
road
before.
--
Regards,
Tom Ogilvy
Stuart wrote in message
...
Yes, qualifying worked.
Thanks also for the test advice.
Regards.
"J.E. McGimpsey" wrote in message
...
For one thing, you're not qualifying Columns(6) with the worksheet.
That shouldn't cause the error if your activesheet column 6 has text
- does it?
Try:
Set rng = .Columns(6).SpecialCells(xlConstants, XLTextValues)
Also, I've found it's nearly always worth testing the results of
SpecialCells"
Set rng = .Columns(6).SpecialCells(xlConstants, XLTextValues)
If rng is Nothing Then
<handle no text found
Else
<your existing code
End If
In article ,
"Stuart" wrote:
I'm trying to build an array where the start of each
element will be defined by a particular value in
Col 6.
In the following code I receive the above error message
on the line marked ***
For Each ws In Workbooks(SourceWorkbook).Worksheets()
With ws
Shtname = ws.Name
Dim i As Long
Dim varr()
Dim rng As Range, Cell As Range, rng1 As Range
ReDim varr(1 To 1)
If Not (UCase(.Name) = "MASTER" Or UCase(.Name) _
= "COVER" Or UCase(.Name) = "CONTENTS" Or _
UCase(.Name) = "SUMMARY") Then
*** Set rng = Columns(6).SpecialCells(xlConstants, xlTextValues)
For Each Cell In rng
If Cell.Value = "£" Then
If IsEmpty(varr(1)) Then
Set varr(1) = Cell
Else
ReDim Preserve varr(1 To UBound(varr) + 1)
Set varr(UBound(varr)) = Cell
End If
End If
Next
etc
As far as I can see with the test workbook, there are instances of
'£' in Col 6 in every worksheet. Why does the code appear to fail to
find '£', please?
Regards.
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