It seems older macros are not working. I have proper securtiy set to run VBA macros but when I have tried a few methods for clearing ALL filters on a worksheet, I get a comp
Here's the one-liner I use. It checks for an auto-filter and if found, removes it.
Unlike some answers, this code won't create an auto-filter if used on a worksheet that is not auto-filtered in the first place.
If Cells.AutoFilter Then Cells.AutoFilter
If the sheet already has a filter on it then:
Sub Macro1()
Cells.AutoFilter
End Sub
will remove it.
I found this workaround to work pretty effectively. It basically removes autofilter from the table and then re-applies it, thus removing any previous filters. From my experience this is not prone to the error handling required with the other methods mentioned here.
Set myTable = YOUR_SHEET.ListObjects("YourTableName")
myTable.ShowAutoFilter = False
myTable.ShowAutoFilter = True
Try something like this:
Sub ClearDataFilters()
'Clears filters on the activesheet. Will not clear filters if the sheet is protected.
On Error GoTo Protection
If ActiveWorkbook.ActiveSheet.FilterMode Or _
ActiveWorkbook.ActiveSheet.AutoFilterMode Then _
ActiveWorkbook.ActiveSheet.ShowAllData
Exit Sub
Protection:
If Err.Number = 1004 And Err.Description = _
"ShowAllData method of Worksheet class failed" Then
MsgBox "Unable to Clear Filters. This could be due to protection on the sheet.", _
vbInformation
End If
End Sub
.FilterMode returns true if the worksheet is in filter mode. (See this for more information.)
See this for more information on .AutoFilter.
And finally, this will provide more information about the .ShowAllData method.
Im using .filtermode if filter is on it returns true
Dim returnValue As Boolean
returnValue = worksheet1.FilterMode
if returnValue Then
worksheet1.ShowAllData
End If
Simply activate the filter headers and run showalldata, works 100%. Something like:
Range("A1:Z1").Activate
ActiveSheet.ShowAllData
Range("R1:Y1").Activate
ActiveSheet.ShowAllData
If you have the field headers in A1:Z1 and R1:Y1 respectively.