You don’t provide the formulas you used in the conditional formatting conditions. I’ve also heard of problems with conditional formatting in 2007. I don’t have enough 2007 CF under my belt to know.

Trying to determine if I’ve uncovered a bug, a bad install of Excell 2007, or operator error.

I’m using the conditional formatting style for a series of investment returns. However, it seems that the internal calculation for one, two, or three standard is incorrect. In fact, regardless of the number of standard deviations that I select, the same number of cells get formatted. Even stranger, it only seems to occur when looking at observations below the mean.

In my case, the mean is 2.55% with a standard deviation of .73% But every cell that is .77% below the mean gets highlighted regardless of the number of standard deviations is selected.

I could write some VBA code to accomplish this, but would rather have the system work as advertised.

Sub FormatPT1()
Dim c As Range

With ActiveSheet.PivotTables(“PivotTable1″)

” reset default formatting
With .TableRange1
.Font.Bold = False
.Interior.ColorIndex = 0
End With

With ActiveSheet.PivotTables(“PivotTable1″)
With Intersect(.PivotFields(“BRAND DESCRIPTION”).PivotItems(“Name1″).DataRange.EntireRow, _
.TableRange1)
.Font.Bold = True
.Interior.ColorIndex = 6

End With
End With
End With
End Sub

Sub FormatPT2()
Dim c As Range

With ActiveSheet.PivotTables(“PivotTable1″)
With Intersect(.PivotFields(“BRAND DESCRIPTION”).PivotItems(“Name2″).DataRange.EntireRow, _
.TableRange1)
.Font.Bold = True
.Interior.ColorIndex = 6

End With
End With
End Sub
Private Sub Worksheet_PivotTableUpdate(ByVal Target As PivotTable)
Select Case Target.Name
Case “PivotTable1″
FormatPT1
FormatPT2
End Select
End Sub

I made a mistake in my post earlier. I am trying to get the same format for either Name 1 or Name 2 (Sorry! I entered Name 2 twice in my previous post). Is there a way I can include this without running into that same error?

Thanks again!!

ActiveSheet.PivotTables(“PivotTable1″)
ActiveSheet.PivotTables(1)

The “With ActiveSheet.PivotTables(1)” block appears twice, the second one inside the first.

I think this will do it:

Sub FormatPT1()

  With ActiveSheet.PivotTables("PivotTable1")
    '' reset default formatting
    With .TableRange1
      .Font.Bold = False
      .Interior.ColorIndex = 0
    End With

    With Intersect(.PivotFields("BRAND DESCRIPTION").PivotItems("NAME 2").DataRange.EntireRow, _
        .TableRange1)
      .Font.Bold = True
      .Interior.ColorIndex = 6
    End With
  End With
End Sub

Running into another issue and would appreciate any help you can provide! My code was working fine for highlighting an entire row when I had it based on one condition only. When I try to just add in another section for a second condition (wanting the same result, the entire row highlighted) I receive an error (specifically with my End Sub).

Sub FormatPT1()
Dim c As Range

With ActiveSheet.PivotTables(“PivotTable1″)

” reset default formatting
With .TableRange1
.Font.Bold = False
.Interior.ColorIndex = 0
End With

With ActiveSheet.PivotTables(1)
With Intersect(.PivotFields(“BRAND DESCRIPTION”).PivotItems(“NAME 2″).DataRange.EntireRow, _
.TableRange1)
.Font.Bold = True
.Interior.ColorIndex = 6

With ActiveSheet.PivotTables(1)
With Intersect(.PivotFields(“BRAND DESCRIPTION”).PivotItems(“NAME 2″).DataRange.EntireRow, _
.TableRange1)
.Font.Bold = True
.Interior.ColorIndex = 6

End With
End With
End With
End Sub

I had finally the opportunity to evolve on top of your nice code example, adapting the data range selection technique to my data, and I must say it was a really good push towards my goal! What a splendid work you’re doing! Please keep on doing it, sharing with everybody these high quality, not easy to find, bits of information! Thank you!

As I told you previously, my real world pivot table gets its data from an Access SQL query, forcing me to delve into some additional issues. But overall this problem is solved, thanks to your invaluable help!

There’s one last problem still unsolved, however. Even though it’s not a “must to have”, any insight would be greatly appreciated. Compact layout tables (Excel 2007) have several advantages that are worth keeping, but with them this code unfortunately doesn’t work. Oddly enough, if we set “Cat2″’s field settings to “Show item labels in outline form”, the ranges returned by the row selection (pi1) loop revert to empty single row, two column ones. These have empty cells if we don’t use category sub-totals at top row level. The real data ranges aren’t returned at all…

Any ideas?

Once again, many thanks!

NL

I have no words for what you just did! You shed some light into my problem, and couldn´t be more responsive and complete than you were. I’m more than grateful.

Later on I hope to try this code. I’ll need eventually to adapt it, as my table comes from an external Access query. Mainly use “VisibleItems” instead. I’ll post my feedback.

Huge thanks!

NL

which produces this pivot table

The code does require using Intersect(), but it’s not really inefficient.

Sub FormatByBlock()   Dim pi1 As PivotItem  Dim pi2 As PivotItem  Dim r As Range  Dim v As Variant  Dim i1 As Long  Dim i2 As Long  Dim vMax As Double  Dim c As Range   With ActiveSheet.PivotTables("PTformatted")     ' reset default formatting    With .DataBodyRange      .Font.Bold = False      .Interior.ColorIndex = 0    End With     ' loop to get ranges    For Each pi1 In .PivotFields("Cat2").PivotItems      For Each pi2 In .PivotFields("Cat3").PivotItems        Set r = Intersect(pi1.DataRange, pi2.DataRange)         ' find max in range        v = r.Value        vMax = v(LBound(v, 1), LBound(v, 2))        For i1 = LBound(v, 1) To UBound(v, 1)          For i2 = LBound(v, 2) To UBound(v, 2)            If v(i1, i2) > vMax Then              vMax = v(i1, i2)            End If          Next        Next         ' highlight max in range        For i1 = LBound(v, 1) To UBound(v, 1)          For i2 = LBound(v, 2) To UBound(v, 2)            If v(i1, i2) = vMax Then              With r.Cells(i1 + 1 - LBound(v, 1), i2 + 1 - LBound(v, 2))                .Font.Bold = True                .Interior.ColorIndex = 6              End With            End If          Next        Next       Next    Next   End WithEnd Sub

Here is the formatted table