You never change the activecell in your code and you never assign a value to
colNum (besides not using Interior as pointed out by others).

Dim RowNum As Integer: Dim ColNum As Integer
Dim InitialsColumn As Integer, cell As Range
Dim sh = Range("INITIALS").Parent
InitialsColumn =Range("INITIALS").Column
For Each cell In Range("All_ASSIGNED_Schedules").
' QUICK VIEW all 4-12 shifts with Lavender
If InStr(1, cell.Value, "4") Then
cell.Interior.ColorIndex = 39
sh.cells(cell.row, InitialsColumn - 1).Interior _
.ColorIndex = 39
End If
Next cell

If the column is adjacent to Table, then the above should work. If there is
a blank column in between, then change -1 to -2.

--
Regards,
Tom Ogilvy



"JimP" wrote in message
om...
To anyone,

I have a single column RANGE called "INITIALS" that is located 1
column to the left of a table of data who's RangeName is
"All_ASSIGNED_Schedules"

What I'm trying to do is find all cells in the data table containing
the value "4" and highlight the cell with color. (This part of the
code works fine) ...
but then, on the SAME row I wish to also highlight the column
containing the individuals INITIALS. When I step thru the code, I Know
I'm at the correct column because it reveals the correct set of
INITIALS, but I can't change the color of that cell?

Please advise what's wrong with the line containing the Comment: '???


Dim RowNum As Integer: Dim ColNum As Integer
Dim InitialsColumn As Integer: InitialsColumn =
Range("INITIALS").Column
Dim cell As Range: Range("All_ASSIGNED_Schedules").Select
For Each cell In Selection
' QUICK VIEW all 4-12 shifts with Lavender
If InStr(1, cell.Value, "4") Then
cell.Interior.ColorIndex = 39
RowNum = cell.Row: ColNum = cell.Column
ActiveCell.Offset(0, InitialsColumn - ColNum).ColorIndex =
39 '???
End If
Next cell


Any help would be appreciated!

Jim Pellechi