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Inserting a column to the right of the currently selected and formatted column, including formatting

Posted on 2011-09-16
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Last Modified: 2012-06-21
I've got a series of spreadsheets (see example attached) into which I need to insert YTD columns.  I've written a subroutine (Insert_YTD_Columns) which starts at column 5, then shifts one column to the right and inserts a new column.  This results in the new column having the same formatting as the currently selected column.  When I get to the last column, the new column doesn't have the formatting of the columns to the left.

Is there a way, rather than inserting a colum to the left of the currently selected column to insert the column to the right, which will retain the same formatting?  I know I can select column 5, copy it, shift to column 6, then use paste special to paste the formatting from column 5 into 6, but is there a more elegant method?

BTW, I know I could use the FillDown method to copy the formula from the top row of each of the YTD columns, but am not doing so because of this same formatting issue.

EE-InsertColumn.xlsm
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Question by:Dale Fye
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nutsch earned 2000 total points
ID: 36553172
That's actually a good way to do it. Here is my take on your code with that.

Thomas
Option Explicit

Public Sub Insert_YTD_Columns()
    
    Dim sht As Worksheet
    Dim intCol As Integer, intRow As Integer, intLoop As Integer
    Dim strYTDRange As String
    
    Set sht = ActiveSheet
    
    'Loop and insert columns
    intCol = 5
    intRow = sht.Range("B500").End(xlUp).Row
    
    Application.ScreenUpdating = False
    Application.Calculation = xlCalculationManual
    Application.DisplayAlerts = False
    
    While sht.Cells(5, intCol) <> ""
    
        sht.Columns(intCol + 1).Insert Shift:=xlToRight
        sht.Cells(4, intCol + 1) = sht.Cells(4, intCol)
        sht.Cells(4, intCol + 1).NumberFormat = "m/d/yy;@"
        sht.Cells(5, intCol + 1) = "YTD Balance"
        sht.Cells(5, intCol) = "Activity"
                
        sht.Range(Cells(6, intCol + 1), Cells(intRow, intCol + 1)).FormulaR1C1 = "=SUbtotal(9,RC5:RC[-1])"
        sht.Range(Cells(6, intCol), Cells(intRow, intCol)).SpecialCells(xlCellTypeBlanks).Offset(, 1).ClearContents
                    
        intCol = intCol + 2
    Wend
    
    Cells(5, Columns.Count).End(xlToLeft).Offset(, -1).EntireColumn.Copy
    Cells(5, Columns.Count).End(xlToLeft).EntireColumn.PasteSpecial Paste:=xlPasteFormats
    
    'Unhide all of the columns in the active sheet
    sht.Range(sht.Columns(5), sht.Columns(intCol - 1)).EntireColumn.Hidden = False
    
    Application.ScreenUpdating = True
    Application.Calculation = xlCalculationAutomatic
    Application.DisplayAlerts = True
    
End Sub

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Author Closing Comment

by:Dale Fye
ID: 36554514
Thomas,

Thanks for the quick feedback.  

The two lines(33 and 34) were what I really needed, but I particularly like the subTotal function and am going to have to play with that as I format the subtotal rows on that worksheet (and others), which is currently in another subroutine.
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LVL 49

Author Comment

by:Dale Fye
ID: 36767381
Thomas,

Had to revise my code a bit and am now inserting in the current column.  Then setting the formula for that new column to refer to the column to the right (and left if intCol > 5)

If intCol = 5 Then
      sht.Range(Cells(6, 5), Cells(intRow, 5)).FormulaR1C1 = "=RC[1]"
Else
      sht.Range(Cells(6, intCol), Cells(intRow, intCol)).FormulaR1C1 = "=RC[1] - RC[-1]"
End If

But column intCol + 1 contains some rows that are blank (for spacing).  I tried to modify row 28 from your code so that it would create these same blanks in intCol as are in intCol + 1, but cannot get the syntax right.  Care to address this as a continuation of this question, or should I create a new one.

sht.Range(Cells(6, intCol), Cells(intRow, intCol)).SpecialCells(xlCellTypeBlanks).Offset(, 1).ClearContents



sht.Range(Cells(6, intCol + 1), Cells(intRow, intCol + 1)).SpecialCells(xlCellTypeBlanks).Offset(, -11).ClearContents
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Expert Comment

by:nutsch
ID: 36813011
Easier to create a new one. I might not get to this for a bit, so you'll get a faster answer that way (from other experts). If you have time, you can wait and I'll address it here. I have the request flagged in my mailbox.

T
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