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Create duplicate rows based on criteria from one column

Posted on 2014-12-11
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Last Modified: 2014-12-11
I have a worksheet with seven columns.  The seventh column has values separated by commas that belong to the same row (columns A - F).  I would to take each value from the seventh column and create a new row adding the values from columns A - F for each row.  I have attached an example with one of the rows.
Excel-Example.xlsx
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Question by:amergts
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by:leonstryker
ID: 40494651
What you need to do would be to read the cells A through G into separate variables. Use the Split function, or even better the SplitEx (http://www.cpearson.com/excel/Split.aspx) function to break up your string into an array. Then write the values back to the sheet (I suggest a different sheet than the one you started on).

Leon
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Expert Comment

by:Rob Henson
ID: 40494654
How big is your true data set? Would it be too cumbersome to insert the additional rows manually?

If not you can then use Text to columns to split your column G values, copy paste transpose to fill the new rows and then the blank cells ( A - F ) can be filled using the filter function to show blank rows and populate with a simple formula linking the cells to the row above.

Are the values in column G consistent in quantity and in length?

Thanks
Rob H
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Accepted Solution

by:
Martin Liss earned 500 total points
ID: 40494912
Here's a working macro.
Sub SplitLines()
Dim lngRow As Long
Dim strParts() As String
Dim lngPart As Long
Dim lngLastRow As Long

With ActiveSheet
    lngLastRow = .Range("A1048576").End(xlUp).Row
    For lngRow = lngLastRow To 2 Step -1
        strParts = Split(.Cells(lngRow, 7), ",")
        .Cells(lngRow, 7) = strParts(UBound(strParts))
        For lngPart = UBound(strParts) - 1 To 0 Step -1
            .Rows(lngRow & ":" & lngRow).Insert Shift:=xlDown, CopyOrigin:=xlFormatFromLeftOrAbove
            .Range("A" & lngRow & ":F" & lngRow).Value = .Range("A" & lngRow + 1 & ":F" & lngRow + 1).Value
            .Cells(lngRow, 7) = strParts(lngPart)
        Next
    Next
End With

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Author Closing Comment

by:amergts
ID: 40495037
Thanks Martin!  Incredible!
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Expert Comment

by:Martin Liss
ID: 40495050
You're welcome and I'm glad I was able to help.

In my profile you'll find links to some articles I've written that may interest you.
Marty - MVP 2009 to 2014
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