VBA code to find first blank cell in a column and activate it

  • Thread starter Thread starter Wes_A
  • Start date Start date
W

Wes_A

MS XP Pro, MS Office 2007.
Can anyone assist with macro to find the first blank cell in a column and
then to make it the active (selected) cell so that data can be pasted to it
and the row in which it is placed.
I cant find a stable and reliable solution with the Find Blank Cell
procedure in the menu.
 
When you say first blank cell, do you mean -
From the top down, the first blank cell with possibly other non blank cells
below that blank cell.
or -
From the bottom up, the cell below the first non-blank cell.

Regards,
Peter T
 
Dim FirstBlankCell as Range
Set FirstBlankCell=Range("A" & rows.Count).end(xlup).offset(1,0)
FirstBlankCell.Activate

Note that you do not need to select a cell to paste data into it, and it is
slowing down your macro.

Look at this:

Worksheets("Sheet2").Range("A2:D2").copy Destination:=FirstBlankCell

Regards,
Per
 
How about:

Sub FindFirstBlank()
Dim r1 As Range, r2 As Range
Set r1 = Intersect(Range("B:B"), Cells.SpecialCells(xlCellTypeBlanks))
Set r2 = Cells(Rows.Count, "B").End(xlUp).Offset(1, 0)
If r1 Is Nothing Then
r2.Select
Else
r1.Select
End If
End Sub
 
This is a correction:

Sub FindFirstBlank()
Dim r1 As Range, r2 As Range
Set r1 = Intersect(Range("B:B"), Cells.SpecialCells(xlCellTypeBlanks))
Set r2 = Cells(Rows.Count, "B").End(xlUp).Offset(1, 0)
If r1 Is Nothing Then
r2.Select
Else
r1(1).Select
End If
End Sub
 
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