ID: Q184378
The information in this article applies to:
The way that Microsoft Excel 98 Macintosh Edition deletes cells in a For Each...Next loop in a Microsoft Visual Basic for Applications Sub procedure differs from the way that earlier versions of Microsoft Excel delete cells. This article describes the differences and includes a Visual Basic Sub procedure example that deletes cells in a loop.
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To use the macro in this article, type the following sample data in a worksheet:
A1: a B1: 1
A2: b B2: 2
A3: x B3: 3
A4: x B4: 4
A5: c B5: 5
A6: x B6: 6
A7: d B7: 7
A8: x B8: 8
A9: x B9: 9
A10: e B10: 10
In a new macro module, type the following macro:
Sub DeleteCells()
'Loop through cells A1:A10 and delete cells that contain an "x."
For Each c in Range("A1:A10")
If c = "x" Then c.EntireRow.Delete
Next
End Sub
When you run the macro DeleteCells in Microsoft Excel 98, only rows 3, 6, and 8 are deleted. Although rows 4 and 9 contain an "x" in column A, the macro does not delete the rows. The results of running the macro in Microsoft Excel 98 are the following:
A1: a B1: 1
A2: b B2: 2
A3: x B3: 4
A4: c B4: 5
A5: d B5: 7
A6: x B6: 9
A7: e B7: 10
When Microsoft Excel deletes row 3, all cells move up one row. For example,
cell A3 assumes the contents of cell A4, cell A4 assumes the contents of
cell A5, and so forth. After the For Each...Next loop evaluates a cell, it
evaluates the next cell; thus, when cells are shifted, they may be skipped
by the loop.
When you run the macro DeleteCells in Microsoft Excel 5.0, the macro deletes all rows that contain an "x." The results of the macro in Microsoft Excel 5.0 are the following:
A1: a B1: 1
A2: b B2: 2
A3: c B3: 5
A4: d B4: 7
A5: e B5: 10
When row 3 is deleted, all cells move up one row; and cell A3 assumes the
contents of cell A4, cell A4 assumes the contents of cell A5, and so forth.
However, unlike the behavior of the loop in Microsoft Excel 98, when the
For Each...Next loop evaluates a cell in Microsoft Excel 5.0, it
reevaluates the cell if it is deleted in the loop; thus, the cells are not
skipped.
Use the following macro when you want to use a loop to delete cells:
Sub DeleteCells2()
Dim rng As Range
Dim i As Integer, counter As Integer
'Set the range to evaluate to rng.
Set rng = Range("A1:A10")
'initialize i to 1
i = 1
'Loop for a count of 1 to the number of rows
'in the range that you want to evaluate.
For counter = 1 To rng.Rows.Count
'If cell i in the range contains an "x",
'delete the row.
'Else increment i
If rng.Cells(i) = "x" Then
rng.Cells(i).EntireRow.Delete
Else
i = i + 1
End If
Next
End Sub
The results of this macro in Microsoft Excel 98 and earlier versions are as
follows:
A1: a B1: 1
A2: b B2: 2
A3: c B3: 5
A4: d B4: 7
A5: e B5: 10
This is an alternate method to the method shown above. It produces the same results.
Sub DeleteCells3()
Dim rng As Range, i As Integer
'Set the range to evaluate to rng.
Set rng = Range("A1:A10")
'Loop backwards through the rows
'in the range that you want to evaluate.
For i = rng.Rows.Count To 1 Step -1
'If cell i in the range contains an "x", delete the entire row.
If rng.Cells(i).Value = "x" Then rng.Cells(i).EntireRow.Delete
Next
End Sub
For more information about looping in a macro, from the Visual Basic Editor, click the Office Assistant, type "looping," click Search, and then click to view "Looping Through Code."
NOTE: If the Assistant is hidden, click the Office Assistant button on the Standard toolbar. If the Assistant is not able to answer your query, please see the following article in the Microsoft Knowledge Base:
ARTICLE-ID: Q176476
TITLE : OFF: Office Assistant Not Answering Visual Basic Questions
Additional query words: 8.0 8.00 XL98
Keywords : kbprg kbdta kbconversion xlvbainfo OffVBA xlvbmigrate
Version : MACINTOSH:98
Platform : MACINTOSH
Issue type : kbprb
Last Reviewed: June 30, 1999