VBA：为什么自动过滤器在手动过滤器时不起作用？ （运行时错误'1004'：对象'Range'的方法'AutoFilter'失败）

Question

Getting a Run time error for the following code.

Working with a data dump that has data in rows 1000, etc for some columns. From what I read on other forums, autofilter won't work if there is no values "near" the cell. However, when I put in test value closer to the filter, it still won't work. Manual filter is working on this same range.

CURrow = ActiveCell.Row
LASTrow = ActiveCell.SpecialCells(xlLastCell).Row
LASTcol = ActiveCell.SpecialCells(xlLastCell).Column
If LASTrow > CURrow Then
    Columns("A").Insert Shift:=xlToRight
    Range(Cells(CURrow + 1, 1), Cells(LASTrow, 1)).Value = "|"
    Columns(1).HorizontalAlignment = xlCenter
    Range(Cells(CURrow, 1), Cells(CURrow, LASTcol - 1)).AutoFilter
    Rows(CURrow + 1).Select
    ActiveWindow.FreezePanes = True
End If

Answer 1

It's not quite clear what you are trying to do and how it does not work.

Your code

• adds a column to the left
• types a pipe to all the rows below the selected cell (row)
• puts on the filter function starting from the selected row and including not the two rightmost columns (the added and -1 makes -2) of the original data
• uses Freeze Pane to keep the selected row visible while scrolling

IF your idea were to filter out the rows you marked with pipes, you'd need a title for the column AND IF the column titles would be on row 1 my modified version would be like this:

Sub MarkForFilterFromHereBelow()

CURrow = ActiveCell.Row
LASTrow = ActiveCell.SpecialCells(xlLastCell).Row
LASTcol = ActiveCell.SpecialCells(xlLastCell).Column

If LASTrow > CURrow Then
Columns("A").Insert Shift:=xlToRight
Range(Cells(CURrow + 1, 1), Cells(LASTrow, 1)).Value = "|"
Columns(1).HorizontalAlignment = xlCenter
Cells(1, 1).Value = "FilterColumn" 'This is the column title
Range(Cells(1, 1), Cells(LASTrow, LASTcol + 1)).AutoFilter
'Here you can add a row to filter the rows marked with |?
Rows(2).Select
ActiveWindow.FreezePanes = True
End If

End Sub

But clearly I have too many ifs to answer to your question. :-)