Thanks, Dave, for the quick response. I do have a Sheet1 in the
workbook. To help trouble-shoot the problem I rewrote the code as
follows:
Sub cleancolumn3()
Dim ws As Worksheet
Dim rsplen As Long
Dim rspstring As String
Set ws = ThisWorkbook.Worksheets("Sheet1")
For i = 1 To ws.Cells(Rows.Count, 14).End(xlUp).Row
If Left(ws.Cells(i, 14), 1) = "'" Then rsplen = Len(ws.Cells(i, 14)) -
1: rspstring = Right(ws.Cells(i, 14), rsplen): ws.Cells(i, 14) =
rspstring
Next i
End Sub

The last statement in the If line viz. "ws.Cells(i,14) = rspstring" is
highlighted when I debug the Application Defined or Object Defined
error. The problem seems to be assigning the string to the cell
value.

Any other way to do this?

Regards,
Raj


On Mar 26, 1:25*pm, "ozgrid.com" wrote:
Do you have a "Sheet1" tab name in "Thisworkbook"?

--
Regards
Dave Hawleywww.ozgrid.com"Raj" wrote in message

...
Yes. But *the error again occurred again. This time the line above the
End If was highlighted in yellow with the same error code displayed.

Regards,
Raj

On Mar 26, 11:35 am, "ozgrid.com" wrote:

Hi Raj

Do you mean?

Sub Cleancolumn1()
With ThisWorkbook.Worksheets("Sheet1")
For i = 1 To .Cells(Rows.Count, 14).End(xlUp).Row
If Left(.Cells(i, 14), 1) = "'" Then
.Cells(i, 14) = Right(.Cells(i, 14), Len(.Cells(i, 14)) - 1)
End If
Next i
End With
End Sub

--
Regards
Dave Hawleywww.ozgrid.com"Raj" wrote in message

...

Hi,

When I run the following code, I am getting the Application or object
defined error. The second line is a single line in my code. The VBE
highlights the portion after "Then" in the second line.

Sub Cleancolumn1()
For i = 1 To ThisWorkbook.Worksheets("Sheet1").Cells(Rows.Count ,
14).End(xlUp).Row
If Left(Cells(i, 14), 1) = "'" Then
ThisWorkbook.Worksheets("Sheet1").Cells(i, 14) = Right(Cells(i, 14),
Len(Cells(i, 14)) - 1)
Next i
End Sub

What is going wrong?

Thanks in advance for the help.

Regards,
Raj