'oTbl is nothing' - you are comparing objects not values.
The 'is' operator is not the same as the '=' operator'. The 'is' operator compares the references to objects, not the content of the objects.
Deleting all rows deletes the word object and leaves oTbl in an indeterminate state.
Thus you can refer to the VBA object but trying to access its information gives an error because there is no object. If the object has been deleted then oTbl will be nothing. Hence you will be comparing the reference to nothing with the reference to nothing which is always true. One of the little joys of working with objects.
The simplest way to deal with this situation is by abstracting the object range (in fact ranges are by far the best way to work in VBA for Word). The following code works without needing an on error statement assuming there is at least one uniform table in the active document.
Code:
Sub Macro2()
Dim oTbl As Word.Table
Dim oTbl_range As Word.Range
Set oTbl = ActiveDocument.Tables(1)
Set oTbl_range = oTbl.Range
Do Until oTbl_range.Tables.Count = 0
Debug.Print oTbl.Range.Rows.Count
oTbl.Rows(1).Delete
Loop
End Sub
Applying the same approach to you code gives
Code:
Dim oTbl_range As Word.Range
Set oTbl_range = oTbl.Range
' Remove empty rows
With oTble_range.Tables(1)
RowCount = .Rows.Count
columnCount = .Columns.Count
For Row = RowCount To 1 Step -1
Delete = True
For Column = 1 To columnCount
If Not (.Cell(Row, Column).Range.Text = Chr(13) & Chr(7)) Then
Delete = False
Exit For
End If
Next Column
If (Delete) Then
' Remove row
Call .Rows(Row).Delete
End If
Next Row
End With
' NOTE: Need to check to see if table still exists here (If all rows were deleted)
If oTbl_range.Tables.Count > 0 Then ' by abstracting the table range you can subsequently test for the presence of a table in the range
' Do other processing steps to table...