Diegodcr

@Edson Luiz Branco Na verdade eu tive que tratar de forma diferente, célula por célula pois o código que você citou acima, não funcionava.
Option Explicit Sub test()     Dim fullName As String     fullName = "c:\users\domenic\desktop\sample.txt"          Dim columnCount As Long     columnCount = GetColumnCountFromTextFile(fullName, vbTab)          If columnCount = 0 Then         MsgBox "No data found!", vbExclamation         Exit Sub     End If          Workbooks.OpenText _         Filename:=fullName, _         DataType:=xlDelimited, _         Tab:=True, _         FieldInfo:=GetFieldInfo(columnCount)           End Sub Private Function GetColumnCountFromTextFile(ByVal PathAndFilename As String, Optional ByVal Delim As String = ",") As Long '   Returns the number of columns contains in a text file '   First non-blank line used to determine number of columns based on delimiter     Dim textLine As String     Dim fileNumber As Long     Dim columnCount As Long          columnCount = 0     fileNumber = FreeFile()     Open PathAndFilename For Input As #fileNumber         Do Until EOF(fileNumber)             Line Input #fileNumber, textLine             If Len(textLine) > 0 Then                 columnCount = UBound(Split(textLine, Delim)) + 1                 Exit Do             End If         Loop     Close #fileNumber          GetColumnCountFromTextFile = columnCount End Function Private Function GetFieldInfo(ByVal columnCount As Long) As Variant '   Returns a two-dimensional array containing a column number and data type pair '   Sets all columns to text format     ReDim fieldInfoArray(1 To columnCount, 1 To 2)          Dim i As Long     For i = LBound(fieldInfoArray, 1) To UBound(fieldInfoArray, 1)         fieldInfoArray(i, 1) = i         fieldInfoArray(i, 2) = xlTextFormat     Next i          GetFieldInfo = fieldInfoArray()      End Function  

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