Это сводило меня с ума весь день (поскольку на самом деле вы не можете контролировать типы столбцов Excel до открытия файла CSV), и это сработало для меня, используя VB.NET и Excel Interop:
'Convert .csv file to .txt file.
FileName = ConvertToText(FileName)
Dim ColumnTypes(,) As Integer = New Integer(,) {{1, xlTextFormat}, _
{2, xlTextFormat}, _
{3, xlGeneralFormat}, _
{4, xlGeneralFormat}, _
{5, xlGeneralFormat}, _
{6, xlGeneralFormat}}
'We are using OpenText() in order to specify the column types.
mxlApp.Workbooks.OpenText(FileName, , , Excel.XlTextParsingType.xlDelimited, , , True, , True, , , , ColumnTypes)
mxlWorkBook = mxlApp.ActiveWorkbook
mxlWorkSheet = CType(mxlApp.ActiveSheet, Excel.Worksheet)
Private Function ConvertToText(ByVal FileName As String) As String
'Convert the .csv file to a .txt file.
'If the file is a text file, we can specify the column types.
'Otherwise, the Codes are first converted to numbers, which loses trailing zeros.
Try
Dim MyReader As New StreamReader(FileName)
Dim NewFileName As String = FileName.Replace(".CSV", ".TXT")
Dim MyWriter As New StreamWriter(NewFileName, False)
Dim strLine As String
Do While Not MyReader.EndOfStream
strLine = MyReader.ReadLine
MyWriter.WriteLine(strLine)
Loop
MyReader.Close()
MyReader.Dispose()
MyWriter.Close()
MyWriter.Dispose()
Return NewFileName
Catch ex As Exception
MsgBox(ex.Message)
Return ""
End Try
End Function