vb6 读取*.csv文件的最后一行数据问题

Private Sub Command1_Click()

MsgBox Replace(getchar(GetFileLastLine(App.Path & "\9056_020.csv"), ",", 3, 4), "Operation=", "")

End Sub



Function GetFileLastLine(fName As String) As String

    Dim hFile As Long

    Dim recNumber1 As Long, recNumber2 As Long

    Dim Char As String * 1

    Dim strTmp As String

     

    recNumber2 = FileLen(fName)    

    hFile = FreeFile

    Open fName For Binary As hFile

        Do

            Get #hFile, recNumber2, Char

            recNumber2 = recNumber2 - 1

        Loop Until Char Like "[0-9a-zA-Z]"

        recNumber1 = recNumber2

        Do

            Get #hFile, recNumber1, Char

            recNumber1 = recNumber1 - 1

        Loop Until Char = vbCr Or Char = vbLf

        strTmp = Space(recNumber2 - recNumber1 + 1)

        Get #hFile, recNumber1 + 1, strTmp

    Close

    GetFileLastLine = strTmp

 

End Function



Private Function getchar(mstr As String, charx As String, bofnum As Integer, eofnum As Integer) As String

Dim x As Integer, strartnum As Integer, endnum As Integer

For i = 1 To Len(mstr)

    If x = eofnum Then Exit For

        If Mid(mstr, i, 1) = charx Then

            x = x + 1

        If x = bofnum Then

            strartnum = i

        ElseIf x = eofnum Then

            endnum = i

        End If

    End If

Next i

getchar = Mid(mstr, strartnum + 1, endnum - strartnum - 1)

End Function