To : "since1990(level)" and "lihonggen0(李洪根,用.NET,标准答案来了) "
难道用SMTP 发送附件时一定要编码吗(UUEncode或Base64Encode)?
小弟发送文本格式的附件时就没用编码,也能打开附件。但是如果用两位说的函数进行编码,打开附件(在网站上)时就是乱码了?????
For i = 0 To lstAttachments.ListCount - 1
lstAttachments.ListIndex = i
m_strEncodedFiles = m_strEncodedFiles & _
UUEncodeFile(lstAttachments.Text) & vbCrLf
Next i
上面的代码将附件的路径作为参数传递给UUEncodeFile函数。该函数的作用是按照我们前面所讲的算法对字符进行编码。编码后的数据被保存在一个模块级变量m_strEncodedFile中。然后该变量的内容被添加到邮件正文中:
Public Function UUEncodeFile(strFilePath As String) As String
Dim intFile As Integer 'file handler
Dim intTempFile As Integer 'temp file
Dim lFileSize As Long 'size of the file
Dim strFileName As String 'name of the file
Dim strFileData As String 'file data chunk
Dim lEncodedLines As Long 'number of encoded lines
Dim strTempLine As String 'temporary string
Dim i As Long 'loop counter
Dim j As Integer 'loop counter
Dim strResult As String
'
'Get file name
strFileName = Mid$(strFilePath, InStrRev(strFilePath, "\") + 1)
'
'Insert first marker: "begin 664 ..."
strResult = "begin 664 " + strFileName + vbLf
'
'Get file size
lFileSize = FileLen(strFilePath)
lEncodedLines = lFileSize \ 45 + 1
'
'Prepare buffer to retrieve data from
'the file by 45 symbols chunks
strFileData = Space(45)
'
intFile = FreeFile
'
Open strFilePath For Binary As intFile
For i = 1 To lEncodedLines
'Read file data by 45-bytes cnunks
'
If i = lEncodedLines Then
'Last line of encoded data often is not
'equal to 45, therefore we need to change
'size of the buffer
strFileData = Space(lFileSize Mod 45)
End If
'Retrieve data chunk from file to the buffer
Get intFile, , strFileData
'Add first symbol to encoded string that informs
'about quantity of symbols in encoded string.
'More often "M" symbol is used.
strTempLine = Chr(Len(strFileData) + 32)
'
If i = lEncodedLines And (Len(strFileData) Mod 3) Then
'If the last line is processed and length of
'source data is not a number divisible by 3,
'add one or two blankspace symbols
strFileData = strFileData + Space(3 - (Len(strFileData) Mod 3))
End If
For j = 1 To Len(strFileData) Step 3
'Breake each 3 (8-bits) bytes to 4 (6-bits) bytes
'
'1 byte
strTempLine = strTempLine + Chr(Asc(Mid(strFileData, j, 1)) \ 4 + 32)
'2 byte
strTempLine = strTempLine + _
Chr((Asc(Mid(strFileData, j, 1)) Mod 4) * 16 + Asc(Mid(strFileData, j + 1, 1)) \ 16 + 32)
'3 byte
strTempLine = strTempLine + _
Chr((Asc(Mid(strFileData, j + 1, 1)) Mod 16) * 4 + Asc(Mid(strFileData, j + 2, 1)) \ 64 + 32)
'4 byte
strTempLine = strTempLine + _
Chr(Asc(Mid(strFileData, j + 2, 1)) Mod 64 + 32)
Next j
'add encoded line to result buffer
strResult = strResult + strTempLine + vbLf
'reset line buffer
strTempLine = ""
Next i
Close intFile
'add the end marker
strResult = strResult & "'" & vbLf + "end" + vbLf
'asign return value
UUEncodeFile = strResult