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"Upload Files with HTML Forms and Pure ASP"
by Steven W. Disbrow
This artical lets How to Upload files thur Pure ASP
Use multipart/form-data encoding scheme  in HTML Form to Upload File
Below Code is given Paster from MSDN for ur Reference ....
for details search the above artical...

<form action="fileuploader.asp"
         enctype="multipart/form-data" method="post"
         name="form1" id="form1">


<%
'Adjust this depending on the size of the files you'll
`be expecting; longer timeout for larger files!
Server.ScriptTimeout = 5400

Const ForWriting = 2
Const TristateTrue = -1
CrLf = Chr(13) & Chr(10)

'This function retreives a field's name
Function GetFieldName(infoStr)
      sPos = InStr(infoStr, "name=")
      EndPos = InStr(sPos + 6, infoStr, Chr(34) & ";")
      If EndPos = 0 Then
            EndPos = inStr(sPos + 6, infoStr, Chr(34))
      End If
      GetFieldName = Mid(infoStr, sPos + 6, endPos - _
            (sPos + 6))
End Function

'This function retreives a file field's filename
Function GetFileName(infoStr)
      sPos = InStr(infoStr, "filename=")
      EndPos = InStr(infoStr, Chr(34) & CrLf)
      GetFileName = Mid(infoStr, sPos + 10, EndPos - _
            (sPos + 10))
End Function

'This function retreives a file field's MIME type
Function GetFileType(infoStr)
      sPos = InStr(infoStr, "Content-Type: ")
      GetFileType = Mid(infoStr, sPos + 14)
End Function

'Yank the file (and anything else) that was posted
PostData = ""
Dim biData
biData = Request.BinaryRead(Request.TotalBytes)
'Careful! It's binary! So, let's change it into
`something a bit more manageable.
For nIndex = 1 to LenB(biData)
      PostData = PostData & Chr(AscB(MidB(biData,nIndex,1)))
Next

'Having used BinaryRead, the Request.Form collection is
`no longer available to us. So, we have to parse the
`request variables ourselves!
'First, let's find that encoding type!
ContentType = Request.ServerVariables( _
      "HTTP_CONTENT_TYPE")
ctArray = Split(ContentType, ";")
'File posts only work well when the encoding is
'"multipart/form-data", so let's check for that!
If Trim(ctArray(0)) = "multipart/form-data" Then
      ErrMsg = ""
      ' grab the form boundary...
      bArray = Split(Trim(ctArray(1)), "=")
      Boundary = Trim(bArray(1))
      'Now use that to split up all the variables!
      FormData = Split(PostData, Boundary)
      'Extract the information for each variable and its data
      Dim myRequest, myRequestFiles(9, 3)
      Set myRequest = CreateObject("Scripting.Dictionary")
      FileCount = 0
      For x = 0 to UBound(FormData)
            'Two CrLfs mark the end of the information about
            `this field; everything after that is the value
            InfoEnd = InStr(FormData(x), CrLf & CrLf)
            If InfoEnd > 0 Then
                  'Get info for this field, minus stuff at the end
                  varInfo = Mid(FormData(x), 3, InfoEnd - 3)
                  'Get value for this field, being sure to skip
                  `CrLf pairs at the start and the CrLf at the end
                  varValue = Mid(FormData(x), InfoEnd + 4, _
                        Len(FormData(x)) - InfoEnd - 7)
                  'Is this a file?
                  If (InStr(varInfo, "filename=") > 0) Then
                        'Place it into our files array
                        '(While this supports more than one file
                        `uploaded at a time we only consider the
                        `single file case in this example)
                        myRequestFiles(FileCount, 0) = GetFieldName( _
                              varInfo)
                        myRequestFiles(FileCount, 1) = varValue
                        myRequestFiles(FileCount, 2) = GetFileName( _
                              varInfo)
                        myRequestFiles(FileCount, 3) = GetFileType( _
                              varInfo)
                        FileCount = FileCount + 1
                  Else
                        'It's a regular field
                        myRequest.add GetFieldName(varInfo), varValue
                  End If
            End If
      Next
Else
      ErrMsg = "Wrong encoding type!"
End If

'Save the actual posted file
'If supporting more than one file, turn this into a loop!

Set lf = server.createObject("Scripting.FileSystemObject")
If myRequest("filename") = "original" Then
      'Use the filename that came with the file
      'At this point, you need to determine what sort of
      `client sent the file. Macintoshes only send the file
      `name, with no path information, while Windows clients
      `send the entire path of the file that was selected
      BrowserType = UCase(Request.ServerVariables( _
            "HTTP_USER_AGENT"))
      If (InStr(BrowserType, "WIN") > 0) Then
            'It's Windows; yank the filename off the end!
            sPos = InStrRev(myRequestFiles(0, 2), "\")
            fName = Mid(myRequestFiles(0, 2), sPos + 1)
      End If
      If (InStr(BrowserType, "MAC") > 0) Then
            'It's a Mac. Simple.
            '(Mac filenames can contain characters that are
            'illegal under Windows, so look out for that!)
            fName = myRequestFiles(0, 2)
      End If
      'If your upload path is different, set that here
      FilePath = "./" & fName
Else
      'Use the user-specified filename instead
      'If your upload path is different, set that here
      FilePath = "./" & myRequest("userSpecifiedName")
End If
SavePath = Server.MapPath(FilePath)
Set SaveFile = lf.CreateTextFile(SavePath, True)
SaveFile.Write(myRequestFiles(0, 1))
SaveFile.Close

'IIS may hang if you don't explicitly return SOMETHING.
'So, redirect to another page or provide some kind of
`feedback below...
%>
<html>
<body>
<% If ErrMsg = "" Then %>
      Thanks for the file! It was yummy!
<% Else %>
      <%= ErrMsg %>
<% End If %>
</body>
</html>