Classic ASP
REST File Upload (multipart/form-data)

Demonstrates how to upload a file using multipart/form-data.
Chilkat Classic ASP Downloads

Classic ASP
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8">
</head>
<body>
<%
success = 0

' This example requires the Chilkat API to have been previously unlocked.
' See Global Unlock Sample for sample code.

set rest = Server.CreateObject("Chilkat.Rest")

' Connect to the HTTP server using TLS.
bTls = 1
port = 443
bAutoReconnect = 1
' Make sure to replace "www.chilkatsoft.com" with your domain..
success = rest.Connect("www.chilkatsoft.com",port,bTls,bAutoReconnect)
If (success <> 1) Then
    Response.Write "<pre>" & Server.HTMLEncode( rest.LastErrorText) & "</pre>"
    Response.End
End If

set fileStream = Server.CreateObject("Chilkat.Stream")
fileStream.SourceFile = "qa_data/jpg/starfish.jpg"

success = rest.AddHeader("Content-Type","multipart/form-data")

rest.PartSelector = "1"
success = rest.AddHeader("Content-Type","image/jpg")
success = rest.AddHeader("Content-Disposition","form-data; name=""filename""; filename=""starfish.jpg""")
success = rest.SetMultipartBodyStream(fileStream)
rest.PartSelector = "0"

responseBody = rest.FullRequestMultipart("POST","/the_uri_path_to_receive_the_upload")
If (rest.LastMethodSuccess <> 1) Then
    Response.Write "<pre>" & Server.HTMLEncode( rest.LastErrorText) & "</pre>"
    Response.End
End If

If (rest.ResponseStatusCode <> 200) Then
    Response.Write "<pre>" & Server.HTMLEncode( "Received error response code: " & rest.ResponseStatusCode) & "</pre>"
    Response.End
End If

Response.Write "<pre>" & Server.HTMLEncode( "File uploaded") & "</pre>"

%>
</body>
</html>