Classic ASP
HTTP POST (Duplicate Simple HTML Form POST)

Demonstrates how to duplicate a simple HTML form POST.
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 req = Server.CreateObject("Chilkat.HttpRequest")
set http = Server.CreateObject("Chilkat.Http")

' This example simulates this FORM:
' <form action="echoPost.asp" method="post">
' First name: <input type="text" name="firstName"><br />
' Last name: <input type="text" name="lastName"><br />
' <input type="submit" value="Submit">
' </form>
' The online FORM is found at this URL:
' https://www.chilkatsoft.com/simpleForm.html

' Build an HTTP POST Request:
req.HttpVerb = "POST"

' The FORM target is http://www.chilkatsoft.com/processPost.asp
' An easy way of filling out most of the HTTP request object
' is to call SetFromUrl:
req.SetFromUrl "https://www.chilkatsoft.com/echoPost.asp"

' Send form params using application/x-www-form-urlencoded
req.ContentType = "application/x-www-form-urlencoded"

' The only remaining task is to add the params to the 
' HTTP request object:
req.AddParam "firstName","Matt"
req.AddParam "lastName","Jones"

' Send the HTTP POST and get the response.
' The POST is being sent to chilkatsoft.com, on port 443 (using TLS)
domain = "chilkatsoft.com"
port = 443
tls = 1

' The HTTP POST is sent here:
set resp = Server.CreateObject("Chilkat.HttpResponse")
success = http.HttpSReq(domain,port,tls,req,resp)
If (success = 0) Then
    Response.Write "<pre>" & Server.HTMLEncode( http.LastErrorText) & "</pre>"
    Response.End
End If

' Display the HTML source of the page returned.
Response.Write "<pre>" & Server.HTMLEncode( resp.BodyStr) & "</pre>"

%>
</body>
</html>