(Visual Basic 6.0) HTTP POST (Duplicate Simple HTML Form POST)

Demonstrates how to duplicate a simple HTML form POST.

Chilkat ActiveX Downloads ActiveX for 32-bit and 64-bit Windows

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

Dim req As New ChilkatHttpRequest
Dim http As New ChilkatHttp

Dim success As Long

'  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)
Dim domain As String
domain = "chilkatsoft.com"
Dim port As Long
port = 443
Dim tls As Long
tls = 1

'  The HTTP POST is sent here:
Dim resp As ChilkatHttpResponse
Set resp = http.SynchronousRequest(domain,port,tls,req)
If (http.LastMethodSuccess = 0) Then
    Debug.Print http.LastErrorText
    Exit Sub
End If

'  Display the HTML source of the page returned.
Debug.Print resp.BodyStr