VB.NET

HTTP POST (Duplicate Simple HTML Form POST)

Demonstrates how to duplicate a simple HTML form POST. VB.NET

Dim success As Boolean = False

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

Dim req As New Chilkat.HttpRequest
Dim http As New 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)
Dim domain As String = "chilkatsoft.com"
Dim port As Integer = 443
Dim tls As Boolean = True

' The HTTP POST is sent here:
Dim resp As New Chilkat.HttpResponse
success = http.HttpSReq(domain,port,tls,req,resp)
If (success = False) Then
    Debug.WriteLine(http.LastErrorText)
    Exit Sub
End If


' Display the HTML source of the page returned.
Debug.WriteLine(resp.BodyStr)