PowerShell

HTTP POST (Duplicate Simple HTML Form POST)

Demonstrates how to duplicate a simple HTML form POST. PowerShell

Add-Type -Path "C:\chilkat\ChilkatDotNet47-x64\ChilkatDotNet47.dll"

$success = $false

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

$req = New-Object Chilkat.HttpRequest
$http = New-Object 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 = $true

# The HTTP POST is sent here:
$resp = New-Object Chilkat.HttpResponse
$success = $http.HttpSReq($domain,$port,$tls,$req,$resp)
if ($success -eq $false) {
    $($http.LastErrorText)
    exit
}

# Display the HTML source of the page returned.
$($resp.BodyStr)