Xojo Plugin

multipart/form-data HTTP POST

Demonstrates how to construct and send a multipart/form-data HTTP POST. Xojo Plugin

Dim success As Boolean
success = False

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

// Sends the following multipart/form-data POST:

// POST /xyz/something HTTP/1.1
// Host: domain
// Content-Type: multipart/form-data; boundary=------------090708030009010000030901
// Content-Length: 2220
// 
// --------------090708030009010000030901
// Content-Disposition: form-data; name="param1"
// 
// value1
// --------------090708030009010000030901
// Content-Disposition: form-data; name="param2"
// 
// value2
// --------------090708030009010000030901
// Content-Disposition: form-data; name="starfish20"; filename="starfish20.jpg"
// Content-Type: image/jpeg
// 
// JPEG DATA HERE...
// --------------090708030009010000030901
// Content-Disposition: form-data; name="helloWorld"; filename="helloWorld.pdf"
// Content-Type: application/pdf
// 
// PDF DATA HERE...
// --------------090708030009010000030901
// Content-Disposition: form-data; name="tinyA"; filename="tinyA.xml"
// Content-Type: text/xml
// 
// XML DATA HERE...
// --------------090708030009010000030901--
// 

Dim req As New Chilkat.HttpRequest
req.HttpVerb = "POST"
req.ContentType = "multipart/form-data"
req.Path = "/xyz/something"

req.AddParam "param1","value1"
req.AddParam "param2","value2"

// Add some small files to the request. (Small so we can see what the full request looks like without too much data..)
Dim bd As New Chilkat.BinData
success = bd.LoadFile("qa_data/jpg/starfish20.jpg")
success = req.AddBdForUpload("starfish20","starfish20.jpg",bd,"image/jpeg")

success = bd.LoadFile("qa_data/pdf/helloWorld.pdf")
success = req.AddBdForUpload("helloWorld","helloWorld.pdf",bd,"application/pdf")

success = bd.LoadFile("qa_data/xml/tinyA.xml")
success = req.AddBdForUpload("tinyA","tinyA.xml",bd,"text/xml")

Dim http As New Chilkat.Http

Dim resp As New Chilkat.HttpResponse
success = http.HttpSReq("example.com",443,True,req,resp)
If (success = False) Then
    System.DebugLog(http.LastErrorText)
    Return
End If

System.DebugLog("HTTP response status: " + Str(resp.StatusCode))

System.DebugLog("Received:")
System.DebugLog(resp.BodyStr)