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Visual Basic 6.0

multipart/form-data HTTP POST

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

Chilkat Visual Basic 6.0 Downloads

Visual Basic 6.0
Dim success As Long
success = 0

' 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 ChilkatHttpRequest
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 ChilkatBinData
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 ChilkatHttp

Dim resp As New ChilkatHttpResponse
success = http.HttpSReq("example.com",443,1,req,resp)
If (success = 0) Then
    Debug.Print http.LastErrorText
    Exit Sub
End If

Debug.Print "HTTP response status: " & resp.StatusCode

Debug.Print "Received:"
Debug.Print resp.BodyStr