What is the correct places for parameters in OKHttp in a multipart form call?

Question

I'm trying to replicate a jpg file upload using OKHttp to a multipart form on a PHP server. I believe, I got some of the parameters in the wrong place, I don't have the familiarity with multipart forms in http and the nomenclature.

Here's what I'm trying to accomplish

Post the parameters (name value pairs): myuser, token, types to https://www.somesite.com/jpgphotoupload.php

I'm then making a multipart form request with method POST, with the following: path: https://www.somesite.com/jpgphotoupload.php

file data: JPEG compressed image data of size 480 x 640 (This I understand)

mimeType: image/jpeg (This I understand)

Not sure where the following name value pairs should be placed as part of the multipart form request, tried addFormDataPart

parameters: again the parameter form above, (myuser, token, types)

name: imagefile

fileName: myname.jpg

In addition, here's what else may be pertinent

            "Connection" , "Keep-Alive"
            "ENCTYPE", "multipart/form-data"
            "Content-Type", "multipart/form-data"

Here's the code that I have currently.

MediaType MEDIA_TYPE_JPG = MediaType.parse("image/jpg");

            OkHttpClient client = new OkHttpClient();
                RequestBody requestBody = new MultipartBuilder()
                        .type(MultipartBuilder.FORM)
                        .addPart(
                                Headers.of("Content-Disposition", "form-data; name=\"imagefile\""),
                                RequestBody.create(MEDIA_TYPE_JPG, new File("/storage/emulated/0/download/camerapic.jpg")))
                        .addFormDataPart("myuser", getprefmyuser(getBaseContext()))
                        .addFormDataPart("token", getpreftoken(getBaseContext()))
                        .addFormDataPart("types", "type1")
                        .addFormDataPart("fileName", "myname.jpg")
                        .build();

                Request request = new Request.Builder()
                        .header("myuser", getprefmyuser(getBaseContext()))
                        .header("token", getpreftoken(getBaseContext()))
                        .header("type", "car")
                        .url("https://www.somesite.com/jpgphotoupload.php")
                        .post(requestBody)
                        .build();

            Response response = null;
            try {
                response = client.newCall(request).execute();
            } catch (IOException e) {
                e.printStackTrace(); 
        ... 

        return null;
        }

Answer 1

In my case, I needed to upload a video to an Amazon S3 bucket. This is what worked for me.

File sourceFile = new File(myUri);

RequestBody requestBody = new MultipartBuilder()
                    .type(MultipartBuilder.FORM)
                    .addFormDataPart("keyOne", "valueOne")
                    .addFormDataPart("keyTwo", "valueTwo")
                    .addFormDataPart("file", "myFileName", RequestBody.create(MediaType.parse("video/quicktime"), sourceFile))
                    .build();