Dealing With Files in REST API - White Label Developers

We often come across scenarios where we need to deal with files as a part of the payload of an HTTP request or as a response. Though it is easy to do it manually in postman or swagger, it is a challenge when it comes to automation. In this article, let’s see how to automate such api endpoints by sending files as a part of requests and how to download if a response is also a file.

Attach File to Request

Scalaj-HTTP is a wrapper for an HTTP client in Scala. It provides a method — postMulti to attach files as a part of a request.

Let’s say there is an endpoint that takes only a file as a payload. (Note, the file should be a multipart of type.)

MultiPart(filename, filepath, mimetype, byteArrayInputStream)

Define your file as a multipart like above. Wonder what is last argument byteArrayInputStream? you should pass the file as a byte array to the request. For example, if you are going to attach an image (png) it will be like below.

MultiPart("image", "src/image.png", "image/png", getByteArray("src/image.png"))

Now, if you are going to use the post method, the entire request goes like this.

val res = Http(url)
        .postMulti(MultiPart(filename, filepath, mimetype, data))

        .method("POST")
        .asString

That’s it, this will do the process of attaching files to your request.

Download the File From the Response

Let’s say I am getting a file as a response from an endpoint, and I need to download the file. Follow the below 2 steps to do it.

1. Get the Response Using scalaj-HTTP Client

Let’s assume we have a simple GET endpoint that gives an image as a response. By using the following code, you will be getting the response.

val res = Http(url)
    .header("accept", "image/png")
    .method("GET")
    .option(HttpOptions.connTimeout(100000))
    .option(HttpOptions.readTimeout(100000))

Remember, how did we send the image as a byteArray to request? Now, it is a reverse to this.

Get the response as a byteArrayInputStream and write it to a file.


val file = new File("src/image/image.png")
file.createNewFile()

val bis = new ByteArrayInputStream(res.asBytes.body)
IOUtils.copy(bis, new FileOutputStream(file))

That’s it. You will be able to see the image in the desired path.

src dzone.com

author Gowthamraj Palani

img helpdeskgeek.com/