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Sending a file to the api server using Spring Boot RestTemplate (MultipartFile, form)

  • Files can be transmitted from the client using Html Form, but the SpringBoot server can also directly transmit files to another server.
  • Let’s learn how to transfer a file using RestTemplate.
  • Let’s briefly look at only the most important logic.
    • Create Header. Set Content-Type: multipart/form-data in Header.
    • When sending a file, you do not send the file right away, but you need to change it to ByteArrayResource. Convert it to ByteArrayResource and add it to the body.
    • You can create a Request object for file transfer or send it using map.
  • Personally, I was not sure what format to send the file in, but I learned how to send the file using ByteArrayResource.
fun uploadToOtherServer() {
    // 1. Create headers
    val headers = HttpHeaders()
    headers.contentType = MediaType.MULTIPART_FORM_DATA

    // 2. File -> ByteArrayResource
    val contentsAsResource = object : ByteArrayResource(uploadRequest.file.readBytes()) {
        override fun getFilename(): String {
            return uploadRequest.file.name
        }
    }

    // 3. Create a MultiValueMap to set the form parameters
    // Add other information if you need
    val body: MultiValueMap<String, Any> = LinkedMultiValueMap()
    body.add("file", contentsAsResource)
    body.add("createdAt", uploadRequest.createdAt)

    // 4. Send the request
    val requestEntity: HttpEntity<MultiValueMap<String, Any>> = HttpEntity(body, headers)
    val responseEntity = restTemplate.postForEntity<Any>(
        "http://endpoint-url.com/upload", // endpoint url
        requestEntity
    )

    // Add your logic here
}