Java - handle file upload and create a new http request efficiently

Question

In my custom application I accept files through a file upload form and get MultipartFile object. Then according to the other parameters in the upload form I make some processing and I prepare another post request including the file and send the request to another server.

The code I use is as follows :

In the controller part I take the input steram

             MultipartFile file = uploadedFile.getFile();
             InputStream inputStream = file.getInputStream();

The stream is passed to service method and the new request is prepared as below

            CloseableHttpClient closeableHttpClient = HttpClients.createDefault();

            HttpPost uploadFile = new HttpPost(url + port + path);

            MultipartEntityBuilder multipartEntityBuilder = MultipartEntityBuilder.create();

            .... other parameters set here....


            multipartEntityBuilder.addBinaryBody("attachment", IOUtils.toByteArray(inputStream), ContentType.APPLICATION_OCTET_STREAM, fileName);

            HttpEntity httpEntity = multipartEntityBuilder.build();

            uploadFile.setEntity(httpEntity);

            HttpResponse httpResponse = closeableHttpClient.execute(uploadFile);

Is there a more efficient way of doing this, is the transformation of

     file.getInputStream()

to

     byte[] // by IOUtils.toByteArray(inputStream)

really needed?

Any other advice for the method is also appreciated.

mcvkr · Accepted Answer · 2014-08-26 06:38:27Z

1

We can call the multipartEntityBuilder.addBinaryBody method directly with inputStream as below:

multipartEntityBuilder.addBinaryBody("attachment", IOUtils.toByteArray(inputStream), ContentType.APPLICATION_OCTET_STREAM, fileName);

So there will be a great increase in memory usage that was caused by IOUtils.toByteArray() method and file will be sent via streaming.

answered Aug 26, 2014 at 6:38

mcvkr

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