I have a web service call through which zip files can be uploaded. The files are then forwarded to another service for storage, unzipping, etc. For now the file is stored on the file system, then a FileSystemResource is built.
Resource zipFile = new FileSystemResource(tempFile.getAbsolutePath());
I could use a ByteStreamResource in order to save time(the saving of the file on disk is not needed before forwarding) but for that i need to build a byte array. In case of large files I will get an "OutOfMemory : java heap space" error.
ByteArrayResource r = new ByteArrayResource(inputStream.getBytes());
Any solutions to forwarding files without getting an OutOfMemory error using RestTemplate?
Edit: The other answers are better (use Resource
) https://stackoverflow.com/a/36226006/116509
My original answer:
You can use execute
for this kind of low-level operation. In this snippet I've used Commons IO's copy
method to copy the input stream. You would need to customize the HttpMessageConverterExtractor
for the kind of response you're expecting.
final InputStream fis = new FileInputStream(new File("c:\\autoexec.bat")); // or whatever
final RequestCallback requestCallback = new RequestCallback() {
@Override
public void doWithRequest(final ClientHttpRequest request) throws IOException {
request.getHeaders().add("Content-type", "application/octet-stream");
IOUtils.copy(fis, request.getBody());
}
};
final RestTemplate restTemplate = new RestTemplate();
SimpleClientHttpRequestFactory requestFactory = new SimpleClientHttpRequestFactory();
requestFactory.setBufferRequestBody(false);
restTemplate.setRequestFactory(requestFactory);
final HttpMessageConverterExtractor<String> responseExtractor =
new HttpMessageConverterExtractor<String>(String.class, restTemplate.getMessageConverters());
restTemplate.execute("http://localhost:4000", HttpMethod.POST, requestCallback, responseExtractor);
(Thanks to Baz for pointing out you need to call setBufferRequestBody(false)
or it will defeat the point)