如何使用RESTTemplate转发大文件？

Question

Resource zipFile = new FileSystemResource(tempFile.getAbsolutePath());

ByteArrayResource r = new ByteArrayResource(inputStream.getBytes());

使用RESTTemplate转发文件而不得到OutOfMemory错误的解决方案吗？

Answer 1

编辑：其他答案更好（使用资源)https://stackoverflow.com/a/36226006/116509

我原来的答案是：

对于这种低级操作，可以使用execute。在这个片段中，我使用了Commons IO的copy方法来复制输入流。您需要自定义HttpMessageConverterExtractor以获得所需的响应类型。

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);