Question

我有以下代码用于将文件发布到服务，它工作正常。我唯一的问题是，我必须编写一个临时文件来获取FileSystemResource以使用restTemplate发布对象

无论如何，我可以调整以下代码，以便我不必编写临时文件吗？

    public String postNewIcon2(Integer fileId, MultipartFile multiPartfile) {
    LOG.info("Entered postNewIcon");

    Map<String, Object> params = getParamsWithAppKey();
    params.put("fileId", fileId);

    String result = null;
    File tempFile = null;
    try {

        String originalFileNameAndExtension = multiPartfile.getOriginalFilename();

        String tempFileName = "c:\\temp\\image";
        String tempFileExtensionPlusDot = ".png";

        tempFile = File.createTempFile(tempFileName, tempFileExtensionPlusDot);
        multiPartfile.transferTo(tempFile);
        FileSystemResource fileSystemResource = new FileSystemResource(tempFile);

        // URL Parameters
        MultiValueMap<String, Object> parts = new LinkedMultiValueMap<String, Object>();
        parts.add("file", fileSystemResource);

        // Post
        result = restTemplate.postForObject(getFullURLAppKey(URL_POST_NEW_ICON), parts, String.class, params);

    } catch (RestClientException restClientException) {
        System.out.println(restClientException);
    } catch (IOException ioException) {
        System.out.println(ioException);
    } finally {
        if (tempFile != null) {
            boolean deleteTempFileResult = tempFile.delete();
            LOG.info("deleteTempFileResult: {}", deleteTempFileResult);
        }
    }
    return result;
}

谢谢

Answer 1

在Kresimir Nesek的帮助下回答此链接 Sending Multipart File as POST parameters with RestTemplate requests

以下代码完成了这一操作 - 现在不需要临时文件

    MultiValueMap<String, Object> map = new LinkedMultiValueMap<String, Object>();
final String filename="somefile.txt";
map.add("name", filename);
map.add("filename", filename);
ByteArrayResource contentsAsResource = new ByteArrayResource(content.getBytes("UTF-8")){
            @Override
            public String getFilename(){
                return filename;
            }
        };
map.add("file", contentsAsResource);
String result = restTemplate.postForObject(urlForFacade, map, String.class);

Answer 2

MultipartFile需要有一些临时位置。请尝试此代码，以获取物理文件：

private File getTempFile(MultipartFile attachment){
    CommonsMultipartFile commonsMultipartFile = (CommonsMultipartFile) attachment;
    DiskFileItem diskFileItem = (DiskFileItem) commonsMultipartFile.getFileItem();
    return diskFileItem.getStoreLocation();
}

春天休息 - 发布文件

2 个答案: