如何将BufferedImage转换为MultiPart文件而不将文件保存到磁盘？

I want to make a service call that takes in the following Request Parameter in a rest service. This method uploads file to an image server.

我想进行一个服务调用,在休息服务中接受以下请求参数。此方法将文件上载到图像服务器。

 @RequestMapping(value = "/upload", method = RequestMethod.POST)
    public ResponseEntity<String> uploadFile(
            @RequestParam("file") MultipartFile file) {

Following is the code where the call to the service is being made - I read an image in a BufferedImage object from a image url

以下是调用服务的代码 - 我从图像网址读取BufferedImage对象中的图像

     BufferedImage subImage= ImageIO.read(new URL(<some image url goes here>));
     File outputFile = new File("C:\\" + "myimage" + ".jpg");

    ImageIO.write(subImage, "jpg", outputFile);

    MultiValueMap<String, Object> body = new LinkedMultiValueMap<String, Object>();
    String url="http://serviceurl/upload";

    body.add("file", outputFile);

    HttpHeaders headers = new HttpHeaders();
    headers.setContentType(MediaType.MULTIPART_FORM_DATA);

    HttpEntity<MultiValueMap<String, Object>> entity = new HttpEntity<MultiValueMap<String, Object>>(body, headers);

    restTemplate.exchange(url, HttpMethod.POST, entity, String.class);

AS you can see, a file is first being created and saved to the disk. How can I avoid this step and just use BufferedImage object (I do not want to save the file to the local disk).

如您所见,首先创建文件并将其保存到磁盘。如何避免此步骤并只使用BufferedImage对象(我不想将文件保存到本地磁盘)。

Tried the solution below but it seems to me that you cannot achieve this without saving file on your disk. Is that true?

尝试下面的解决方案,但在我看来,如果不保存磁盘上的文件,你无法实现这一点。真的吗?

1 个解决方案

#1

you can do it in this way.. I have created an implementation class of MultipartFile and I am using that newly created class to create a MultipartFile file.

你可以这样做..我已经创建了一个MultipartFile的实现类,我正在使用新创建的类来创建一个MultipartFile文件。

MultipartFile Implementation

public class MultipartImage implements MultipartFile {


private byte[] bytes;
String name;
String originalFilename;
String contentType;
boolean isEmpty;
long size;

public MultipartImage(byte[] bytes, String name, String originalFilename, String contentType,
        long size) {
    this.bytes = bytes;
    this.name = name;
    this.originalFilename = originalFilename;
    this.contentType = contentType;
    this.size = size;
    this.isEmpty = false;
}

@Override
public String getName() {
    return name;
}

@Override
public String getOriginalFilename() {
    return originalFilename;
}

@Override
public String getContentType() {
    return contentType;
}

@Override
public boolean isEmpty() {
    return isEmpty;
}

@Override
public long getSize() {
    return size;
}

@Override
public byte[] getBytes() throws IOException {
    return bytes;
}

@Override
public InputStream getInputStream() throws IOException {
    // TODO Auto-generated method stub
    return null;
}

@Override
public void transferTo(File dest) throws IOException, IllegalStateException {
    // TODO Auto-generated method stub

}
}

Converting Jpg to MultipartFile

将Jpg转换为MultipartFile

BufferedImage originalImage = ImageIO.read(new File("path to file"));
ByteArrayOutputStream baos = new ByteArrayOutputStream();
ImageIO.write( originalImage, "jpg", baos );
baos.flush();

MultipartFile multipartFile = new MultipartImage(baos.toByteArray());

if you dont want to create an implementation of MultipartFile class your own you can use org.springframework.mock.web.MockMultipartFile from spring

如果您不想创建自己的MultipartFile类的实现,可以使用org.springframework.mock.web.MockMultipartFile from spring

Example:

 MultipartFile multipartFile = MockMultipartFile(fileName, baos.toByteArray());

#1