Java spring: best way to convert a File to a MultipartFile

Question

I created a file (below) but I want to convert it to MultipartFile, how can I do it?

I already tried this code, without sucess:

File file = new File("text.txt");

FileInputStream input = new FileInputStream(file);
MultipartFile multipartFile = new MockMultipartFile("file", file.getName(), "text/plain", IOUtils.toByteArray(input));

and that one result in a error:

File file = new File("text.txt");
    DiskFileItem fileItem = new DiskFileItem("file", "text/plain", false, file.getName(), (int) file.length() , file.getParentFile());
    fileItem.getOutputStream();
    MultipartFile multipartFile = new CommonsMultipartFile(fileItem);

Error

CommonsMultipartFile (org.apache.commons.fileupload.FileItem) in CommonsMultipartFile cannot be applied to (org.apache.tomcat.util.http.fileupload.disk.DiskFileItem)

Thanks

Answer 1

InputStream stream =  new FileInputStream(file)
multipartFileToSend = new MockMultipartFile("file", file.getName(), MediaType.TEXT_HTML_VALUE, stream);

try this

Answer 2

I propose this method which allows converting a File to a MultipartFile:

public static MultipartFile buildMultipartFile(@NonNull final File file,
    @NonNull final String multipartFileParameterName) throws IOException {
       MultipartFile multipartFile = null;

       try (final FileInputStream input = new FileInputStream(file)) {
           multipartFile = new MockMultipartFile(
                   multipartFileParameterName,
                   file.getName(),
                   Files.probeContentType(file.toPath()),
                   IOUtils.toByteArray(input));
       }

       return multipartFile;
}

Answer 3

In my case, works perfectly so, thank you Parthiban Manickam!!

    InputStream stream = new FileInputStream(zipFile);
    MockMultipartFile multipartFileToSend = new MockMultipartFile("file", zipFile.getName(),
            String.valueOf(MediaType.MULTIPART_FORM_DATA), stream);