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I am a student who has just started to learn Spring boot and java.

I am implementing code that insert information about multipart file list into a one-to-many table as POST.


like this.

enter image description here


For one-to-many relationships, I refer to the tutorial posting below.
https://www.callicoder.com/hibernate-spring-boot-jpa-one-to-many-mapping-example/
Like the tutorial, I want to use @RequestBody, but using multipartFile and @RequestBody results in an error so I am using @RequestParam.

But it doesn't work normally because of my lack of Java basic knowledge.

What I want to do is to bring the epi_storage's id by findById and then save the information of file list with @RequestParam MultipartFile to the sub_storage table.

below is controller. 

 @PostMapping (value ="/newSub", consumes = {MediaType.MULTIPART_FORM_DATA_VALUE})
    public SubStorage NewSub(@PathVariable (value="epiId") int epiId,
                            @RequestParam("subFiles") MultipartFile[] subFiles) {
                     return  subStorageDAO.findById(epiId).map( -> {
                         return Arrays.asList(subFiles)
                                    .stream()
                                    .map(subFile->uploadFile(subFile))
                                    .collect(Collectors.toList());
                    }).orElseThrow(() -> new ResourceNotFoundException("EpiId " + epiId + " not found"));

                            }


    public SubStorage uploadFile(@RequestParam("subFile") MultipartFile subFile) {
        SubStorage subStorage = subStorageService.storeSubFile(epiId, subFile);

        return subStorage;
    }



below is service's storeSubFile()


 public SubStorage storeSubFile(int epiId, MultipartFile subFile) {

        EpiStorage epiStorage = SubStorage.setEpiStorage(epiId);

        // Normalize file name
        String subFileName = StringUtils.cleanPath(subFile.getOriginalFilename());

        String subFileUri = ServletUriComponentsBuilder.fromCurrentContextPath()
                .path("/uploads/")
                .path(subFileName)
                .toUriString();

        try {

            if(subFileName.contains("..")) {
                throw new FileStorageException ("There is a wrong word in " + subFileName);
            }

            //Copy file to the target location (Replacing existing file with the same name)
            Path targetLocation = this.fileStorageLocation.resolve(subFileName);

            Files.copy(subFile.getInputStream(), targetLocation, StandardCopyOption.REPLACE_EXISTING);

            SubStorage subStorage = new SubStorage(subFileName, subFileUri, subFile.getContentType(), subFile.getSize(), epiStorage);

            subStorageDAO.save(subStorage);

            return subStorage;

        } catch (IOException ex) {
            throw new FileStorageException("Could not store file " +subFileName + ". Please try again!", ex);
        }
    }



sorry for my bad code...
Maybe I'm a lot wrong with Java grammar. But I don't know where to fix it.
I'd really appreciate for your help.

J H
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  • it is complex to understand, you can share your code repo, and describe the error you come accross. – suiwenfeng Nov 20 '19 at 10:00
  • What kind of error did you encountered by using `@RequestBody` with `MultipartFile`? BTW, for method `NewSub` is missing `/{epiId}` in URL and for method `uploadFile`, you don't have to add `@RequestParam`. – LHCHIN Nov 20 '19 at 10:05
  • I just referred to this post about ```@RequestBody``` and ```multipartFile``` https://stackoverflow.com/questions/23533355/spring-controller-requestbody-with-file-upload-is-it-possible – J H Nov 20 '19 at 10:13
  • thanks. I forgot ```/{epiId}```. I cant run because of the red line at ```.map( -> ``` and ```subFile``` args at storeSubFile() from controller. I think there is wrong at java grammar. is the way right that define setter epiId at service? – J H Nov 20 '19 at 10:18

0 Answers0