La requête a été rejetée car aucune limite multi-parties n'a été trouvée dans springboot

Que je suis en train d'essayer cela avec le printemps de démarrage et de webservices avec facteur de chrome add-ons.

En facteur content-type="multipart/form-data" et j'en suis la ci-dessous exception.

HTTP Status 500 - Request processing failed; 
nested exception is org.springframework.web.multipart.MultipartException: Could not parse multipart servlet request; 
nested exception is java.io.IOException: 
org.apache.tomcat.util.http.fileupload.FileUploadException: the request was rejected because no multipart boundary was found

Dans le Contrôleur, j'ai spécifié le code ci-dessous

@ResponseBody
@RequestMapping(value = "/file", headers = "Content-Type= multipart/form-data", method = RequestMethod.POST)

public String upload(@RequestParam("name") String name,
        @RequestParam(value = "file", required = true) MultipartFile file)
//@RequestParam ()CommonsMultipartFile[] fileUpload
{
    //@RequestMapping(value="/newDocument", , method = RequestMethod.POST)
    if (!file.isEmpty()) {
        try {
            byte[] fileContent = file.getBytes();
            fileSystemHandler.create(123, fileContent, name);
            return "You successfully uploaded " + name + "!";
        } catch (Exception e) {
            return "You failed to upload " + name + " => " + e.getMessage();
        }
    } else {
        return "You failed to upload " + name + " because the file was empty.";
    }
}

Ici, je spécifier le gestionnaire de fichier de code

public String create(int jonId, byte[] fileContent, String name) {
    String status = "Created file...";
    try {
        String path = env.getProperty("file.uploadPath") + name;
        File newFile = new File(path);
        newFile.createNewFile();
        BufferedOutputStream stream = new BufferedOutputStream(new FileOutputStream(newFile));
        stream.write(fileContent);
        stream.close();
    } catch (IOException ex) {
        status = "Failed to create file...";
        Logger.getLogger(FileSystemHandler.class.getName()).log(Level.SEVERE, null, ex);
    }
    return status;
}

source d'informationauteur Mohammed Javed | 2016-03-15